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Math 440, Fall 2014, Assignment 9

2014-11-04T18:17:24Z

Vincent.Luczkow: /* Problems: */

__NOTOC__
''Mathematical proofs, like diamonds, are hard as well as clear, and will be touched with nothing but strict reasoning.''

: - John Locke, ''Second Reply to the Bishop of Worcester''

==Carefully define the following terms, then give one example and one non-example of each:==

# Normal space.
# $T_4$-space.
# Urysohn function (for a pair of subsets $A,B\subseteq X$).
# Second countable space.
# Separable space.
# Topological manifold.

==Carefully state the following theorems (you need not prove them):==

# Urysohn's Lemma.
# Tietze Extension Theorem.
# Theorem relating second countable $T_3$-spaces to $T_4$-spaces.

==Solve the following problems:==

# Problems 16A(1) and 16B(1).
# Show that any open ball in $\mathbb{R}^n$ is homeomorphic to all of $\mathbb{R}^n$. ''(Hint: first use translation and dilation maps to show that any open ball is homeomorphic to the open unit ball centered at the origin. Then make a homeomorphism from the open unit ball to all of $\mathbb{R}^n$ by multiplying each point by a certain scalar, depending on the point's distance from the origin.)''
# Show that any open subset of a topological manifold is again a topological manifold in the subspace topology.
# Recall that $GL_n(\mathbb{R})$ denotes the set of invertible $n\times n$ matrices with real entries. Show that $GL_n(\mathbb{R})$ is a topological manifold with respect to the subspace topology it inherits from $\mathbb{R}^{\left(n^2\right)}$.

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Normal Space:A space $X$ is normal iff whenever $A$ and $B$ are disjoint closed sets in $X$, there are disjoint open sets $U$ and $V$ with $A \subset U$ and $B \subset V$.Example:Non-Example:
#$T_4$-Space:A $T_4$ space is a normal $T_1$ space.Example:Non-Example:
#Urysohn Function:A Urysohn function is a continuous function on $X$ created by an application of Urysohn's lemma.Example:Non-Example:
#Second Countable Space:A space $X$ is second countable iff it has a countable base.Example:Non-Example:
#Separable Space:A space is separable iff it is has a countable dense subset.Example:Non-Example:
#Topological Manifold:A topological manifold is a second countable Hausdorff space $X$ for which every point $x \in X$ has a neighborhood homeomorphic to $\mathbb{R}^n$, for some fixed $n$ (the same $n$ at all points).Example: $\mathbb{R}^n$ itself.Non-Examples: The line with two origins is second countable and locally Euclidean, but it is not Hausdorff. The disjoint union of uncountably many copies of $\mathbb{R}$ is Hausdorff and locally Euclidean, but not second countable. The union of the two coordinate axes in $\mathbb{R}^2$ is Hausdorff and second countable, but not locally Euclidean (at the origin).

====Theorems:====
#Urysohn's LemmaA space is normal iff whenever $A$ and $B$ are disjoint closed sets in $X$, there is a continuous function $f: X \rightarrow [0,1]$ with $f(A) = 0$ and $f(B) = 1$.
#Tietze Extension Theorem$X$ is normal iff whenever $A$ is a closed subset of $X$ and $f:A\rightarrow \mathbb{R}$ is continuous, there is an extension of $f$ to all of $X$.
#Theorem Relating Second Countable $T_3$-spaces to $T_4$-spaces.A second countable $T_3$ space is $T_4$.
====Problems:====
#A. Prove that every subspace of a first countable space is first countable.Suppose $X$ is first countable, and $A$ is a subset of $X$. Suppose $a \in A$. Then $a$ has a countable neighborhood base $\mathscr{U}_a$. Define $\mathscr{V}_a = \{U \cap A| U \in \mathscr{U}_a\}$. Suppose $L$ is any neighborhood of $a$ in $A$. Then $L = M \cap A$ for some $M$. Clearly $M$ is a nhood of $a$ in $X$, so there is a $U \in \mathscr{U}_a$ such that $U \subset M$. Then $V = U \cap A \subset M \cap A = L$, so every nhood of $a$ contains an element of $\mathscr{V}_x$, so $\mathscr{V}_a$ is a nhood base of $a$.B.
#The ball is centered at $\vec{x}$. Translate it by $-\vec{x}$. It is of radius $r$. Scale it by $1/r$. Then scale each point by $1/1-|\vec{v}|$.
#An open subset of a topological manifold is homeomorphic to an open subset of $\mathbb{R}^n$. Any point in this subspace then has a neighborhood homeomorphic to an $n$-ball. Thus, by the previous problem, it has a neighborhood homeomorphic to $\mathbb{R}^n$.

Math 440, Fall 2014, Assignment 9

2014-11-04T18:01:50Z

Vincent.Luczkow: /* Theorems: */

__NOTOC__
''Mathematical proofs, like diamonds, are hard as well as clear, and will be touched with nothing but strict reasoning.''

: - John Locke, ''Second Reply to the Bishop of Worcester''

==Carefully define the following terms, then give one example and one non-example of each:==

# Normal space.
# $T_4$-space.
# Urysohn function (for a pair of subsets $A,B\subseteq X$).
# Second countable space.
# Separable space.
# Topological manifold.

==Carefully state the following theorems (you need not prove them):==

# Urysohn's Lemma.
# Tietze Extension Theorem.
# Theorem relating second countable $T_3$-spaces to $T_4$-spaces.

==Solve the following problems:==

# Problems 16A(1) and 16B(1).
# Show that any open ball in $\mathbb{R}^n$ is homeomorphic to all of $\mathbb{R}^n$. ''(Hint: first use translation and dilation maps to show that any open ball is homeomorphic to the open unit ball centered at the origin. Then make a homeomorphism from the open unit ball to all of $\mathbb{R}^n$ by multiplying each point by a certain scalar, depending on the point's distance from the origin.)''
# Show that any open subset of a topological manifold is again a topological manifold in the subspace topology.
# Recall that $GL_n(\mathbb{R})$ denotes the set of invertible $n\times n$ matrices with real entries. Show that $GL_n(\mathbb{R})$ is a topological manifold with respect to the subspace topology it inherits from $\mathbb{R}^{\left(n^2\right)}$.

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Normal Space:A space $X$ is normal iff whenever $A$ and $B$ are disjoint closed sets in $X$, there are disjoint open sets $U$ and $V$ with $A \subset U$ and $B \subset V$.Example:Non-Example:
#$T_4$-Space:A $T_4$ space is a normal $T_1$ space.Example:Non-Example:
#Urysohn Function:A Urysohn function is a continuous function on $X$ created by an application of Urysohn's lemma.Example:Non-Example:
#Second Countable Space:A space $X$ is second countable iff it has a countable base.Example:Non-Example:
#Separable Space:A space is separable iff it is has a countable dense subset.Example:Non-Example:
#Topological Manifold:A topological manifold is a space $X$ for which every point $x \in X$ has a neighborhood homeomorphic to $\mathbb{R}^n$, for some $n$.Example:Non-Example:

====Theorems:====
#Urysohn's LemmaA space is normal iff whenever $A$ and $B$ are disjoint closed sets in $X$, there is a continuous function $f: X \rightarrow [0,1]$ with $f(A) = 0$ and $f(B) = 1$.
#Tietze Extension Theorem$X$ is normal iff whenever $A$ is a closed subset of $X$ and $f:A\rightarrow \mathbb{R}$ is continuous, there is an extension of $f$ to all of $X$.
#Theorem Relating Second Countable $T_3$-spaces to $T_4$-spaces.A second countable $T_3$ space is $T_4$.
====Problems:====
#A. Prove that every subspace of a first countable space is first countable.Suppose $X$ is first countable, and $A$ is a subset of $X$. Suppose $a \in A$. Then $a$ has a countable neighborhood base $\mathscr{U}_a$. Define $\mathscr{V}_a = \{U \cap A| U \in \mathscr{U}_a\}$. Suppose $L$ is any neighborhood of $a$ in $A$. Then $L = M \cap A$ for some $M$. Clearly $M$ is a nhood of $a$ in $X$, so there is a $U \in \mathscr{U}_a$ such that $U \subset M$. Then $V = U \cap A \subset M \cap A = L$, so every nhood of $a$ contains an element of $\mathscr{V}_x$, so $\mathscr{V}_a$ is a nhood base of $a$.B.

Math 440, Fall 2014, Assignment 9

2014-11-04T17:53:55Z

Vincent.Luczkow: /* Solutions: */

Math 440, Fall 2014, Assignment 4

2014-09-30T16:02:49Z

Vincent.Luczkow: /* Theorems: */

__NOTOC__
''No doubt many people feel that the inclusion of mathematics among the arts is unwarranted. The strongest objection is that mathematics has no emotional import. Of course this argument discounts the feelings of dislike and revulsion that mathematics induces....''

: - Morris Kline, ''Mathematics in Western Culture''

==Carefully define the following terms, then give one example and one non-example of each:==

# Subspace topology (on a subset $A$ of a topological space $X$).
# Function continuous at a given point.
# Continuous function.
# Homeomorphism (from one space top another).
# Homeomorphic (spaces).
# Topological property.

==Carefully state the following theorems (you do not need to prove them):==

# Theorem characterizing continuous functions (Theorem 7.2).
# Theorem on composition of continuous functions.
# Theorem on restriction of continuous functions (Theorem 7.5).
# Theorem concerning continuity of functions defined on unions (Theorem 7.6).
# Theorem concerning restriction of codomains (Theorem 7.7).

==Solve the following problems:==

# Problems 6C, 7A, 7C, and 7G.

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Subspace Topology:Let $(X,\tau)$ be a topological space, and let $A \subseteq X$. The subspace topology of $X$ on $A$ is the topological space $(A,\tau_A)$, where $\tau_A = \{A \cap t| t\in \tau\}$.Example:Non-Example:
#Function Continuous at a Given Point:Let $f:X \rightarrow Y$ be a function between topological spaces. $f$ is continuous at $x \in X$ if for every $V \subseteq Y$ that contains $f(x)$ (i.e. $f(x) \in V$, then $f^{-1}(V)$ is open in $X$.Example:Non-Example:
#Continuous Function:Let $X$ and $Y$ be topological spaces. A function $f : X \rightarrow Y$ is continuous if it is continuous at every point of $X$.Example:Non-Example:
#Homeomorphism:A homeomorphism from $X$ to $Y$ is a continuous bijective function with a continuous inverse.Example:Non-Example:
#Homeomorphic Spaces:Two topological spaces $X$ and $Y$Example:Non-Example:
#Topological Property:A topological property is a property that is preserved on homeomorphism. That is, if $(X,\tau)$ and $(Y,\sigma)$ are homeomorphic spaces, then if $X$ has a property, $Y$ has the same property.Example:Cardinality.Non-Example:
====Theorems:====
#Theorem Characterizing Continuous FunctionsLet $f: X \rightarrow Y$. The following conditions are equivalent: $f$ is continuous For each open set $H$ in $Y$, $f^{-1}(H)$ is open in $X$. For each closed set $K$ in $Y$, $f^{-1}(K)$ is closed in $X$. For each $E \subset X$, $f(Cl_X(E) \subset Cl_Y f(E)$.
#Theorem on Composition of Continuous FunctionsIf $f:X\rightarrow Y$ is continuous, and $g: Y\rightarrow Z$ is continuous, then $g \circ f : X \rightarrow Z$ is continuous.
#Theorem on Restriction of Continuous FunctionsIf $f: X \rightarrow Y$ is continuous, and $A \subseteq X$, then $f | A$ is continuous.
#Theorem Concerning Continuity of Functions Defined on UnionsIf $X = A \cup B$, and $A$ and $B$ are both open or both closed, and $f|A$ and $f|B$ are both continuous, then $f$ is continuous.
#Theorem Concerning Restriction of CodomainsLet $f: X \rightarrow Y$, and $Y \subset Z$. Then $f$ from $X$ to $Y$ if and only if $f$ is continuous as a map from $X$ to $Z$.

Math 440, Fall 2014, Assignment 4

2014-09-30T15:56:02Z

Vincent.Luczkow: /* Definitions: */

__NOTOC__
''No doubt many people feel that the inclusion of mathematics among the arts is unwarranted. The strongest objection is that mathematics has no emotional import. Of course this argument discounts the feelings of dislike and revulsion that mathematics induces....''

: - Morris Kline, ''Mathematics in Western Culture''

==Carefully define the following terms, then give one example and one non-example of each:==

# Subspace topology (on a subset $A$ of a topological space $X$).
# Function continuous at a given point.
# Continuous function.
# Homeomorphism (from one space top another).
# Homeomorphic (spaces).
# Topological property.

==Carefully state the following theorems (you do not need to prove them):==

# Theorem characterizing continuous functions (Theorem 7.2).
# Theorem on composition of continuous functions.
# Theorem on restriction of continuous functions (Theorem 7.5).
# Theorem concerning continuity of functions defined on unions (Theorem 7.6).
# Theorem concerning restriction of codomains (Theorem 7.7).

==Solve the following problems:==

# Problems 6C, 7A, 7C, and 7G.

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Subspace Topology:Let $(X,\tau)$ be a topological space, and let $A \subseteq X$. The subspace topology of $X$ on $A$ is the topological space $(A,\tau_A)$, where $\tau_A = \{A \cap t| t\in \tau\}$.Example:Non-Example:
#Function Continuous at a Given Point:Let $f:X \rightarrow Y$ be a function between topological spaces. $f$ is continuous at $x \in X$ if for every $V \subseteq Y$ that contains $f(x)$ (i.e. $f(x) \in V$, then $f^{-1}(V)$ is open in $X$.Example:Non-Example:
#Continuous Function:Let $X$ and $Y$ be topological spaces. A function $f : X \rightarrow Y$ is continuous if it is continuous at every point of $X$.Example:Non-Example:
#Homeomorphism:A homeomorphism from $X$ to $Y$ is a continuous bijective function with a continuous inverse.Example:Non-Example:
#Homeomorphic Spaces:Two topological spaces $X$ and $Y$Example:Non-Example:
#Topological Property:A topological property is a property that is preserved on homeomorphism. That is, if $(X,\tau)$ and $(Y,\sigma)$ are homeomorphic spaces, then if $X$ has a property, $Y$ has the same property.Example:Cardinality.Non-Example:
====Theorems:====
#Theorem Characterizing Continuous FunctionsLet $f: X \rightarrow Y$. The following conditions are equivalent: $f$ is continuous For each open set $H$ in $Y$, $f^{-1}(H)$ is open in $X$. For each closed set $K$ in $Y$, $f^{-1}(K)$ is closed in $X$. For each $E \subset X$, $f(Cl_X(E) \subset Cl_Y f(E)$.
#Theorem on Composition of Continuous FunctionsIf $f:X\rightarrow Y$ is continuous, and $g: Y\ rightarrow Z$ is continuous, then $g \circ f : X \rightarrow Z$ is continuous.
#Theorem on Restriction of Continuous FunctionsIf $f: X \rightarrow Y$ is continuous, and $A \subseteq X$, then $f | A$ is continuous.
#Theorem Concerning Continuity of Functions Defined on UnionsIf $X = A \cup B$, and $A$ and $B$ are both open or both closed, and $f|A$ and $f|B$ are both continuous, then $f$ is continuous.
#Theorem Concerning Restriction of CodomainsLet $f: X \rightarrow Y$, and $Y \subset Z$. Then $f$ from $X$ to $Y$ if and only if $f$ is continuous as a map from $X$ to $Z$.

Math 440, Fall 2014, Assignment 4

2014-09-30T15:26:54Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 4

2014-09-30T15:26:31Z

Vincent.Luczkow: /* Definitions: */

__NOTOC__
''No doubt many people feel that the inclusion of mathematics among the arts is unwarranted. The strongest objection is that mathematics has no emotional import. Of course this argument discounts the feelings of dislike and revulsion that mathematics induces....''

: - Morris Kline, ''Mathematics in Western Culture''

==Carefully define the following terms, then give one example and one non-example of each:==

# Subspace topology (on a subset $A$ of a topological space $X$).
# Function continuous at a given point.
# Continuous function.
# Homeomorphism (from one space top another).
# Homeomorphic (spaces).
# Topological property.

==Carefully state the following theorems (you do not need to prove them):==

# Theorem characterizing continuous functions (Theorem 7.2).
# Theorem on composition of continuous functions.
# Theorem on restriction of continuous functions (Theorem 7.5).
# Theorem concerning continuity of functions defined on unions (Theorem 7.6).
# Theorem concerning restriction of codomains (Theorem 7.7).

==Solve the following problems:==

# Problems 6C, 7A, 7C, and 7G.

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Subspace Topology:Let $(X,\tau)$ be a topological space, and let $A \subseteq X$. The subspace topology of $X$ on $A$ is the topological space $(A,\tau_A)$, where $tau_A = \{A \cap t| t\ in \tau\}$.Example:Non-Example:
#Function Continuous at a Given Point:Let $f:X \rightarrow Y$ be a function between topological spaces. $f$ is continuous at $x \in X$ if for every $V \subseteq Y$ that contains $f(x)$ (i.e. $f(x) \in V$, then $f^{-1}(V)$ is open in $X$.Example:Non-Example:
#Continuous Function:Let $X$ and $Y$ be topological spaces. A function $f : X \rightarrow Y$ is continuous if it is continuous at every point of $X$.Example:Non-Example:
#Homeomorphism:A homeomorphism from $X$ to $Y$ is a continuous bijective function with a continuous inverse.Example:Non-Example:
#Homeomorphic Spaces:Two topological spaces $X$ and $Y$Example:Non-Example:
#Topological Property:A topological property is a property that is preserved on homeomorphism. That is, if $(X,\tau)$ and $(Y,\sigma)$ are homeomorphic spaces, then if $X$ has a property, $Y$ has the same property.Example:Cardinality.Non-Example:

Math 440, Fall 2014, Assignment 4

2014-09-30T15:25:53Z

Vincent.Luczkow: /* Solutions: */

__NOTOC__
''No doubt many people feel that the inclusion of mathematics among the arts is unwarranted. The strongest objection is that mathematics has no emotional import. Of course this argument discounts the feelings of dislike and revulsion that mathematics induces....''

: - Morris Kline, ''Mathematics in Western Culture''

==Carefully define the following terms, then give one example and one non-example of each:==

# Subspace topology (on a subset $A$ of a topological space $X$).
# Function continuous at a given point.
# Continuous function.
# Homeomorphism (from one space top another).
# Homeomorphic (spaces).
# Topological property.

==Carefully state the following theorems (you do not need to prove them):==

# Theorem characterizing continuous functions (Theorem 7.2).
# Theorem on composition of continuous functions.
# Theorem on restriction of continuous functions (Theorem 7.5).
# Theorem concerning continuity of functions defined on unions (Theorem 7.6).
# Theorem concerning restriction of codomains (Theorem 7.7).

==Solve the following problems:==

# Problems 6C, 7A, 7C, and 7G.

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Subspace Topology:Let $(X,\tau)$ be a topological space, and let $A \subseteq X$. The subspace topology of $X$ on $A$ is the topological space $(A,\tau_A)$, where $tau_A = \{A \cap t| t\ in \tau\}$.Example:Non-Example:
#Function Continuous at a Given Point:Let $f:X \rightarrow Y$ be a function between topological spaces. $f$ is continuous at $x \in X$ if for every $V \subseteq Y$ that contains $f(x)$ (i.e. $f(x) \in V$, then $f^{-1}(V)$ is open in $X$.Example:Non-Example:
#Continuous Function:Let $X$ and $Y$ be topological spaces. A function $f : X \rightarrow Y$ is continuous if for every open set $V \subseteq Y$, $f^{-1}(V)$ is open in $X$.Example:Non-Example:
#Homeomorphism:A homeomorphism from $X$ to $Y$ is a continuous bijective function with a continuous inverse.Example:Non-Example:
#Homeomorphic Spaces:Two topological spaces $X$ and $Y$Example:Non-Example:
#Topological Property:A topological property is a property that is preserved on homeomorphism. That is, if $(X,\tau)$ and $(Y,\sigma)$ are homeomorphic spaces, then if $X$ has a property, $Y$ has the same property.Example:Cardinality.Non-Example:

Math 440, Fall 2014, Assignment 3

2014-09-22T14:01:46Z

Vincent.Luczkow: /* Theorems: */

__NOTOC__
''By one of those caprices of the mind, which we are perhaps most subject to in early youth, I at once gave up my former occupations; set down natural history and all its progeny as a deformed and abortive creation; and entertained the greatest disdain for a would-be science, which could never even step within the threshold of real knowledge. In this mood of mind I betook myself to the mathematics, and the branches of study appertaining to that science, as being built upon secure foundations, and so, worthy of my consideration.''

: - Mary Shelley, ''Frankenstein''

==Carefully define the following terms, then give one example and one non-example of each:==

# Nhood (of a point $x$ in a topological space $(X,\tau)$).
# Nhood system (at a point $x$ in a topological space $(X,\tau)$).
# Nhood base (at a point $x$ in a topological spae $(X,\tau)$).
# Basic nhood (of $x$).
# Base (for a topology).
# Basic open set (in $X$).
# Subbase (for a topology).

==Carefully state the following theorems (you do not need to prove them):==

# Theorem characterizing nhood systems (Theorem 4.2).
# Theorem characterizing nhood bases (Theorem 4.5).
# Theorem characterizing bases (Theorem 5.3).
# Theorem characterizing subbases (Theorem 5.6).

==Solve the following problems:==

# Problems 4A, 4D, 4F, 5A, and 5B.

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Nhood of a Point in a Topological Space:Let $(X,\tau)$ be a topological space, and $x \in X$. A neighborhood of $X$ is a set $U\subseteq X$ which contains $x$ and which contains an open set containing $x$. So $x \in V \subseteq U \subseteq X$, and $V \in \tau$.Example:In $\mathbb{R}^2$, the open unit square centered at $(0,0)$ is a neighborhood of $(0,0)$.Non-Example:
#Nhood System at a Point in a Topological Space:Let $(X,\tau)$ be a topological space, and $x \in X$. A neighborhood system of $x$ is the set of all neighborhoods of $x$.Example:In $\mathbb{R}^2$, the neighborhood system of $(0,0)$ is every set $P$ such that $P$ has a non-empty intersection with some open disk centered at $(0,0)$.Non-Example:
#Nhood Base at a Point in a Topological Space:A neighborhood base at $x \in X$ is a collection $B$ of open sets containing $x$, such that every open set containing $x$ is a superset of a member of $B$.$$
\forall S \in \tau: x \in S \rightarrow \exists B_{\lambda} \in B: B_{\lambda} \subseteq S
$$Example:The set of all open disks centered at $(0,0)$ with radius less than 1 is a neighborhood base at $(0,0)$.Non-Example:
#Basic Nhood of $x$:If $B$ is a neighborhood base of $x$, then an element of $B$ is a basic neighborhood of $x$.Example:In the previous example, he open half-disk would be a basic neighborhood of $(0,0)$.Non-Example:
#Base for a Topology:A base for a topological space $(X,\tau)$ is a collection of set $\mathscr{B}$ such that:$$
\bigcup_{B\in\mathscr{B}} B = X
$$If $B_1$ and $B_2$ are in $\mathscr{B}$ and $p\in B_1 \cap B_2$, then there is some $B_3$ such that $p \in B_3 \subseteq B_1 \cap B_2$ If a set is open in $\tau$, then it must be a union of members of $\mathscr{B}$.Example:The set of all open disks of radius less than 1 is a base for $\mathbb{R}^2$ with the Euclidean metric topology (? Proper terminology?)Non-Example:
#Basic Open Set:If $\mathscr{B}$ is a base for $\tau$, then a basic open set is a member of $\mathscr{B}$.Example:Non-Example:
#Subbase for a Topology:A subbase for $\tau$ is a collection of sets which, when closed under finite intersections, is a base for $\tau$.Example:Non-Example:
====Theorems:====
#Theorem Characterizing Nhood SystemsLet $U_x$ be the neighborhood system at $x$. Then: $U \in U_x \rightarrow x \in U$ $U,V \in U_x \rightarrow U \cap V \in U_x$ $U \in U_x$ implies that there is a $V\in U_x$ such that $U$ is a neighborhood of every point of $V$. If $U \in U_x$ and $U \subseteq V$ then $V \in U_x$
#Theorem Characterizing Nhood BasesLet $B_x$ be a neighborhood base of $x$. Then: $V \in B_x \rightarrow x \in V$ $V_1,V_2 \in B_x$ implies there is a $V_3 \in B_x$ such that $V_3 \subseteq V_1 \cap V_2$. If $V \in B_x$ then there is a set $U \in B_x$ such that, for every point $y \in U$, there is a set $W \in B_y$ with $W \subseteq V$.
#Theorem Characterizing BasesIf $\mathscr{B}$ is any collection of subsets of $X$ that satisfies the first two requirements for a base, then it defines a topology by taking the open sets to be unions of elements of $\mathscr{B}$.
#Theorem Characterizing SubbasesIf $X$ is a set, any collection of subsets of $X$ is a subbase for a topology.

Math 440, Fall 2014, Assignment 3

2014-09-22T14:01:22Z

Vincent.Luczkow: /* Definitions: */

__NOTOC__
''By one of those caprices of the mind, which we are perhaps most subject to in early youth, I at once gave up my former occupations; set down natural history and all its progeny as a deformed and abortive creation; and entertained the greatest disdain for a would-be science, which could never even step within the threshold of real knowledge. In this mood of mind I betook myself to the mathematics, and the branches of study appertaining to that science, as being built upon secure foundations, and so, worthy of my consideration.''

: - Mary Shelley, ''Frankenstein''

==Carefully define the following terms, then give one example and one non-example of each:==

# Nhood (of a point $x$ in a topological space $(X,\tau)$).
# Nhood system (at a point $x$ in a topological space $(X,\tau)$).
# Nhood base (at a point $x$ in a topological spae $(X,\tau)$).
# Basic nhood (of $x$).
# Base (for a topology).
# Basic open set (in $X$).
# Subbase (for a topology).

==Carefully state the following theorems (you do not need to prove them):==

# Theorem characterizing nhood systems (Theorem 4.2).
# Theorem characterizing nhood bases (Theorem 4.5).
# Theorem characterizing bases (Theorem 5.3).
# Theorem characterizing subbases (Theorem 5.6).

==Solve the following problems:==

# Problems 4A, 4D, 4F, 5A, and 5B.

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Nhood of a Point in a Topological Space:Let $(X,\tau)$ be a topological space, and $x \in X$. A neighborhood of $X$ is a set $U\subseteq X$ which contains $x$ and which contains an open set containing $x$. So $x \in V \subseteq U \subseteq X$, and $V \in \tau$.Example:In $\mathbb{R}^2$, the open unit square centered at $(0,0)$ is a neighborhood of $(0,0)$.Non-Example:
#Nhood System at a Point in a Topological Space:Let $(X,\tau)$ be a topological space, and $x \in X$. A neighborhood system of $x$ is the set of all neighborhoods of $x$.Example:In $\mathbb{R}^2$, the neighborhood system of $(0,0)$ is every set $P$ such that $P$ has a non-empty intersection with some open disk centered at $(0,0)$.Non-Example:
#Nhood Base at a Point in a Topological Space:A neighborhood base at $x \in X$ is a collection $B$ of open sets containing $x$, such that every open set containing $x$ is a superset of a member of $B$.$$
\forall S \in \tau: x \in S \rightarrow \exists B_{\lambda} \in B: B_{\lambda} \subseteq S
$$Example:The set of all open disks centered at $(0,0)$ with radius less than 1 is a neighborhood base at $(0,0)$.Non-Example:
#Basic Nhood of $x$:If $B$ is a neighborhood base of $x$, then an element of $B$ is a basic neighborhood of $x$.Example:In the previous example, he open half-disk would be a basic neighborhood of $(0,0)$.Non-Example:
#Base for a Topology:A base for a topological space $(X,\tau)$ is a collection of set $\mathscr{B}$ such that:$$
\bigcup_{B\in\mathscr{B}} B = X
$$If $B_1$ and $B_2$ are in $\mathscr{B}$ and $p\in B_1 \cap B_2$, then there is some $B_3$ such that $p \in B_3 \subseteq B_1 \cap B_2$ If a set is open in $\tau$, then it must be a union of members of $\mathscr{B}$.Example:The set of all open disks of radius less than 1 is a base for $\mathbb{R}^2$ with the Euclidean metric topology (? Proper terminology?)Non-Example:
#Basic Open Set:If $\mathscr{B}$ is a base for $\tau$, then a basic open set is a member of $\mathscr{B}$.Example:Non-Example:
#Subbase for a Topology:A subbase for $\tau$ is a collection of sets which, when closed under finite intersections, is a base for $\tau$.Example:Non-Example:
====Theorems:====
#Theorem Characterizing Nhood SystemsLet $U_x$ be the neighborhood system at $x$. Then: $U \in U_x \rightarrow x \in U$ $U,V \in U_x \rightarrow U \cap V \in U_x$ $U \in U_x$ implies that there is a $V\in U_x$ such that $U$ is a neighborhood of every point of $V$. If $U in U_x$ and $U \subseteq V$ then $V \in U_x$
#Theorem Characterizing Nhood BasesLet $B_x$ be a neighborhood base of $x$. Then: $V \in B_x \rightarrow x \in V$ $V_1,V_2 \in B_x$ implies there is a $V_3 \in B_x$ such that $V_3 \subseteq V_1 \cap V_2$. If $V \in B_x$ then there is a set $U \in B_x$ such that, for every point $y \in U$, there is a set $W \in B_y$ with $W \subseteq V$.
#Theorem Characterizing BasesIf $\mathscr{B}$ is any collection of subsets of $X$ that satisfies the first two requirements for a base, then it defines a topology by taking the open sets to be unions of elements of $\mathscr{B}$.
#Theorem Characterizing SubbasesIf $X$ is a set, any collection of subsets of $X$ is a subbase for a topology.

Math 440, Fall 2014, Assignment 3

2014-09-22T14:01:03Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 3

2014-09-22T14:00:23Z

Vincent.Luczkow: /* Definitions: */

__NOTOC__
''By one of those caprices of the mind, which we are perhaps most subject to in early youth, I at once gave up my former occupations; set down natural history and all its progeny as a deformed and abortive creation; and entertained the greatest disdain for a would-be science, which could never even step within the threshold of real knowledge. In this mood of mind I betook myself to the mathematics, and the branches of study appertaining to that science, as being built upon secure foundations, and so, worthy of my consideration.''

: - Mary Shelley, ''Frankenstein''

==Carefully define the following terms, then give one example and one non-example of each:==

# Nhood (of a point $x$ in a topological space $(X,\tau)$).
# Nhood system (at a point $x$ in a topological space $(X,\tau)$).
# Nhood base (at a point $x$ in a topological spae $(X,\tau)$).
# Basic nhood (of $x$).
# Base (for a topology).
# Basic open set (in $X$).
# Subbase (for a topology).

==Carefully state the following theorems (you do not need to prove them):==

# Theorem characterizing nhood systems (Theorem 4.2).
# Theorem characterizing nhood bases (Theorem 4.5).
# Theorem characterizing bases (Theorem 5.3).
# Theorem characterizing subbases (Theorem 5.6).

==Solve the following problems:==

# Problems 4A, 4D, 4F, 5A, and 5B.

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Nhood of a Point in a Topological Space:Let $(X,\tau)$ be a topological space, and $x \in X$. A neighborhood of $X$ is a set $U\subseteq X$ which contains $x$ and which contains an open set containing $x$. So $x \in V \subseteq U \subseteq X$, and $V \in \tau$.Example:In $\mathbb{R}^2$, the open unit square centered at $(0,0)$ is a neighborhood of $(0,0)$.Non-Example:
#Nhood System at a Point in a Topological Space:Let $(X,\tau)$ be a topological space, and $x \in X$. A neighborhood system of $x$ is the set of all neighborhoods of $x$.Example:In $\mathbb{R}^2$, the neighborhood system of $(0,0)$ is every set $P$ such that $P$ has a non-empty intersection with some open disk centered at $(0,0)$.Non-Example:
#Nhood Base at a Point in a Topological Space:A neighborhood base at $x \in X$ is a collection $B$ of open sets containing $x$, such that every open set containing $x$ is a superset of a member of $B$.$$
\forall S \in tau: x \in S \rightarrow \exists B_{lambda} \in B: B_{\lambda} \subseteq S
$$Example:The set of all open disks centered at $(0,0)$ with radius less than 1 is a neighborhood base at $(0,0)$.Non-Example:
#Basic Nhood of $x$:If $B$ is a neighborhood base of $x$, then an element of $B$ is a basic neighborhood of $x$.Example:In the previous example, he open half-disk would be a basic neighborhood of $(0,0)$.Non-Example:
#Base for a Topology:A base for a topological space $(X,\tau)$ is a collection of set $\mathscr{B}$ such that:$$
\bigcup_{B\in\mathscr{B}} B = X
$$If $B_1$ and $B_2$ are in $\mathscr{B}$ and $p\in B_1 \cap B_2$, then there is some $B_3$ such that $p \in B_3 \subseteq B_1 \cap B_2$ If a set is open in $\tau$, then it must be a union of members of $\mathscr{B}$.Example:The set of all open disks of radius less than 1 is a base for $\mathbb{R}^2$ with the Euclidean metric topology (? Proper terminology?)Non-Example:
#Basic Open Set:If $\mathscr{B}$ is a base for $\tau$, then a basic open set is a member of $\mathsrc{B}$.Example:Non-Example:
#Subbase for a Topology:A subbase for $\tau$ is a collection of sets which, when closed under finite intersections, is a base for $\tau$.Example:Non-Example:

Math 440, Fall 2014, Assignment 3

2014-09-21T18:02:08Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 3

2014-09-21T18:00:33Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 3

2014-09-21T18:00:16Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 3

2014-09-21T17:59:43Z

Vincent.Luczkow: /* Solutions: */

Math 440, Fall 2014, Assignment 2

2014-09-12T16:27:30Z

Vincent.Luczkow: /* Definitions: */

__NOTOC__
''We admit, in geometry, not only infinite magnitudes, that is to say, magnitudes greater than any assignable magnitude, but infinite magnitudes infinitely greater, the one than the other. This astonishes our dimension of brains, which is only about six inches long, five broad, and six in depth, in the largest heads.''

: - Voltaire

==Carefully define the following terms, then give one example and one non-example of each:==

# Pseudometric.
# Metric.
# Continuous at $x=x_0$.
# $\epsilon$-disc centered at $x_0$.
# Open set (in a given pseudometric space).
# Closed set (in a given pseudometric space).
# Topology (on a set $X$).
# Stronger (or finer) topology.
# Weaker (or coarser) topology.
# Discrete topology.
# Trivial topology.
# Open set (in a given topological space).
# Closed set (in a given topological space).
# Closure.
# Interior.
# Boundary (or frontier).

==Carefully state the following theorems (you need not prove them):==

# Fundamental properties of open sets (Theorem 2.6).
# Theorem characterizing continuity in terms of open sets (Theorem 2.8).
# Properties of closed sets (Theorem 3.4).
# Properties of the closure operation (Theorem 3.7).
# Properties of the interior operation (Theorem 3.11).
# Theorem relating the boundary of a set to its interior and closure (Theorem 3.14).

==Solve the following problems:==

# Consider the metric space $(X,\rho)$, where $X$ is the three-point set $\{a,b,c\}$ and $\rho$ is the discrete metric. If possible, give three distinct sequences that all converge to $a$, and three distinct sequences that do not. At least one of the latter sequences should have no limit at all. If any of these things is not possible, then explain why not.
# Consider the pseudo-metric space $(X,\rho)$, where $X$ is the three-point set $\{a,b,c\}$ and $\rho$ is the trivial pseudo-metric. If possible, give three distinct sequences that all converge to $a$, and three distinct sequences that do not. At least one of the latter sequences should have no limit at all. If any of these things is not possible, then explain why not.
# Section 2, problems A, B, and D (for B, recall that "$\mathrm{sup}$" is an abbreviation for "supremum," which in turn is a synonym for "least upper bound;" it is a fundamental property of the real number system that every non-empty set which is bounded above has a unique least upper bound).
# Section 3, problems A and F(1).

==Optional problems (not on the quiz or exams, but recommended):==

# Section 2, problem C. (Part 4 of this problem shows how pseudometrics arise in practice, while parts 1 and 2 show how to turn pseudometric spaces into metric spaces. If you ever study functional analysis, you will find yourself relying heavily on this construction. For example, the $L^p$ spaces begin life as pseudometric spaces, but one quickly replaces them by their metric identifications.)

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Pseudometric:Let $X$ be a set. A pseudometric on $X$ is a function $f:X\times X \rightarrow \mathbb{R}_{\geq 0}$ such that: $f(d,d) = 0$ for all $f\in X$. $f(a,b) = f(b,a)$ for all $a,b \in X$. $f(a,b) \leq f(a,c) + f(c,b)$ for all $a,b,c \in X$.Example:Non-Example:If $f$ maps all pairs to 1, then $f$ fails to satisfy the first requirement.
#Metric:A metric on $X$ is a pseudometric on $X$ with the additional requirement that $f(a,b) = 0 \rightarrow a=b$ for all $a,b \in X$.Example:Normal Euclidean distance on $\mathbb{R}$ is a metric.Non-Example:
#Continuous at $x = x_0$:Let $X$ be a metric space. A function $f:X \rightarrow X$ is continuous at $x=x_0$ if for all $\epsilon > 0$ there is a $\delta > 0$ such that if $|x - x_0| \lt \delta$, then $|f(x) - f(x_0)| \lt \epsilon$.Example:Non-Example:
#Topology:A topology on $X$ is a set $T$ of subsets of $X$, such that: $\emptyset \in T$. $X \in T$. For all $X_i \in T$ such that$i \in I$, $\bigcup_{i\in I} X_i \in T$. For all $X_1, X_2 \in T$, $X_1 \cap X_2 \in T$.Example:Non-Example:
#Topology:A topology on $X$ is a set $T$ of subsets of $X$, such that: $\emptyset \in T$. $X \in T$. For all $X_i \in T$ such that$i \in I$, $\bigcup_{i\in I} X_i \in T$. For all $X_1, X_2 \in T$, $X_1 \cap X_2 \in T$.Example:Non-Example:

Math 440, Fall 2014, Assignment 2

2014-09-12T16:25:52Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 2

2014-09-12T15:56:16Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 2

2014-09-12T15:21:05Z

Vincent.Luczkow: /* Definitions: */

__NOTOC__
''We admit, in geometry, not only infinite magnitudes, that is to say, magnitudes greater than any assignable magnitude, but infinite magnitudes infinitely greater, the one than the other. This astonishes our dimension of brains, which is only about six inches long, five broad, and six in depth, in the largest heads.''

: - Voltaire

==Carefully define the following terms, then give one example and one non-example of each:==

# Pseudometric.
# Metric.
# Continuous at $x=x_0$.
# $\epsilon$-disc centered at $x_0$.
# Open set (in a given pseudometric space).
# Closed set (in a given pseudometric space).
# Topology (on a set $X$).
# Stronger (or finer) topology.
# Weaker (or coarser) topology.
# Discrete topology.
# Trivial topology.
# Open set (in a given topological space).
# Closed set (in a given topological space).
# Closure.
# Interior.
# Boundary (or frontier).

==Carefully state the following theorems (you need not prove them):==

# Fundamental properties of open sets (Theorem 2.6).
# Theorem characterizing continuity in terms of open sets (Theorem 2.8).
# Properties of closed sets (Theorem 3.4).
# Properties of the closure operation (Theorem 3.7).
# Properties of the interior operation (Theorem 3.11).
# Theorem relating the boundary of a set to its interior and closure (Theorem 3.14).

==Solve the following problems:==

# Consider the metric space $(X,\rho)$, where $X$ is the three-point set $\{a,b,c\}$ and $\rho$ is the discrete metric. If possible, give three distinct sequences that all converge to $a$, and three distinct sequences that do not. At least one of the latter sequences should have no limit at all. If any of these things is not possible, then explain why not.
# Consider the pseudo-metric space $(X,\rho)$, where $X$ is the three-point set $\{a,b,c\}$ and $\rho$ is the trivial pseudo-metric. If possible, give three distinct sequences that all converge to $a$, and three distinct sequences that do not. At least one of the latter sequences should have no limit at all. If any of these things is not possible, then explain why not.
# Section 2, problems A, B, and D (for B, recall that "$\mathrm{sup}$" is an abbreviation for "supremum," which in turn is a synonym for "least upper bound;" it is a fundamental property of the real number system that every non-empty set which is bounded above has a unique least upper bound).
# Section 3, problems A and F(1).

==Optional problems (not on the quiz or exams, but recommended):==

# Section 2, problem C. (Part 4 of this problem shows how pseudometrics arise in practice, while parts 1 and 2 show how to turn pseudometric spaces into metric spaces. If you ever study functional analysis, you will find yourself relying heavily on this construction. For example, the $L^p$ spaces begin life as pseudometric spaces, but one quickly replaces them by their metric identifications.)

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Pseudometric:Let $X$ be a set. A pseudometric on $X$ is a function $f:X\times X \rightarrow \mathbb{R}_{\geq 0}$ such that: $f(d,d) = 0$ for all $f\in X$. $f(a,b) = f(b,a)$ for all $a,b \in X$. $f(a,b) \leq f(a,c) + f(c,b)$ for all $a,b,c \in X$.Example:Non-Example:If $f$ maps all pairs to 1, then $f$ fails to satisfy the first requirement.
#Metric:A metric on $X$ is a pseudometric on $X$ with the additional requirement that $f(a,b) = 0 \rightarrow a=b$ for all $a,b \in X$.Example:Normal Euclidean distance on $\mathbb{R}$ is a metric.Non-Example:
#Continuous at $x = x_0$:Let $X$ be a metric space. A function $f:\X \rightarrow X$ is continuous at $x=x_0$ if for all $\epsilon > 0$ there is a $\delta > 0$ such that if $|x - x_0| \lt \delta$, then $|f(x) - f(x_0)| \lt \epsilon$.Example:Non-Example:

Math 440, Fall 2014, Assignment 2

2014-09-12T15:19:39Z

Vincent.Luczkow: /* Definitions: */

__NOTOC__
''We admit, in geometry, not only infinite magnitudes, that is to say, magnitudes greater than any assignable magnitude, but infinite magnitudes infinitely greater, the one than the other. This astonishes our dimension of brains, which is only about six inches long, five broad, and six in depth, in the largest heads.''

: - Voltaire

==Carefully define the following terms, then give one example and one non-example of each:==

# Pseudometric.
# Metric.
# Continuous at $x=x_0$.
# $\epsilon$-disc centered at $x_0$.
# Open set (in a given pseudometric space).
# Closed set (in a given pseudometric space).
# Topology (on a set $X$).
# Stronger (or finer) topology.
# Weaker (or coarser) topology.
# Discrete topology.
# Trivial topology.
# Open set (in a given topological space).
# Closed set (in a given topological space).
# Closure.
# Interior.
# Boundary (or frontier).

==Carefully state the following theorems (you need not prove them):==

# Fundamental properties of open sets (Theorem 2.6).
# Theorem characterizing continuity in terms of open sets (Theorem 2.8).
# Properties of closed sets (Theorem 3.4).
# Properties of the closure operation (Theorem 3.7).
# Properties of the interior operation (Theorem 3.11).
# Theorem relating the boundary of a set to its interior and closure (Theorem 3.14).

==Solve the following problems:==

# Consider the metric space $(X,\rho)$, where $X$ is the three-point set $\{a,b,c\}$ and $\rho$ is the discrete metric. If possible, give three distinct sequences that all converge to $a$, and three distinct sequences that do not. At least one of the latter sequences should have no limit at all. If any of these things is not possible, then explain why not.
# Consider the pseudo-metric space $(X,\rho)$, where $X$ is the three-point set $\{a,b,c\}$ and $\rho$ is the trivial pseudo-metric. If possible, give three distinct sequences that all converge to $a$, and three distinct sequences that do not. At least one of the latter sequences should have no limit at all. If any of these things is not possible, then explain why not.
# Section 2, problems A, B, and D (for B, recall that "$\mathrm{sup}$" is an abbreviation for "supremum," which in turn is a synonym for "least upper bound;" it is a fundamental property of the real number system that every non-empty set which is bounded above has a unique least upper bound).
# Section 3, problems A and F(1).

==Optional problems (not on the quiz or exams, but recommended):==

# Section 2, problem C. (Part 4 of this problem shows how pseudometrics arise in practice, while parts 1 and 2 show how to turn pseudometric spaces into metric spaces. If you ever study functional analysis, you will find yourself relying heavily on this construction. For example, the $L^p$ spaces begin life as pseudometric spaces, but one quickly replaces them by their metric identifications.)

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Pseudometric:Let $X$ be a set. A pseudometric on $X$ is a function $f:X\times X \rightarrow \mathbb{R}_{\geq 0}$ such that: $f(d,d) = 0$ for all $f\in X$. $f(a,b) = f(b,a)$ for all $a,b \in X$. $f(a,b) \leq f(a,c) + f(c,b)$ for all $a,b,c \in X$.Example:Non-Example:If $f$ maps all pairs to 1, then $f$ fails to satisfy the first requirement.
#Metric:A metric on $X$ is a pseudometric on $X$ with the additional requirement that $f(a,b) = 0 \rightarrow a=b$ for all $a,b \in X$.Example:Normal Euclidean distance on $\mathbb{R}$ is a metric.Non-Example:
#Continuous at $x = x_0$:Let $X$ be a metric space. A function $f:\X \rightarrow X$ is continuous at $x=x_0$ if for all $\epsilon > 0$ there is a $\delta > 0$ such that if $|f(x) - f(x_0)| \lt \epsilon$, then $|x - x_0| \lt \delta$.Example:Non-Example:

Math 440, Fall 2014, Assignment 2

2014-09-12T15:04:54Z

Vincent.Luczkow: /* Solutions: */

__NOTOC__
''We admit, in geometry, not only infinite magnitudes, that is to say, magnitudes greater than any assignable magnitude, but infinite magnitudes infinitely greater, the one than the other. This astonishes our dimension of brains, which is only about six inches long, five broad, and six in depth, in the largest heads.''

: - Voltaire

==Carefully define the following terms, then give one example and one non-example of each:==

# Pseudometric.
# Metric.
# Continuous at $x=x_0$.
# $\epsilon$-disc centered at $x_0$.
# Open set (in a given pseudometric space).
# Closed set (in a given pseudometric space).
# Topology (on a set $X$).
# Stronger (or finer) topology.
# Weaker (or coarser) topology.
# Discrete topology.
# Trivial topology.
# Open set (in a given topological space).
# Closed set (in a given topological space).
# Closure.
# Interior.
# Boundary (or frontier).

==Carefully state the following theorems (you need not prove them):==

# Fundamental properties of open sets (Theorem 2.6).
# Theorem characterizing continuity in terms of open sets (Theorem 2.8).
# Properties of closed sets (Theorem 3.4).
# Properties of the closure operation (Theorem 3.7).
# Properties of the interior operation (Theorem 3.11).
# Theorem relating the boundary of a set to its interior and closure (Theorem 3.14).

==Solve the following problems:==

# Consider the metric space $(X,\rho)$, where $X$ is the three-point set $\{a,b,c\}$ and $\rho$ is the discrete metric. If possible, give three distinct sequences that all converge to $a$, and three distinct sequences that do not. At least one of the latter sequences should have no limit at all. If any of these things is not possible, then explain why not.
# Consider the pseudo-metric space $(X,\rho)$, where $X$ is the three-point set $\{a,b,c\}$ and $\rho$ is the trivial pseudo-metric. If possible, give three distinct sequences that all converge to $a$, and three distinct sequences that do not. At least one of the latter sequences should have no limit at all. If any of these things is not possible, then explain why not.
# Section 2, problems A, B, and D (for B, recall that "$\mathrm{sup}$" is an abbreviation for "supremum," which in turn is a synonym for "least upper bound;" it is a fundamental property of the real number system that every non-empty set which is bounded above has a unique least upper bound).
# Section 3, problems A and F(1).

==Optional problems (not on the quiz or exams, but recommended):==

# Section 2, problem C. (Part 4 of this problem shows how pseudometrics arise in practice, while parts 1 and 2 show how to turn pseudometric spaces into metric spaces. If you ever study functional analysis, you will find yourself relying heavily on this construction. For example, the $L^p$ spaces begin life as pseudometric spaces, but one quickly replaces them by their metric identifications.)

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
====Definitions:====
#Pseudometric:Let $X$ be a set. A pseudometric on $X$ is a function $f:X\times X \rightarrow \mathbb{R}_{\geq 0}$ such that: $f(d,d) = 0$ for all $f\in X$. $f(a,b) = f(b,a)$ for all $a,b \in X$.Example:Non-Example:

Math 440, Fall 2014, Assignment 1

2014-09-08T17:25:11Z

Vincent.Luczkow: /* Theorems: */

__NOTOC__
''The beginner ... should not be discouraged if ... he finds that he does not have the prerequisites for reading the prerequisites.''

: - P. Halmos

==Carefully define the following terms, then give one example and one non-example of each:==

# Cartesian product (of two sets).
# Power set (of a set).
# Equinumerous (sets).
# Countable set.
# Uncountable set.
# Cardinality of the continuum.
# Partial order.
# Maximal element (of a partially ordered set).
# Largest element (of a partially ordered set).
# Chain (in a partially ordered set).

==Carefully state the following theorems (you need not prove them):==

# Cantor-Bernstein Theorem.
# Cantor's Theorem.
# Continuum Hypothesis (of course this is not a theorem, though it is sometimes taken as an axiom).
# Axiom of Choice (see above).
# Zorn's Lemma.

==Solve the following problems:==

# Prove Cantor's Theorem (exercise 1I.1 contains many hints).
# Problems 1E and 1H (you will use the results of 1H incessantly for the rest of the semester).

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
#Cartesian Product of Two Sets:Let $A$ and $B$ be sets. The cartesian product of $A$ and $B$, $A \times B$, is the set:$$
A \times B = \{(a,b)| a\in A , b \in B\}
$$Example:The cartesian product of $\{1,2\}$ and $\{a,b\}$ is $\{(1,a), (1,b), (2,a), (2,b)\}$Non-Example:Remember that the Cartesian product is a set of tuples. Don't confuse it with the union of two sets, which would be $\{1,2,a,b\}$ for the above example.

#Power Set of a Set:Let $A$ be a set. The power set of $A$ is the set of all subsets of $A$: $$
\mathscr{P}(A) = \{B | B \subseteq A\}
$$Example:The power set of $\{1,2\}$ is $\{\emptyset,\{1\}, \{2\}, \{1,2\}\}$Non-Example:
#Equinumerous Sets:Two sets $A$ and $B$ are equinumerous if there is a bijection between the two sets.Example:$\mathbb{Z}$ and $\mathbb{Q}$ are equinumerous.Non-Example:$\mathbb{Z}$ and $\mathbb{R}$ are not equinumerous.
#Countable Set:A set $S$ is countable if there is an injection from $S$ to $\mathbb{Z}$ (or any other countable set.)Example:$\{a,b,c\}$ is countable - it can easily be injected to $\mathbb{Z}$.Non-Example:$\mathbb{R}$
#Uncountable Set:A set is uncountable if there is no injection from it to $\mathbb{Z}$.Example:$\mathbb{R}$.Non-Example:
#Cardinality of the Continuum:The cardinality of the continuum, $\mathbb{c}$, is the cardinality of the real numbers.Example:$\mathbb{R}$.Non-Example:$\mathbb{Z}$
#Partial Order:A partial ordering on a set $A$ is a relation that is reflexive, transitive, and anti-symmetric. A set with a partial order is also called a poset.Example:$(\mathbb{R},\leq)$ is a partial ordering of the reals.Non-Example:$(\mathbb{R},\lt)$ is not a partial ordering - it is not reflexive.
#Maximal Element of a Poset:If $(A,\leq)$ is a poset, then an element $b \in A$ is maximal if there is no $c \in A$ such that $b \leq c$.Example:a is maximal in the alphabet, by alphabetical order.Non-Example:There is no maximal natural number.
#Largest Element of a Poset:If $(A,\leq)$ is a poset, then an element $b \in A$ is the largest element of $A$ if for all $c\in A$ $b \geq c$.Example:3 is the largest element of the set $\{1,2,3\}$.Non-Example:The set $(0,1)$ has no largest element (within itself).
#Chain in a Poset:Given a poset $(A,\leq)$, a chain in $A$ is a totally ordered subset $B$. Meaning for every two elements $a,b \in B$, either $(a,b) \in \leq$ or $(b,a) \in \leq$.Example:The integers are a chain in the real numbers with respect to $\leq$.Non-Example:

====Theorems:====
#Cantor-Bernstein TheoremLet $A$ and $B$ be sets. If there are injections $f:A \rightarrow B$ and $g:B \rightarrow A$, then $|A| = |B|$. (Does not require the axiom of choice).
#Cantor's TheoremFor all sets $A$, $\mathscr{P}(A)$ has strictly greater cardinality than $A$.
#Continuum HypothesisThere does not exist a set $A$ such that $|\mathbb{Z}| \lt |A| \lt \mathfrak{c}$.
#Axiom of ChoiceLet $X$ be a set of sets. Then there exists a function $f$ such that $f(x) \in x$ for all $x \in X$. Less formally, if we have a collection of sets, we have a way to choose one element from each of those sets.
#Zorn's LemmaLet $A$ be a poset. If every chain in $A$ has a maximal element, then $A$ has a maximal element.

====Book Problems:====

Math 440, Fall 2014, Assignment 1

2014-09-08T17:24:51Z

Vincent.Luczkow: /* Theorems: */

__NOTOC__
''The beginner ... should not be discouraged if ... he finds that he does not have the prerequisites for reading the prerequisites.''

: - P. Halmos

==Carefully define the following terms, then give one example and one non-example of each:==

# Cartesian product (of two sets).
# Power set (of a set).
# Equinumerous (sets).
# Countable set.
# Uncountable set.
# Cardinality of the continuum.
# Partial order.
# Maximal element (of a partially ordered set).
# Largest element (of a partially ordered set).
# Chain (in a partially ordered set).

==Carefully state the following theorems (you need not prove them):==

# Cantor-Bernstein Theorem.
# Cantor's Theorem.
# Continuum Hypothesis (of course this is not a theorem, though it is sometimes taken as an axiom).
# Axiom of Choice (see above).
# Zorn's Lemma.

==Solve the following problems:==

# Prove Cantor's Theorem (exercise 1I.1 contains many hints).
# Problems 1E and 1H (you will use the results of 1H incessantly for the rest of the semester).

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
#Cartesian Product of Two Sets:Let $A$ and $B$ be sets. The cartesian product of $A$ and $B$, $A \times B$, is the set:$$
A \times B = \{(a,b)| a\in A , b \in B\}
$$Example:The cartesian product of $\{1,2\}$ and $\{a,b\}$ is $\{(1,a), (1,b), (2,a), (2,b)\}$Non-Example:Remember that the Cartesian product is a set of tuples. Don't confuse it with the union of two sets, which would be $\{1,2,a,b\}$ for the above example.

#Power Set of a Set:Let $A$ be a set. The power set of $A$ is the set of all subsets of $A$: $$
\mathscr{P}(A) = \{B | B \subseteq A\}
$$Example:The power set of $\{1,2\}$ is $\{\emptyset,\{1\}, \{2\}, \{1,2\}\}$Non-Example:
#Equinumerous Sets:Two sets $A$ and $B$ are equinumerous if there is a bijection between the two sets.Example:$\mathbb{Z}$ and $\mathbb{Q}$ are equinumerous.Non-Example:$\mathbb{Z}$ and $\mathbb{R}$ are not equinumerous.
#Countable Set:A set $S$ is countable if there is an injection from $S$ to $\mathbb{Z}$ (or any other countable set.)Example:$\{a,b,c\}$ is countable - it can easily be injected to $\mathbb{Z}$.Non-Example:$\mathbb{R}$
#Uncountable Set:A set is uncountable if there is no injection from it to $\mathbb{Z}$.Example:$\mathbb{R}$.Non-Example:
#Cardinality of the Continuum:The cardinality of the continuum, $\mathbb{c}$, is the cardinality of the real numbers.Example:$\mathbb{R}$.Non-Example:$\mathbb{Z}$
#Partial Order:A partial ordering on a set $A$ is a relation that is reflexive, transitive, and anti-symmetric. A set with a partial order is also called a poset.Example:$(\mathbb{R},\leq)$ is a partial ordering of the reals.Non-Example:$(\mathbb{R},\lt)$ is not a partial ordering - it is not reflexive.
#Maximal Element of a Poset:If $(A,\leq)$ is a poset, then an element $b \in A$ is maximal if there is no $c \in A$ such that $b \leq c$.Example:a is maximal in the alphabet, by alphabetical order.Non-Example:There is no maximal natural number.
#Largest Element of a Poset:If $(A,\leq)$ is a poset, then an element $b \in A$ is the largest element of $A$ if for all $c\in A$ $b \geq c$.Example:3 is the largest element of the set $\{1,2,3\}$.Non-Example:The set $(0,1)$ has no largest element (within itself).
#Chain in a Poset:Given a poset $(A,\leq)$, a chain in $A$ is a totally ordered subset $B$. Meaning for every two elements $a,b \in B$, either $(a,b) \in \leq$ or $(b,a) \in \leq$.Example:The integers are a chain in the real numbers with respect to $\leq$.Non-Example:

====Theorems:====
#Cantor-Bernstein TheoremLet $A$ and $B$ be sets. If there are injections $f:A \rightarrow B$ and $g:B \rightarrow A$, then $|A| = |B|$. (Does not require the axiom of choice).
#Cantor's TheoremFor all sets $A$, $\mathscr{P}(A)$ has strictly greater cardinality than $A$.
#Continuum HypothesisThere does not exist a set $A$ such that $|\mathbb{Z}| \lt |A| \lt \mathbb{c}$.
#Axiom of ChoiceLet $X$ be a set of sets. Then there exists a function $f$ such that $f(x) \in x$ for all $x \in X$. Less formally, if we have a collection of sets, we have a way to choose one element from each of those sets.
#Zorn's LemmaLet $A$ be a poset. If every chain in $A$ has a maximal element, then $A$ has a maximal element.

====Book Problems:====

Math 440, Fall 2014, Assignment 1

2014-09-08T17:24:12Z

Vincent.Luczkow: /* Theorems: */

__NOTOC__
''The beginner ... should not be discouraged if ... he finds that he does not have the prerequisites for reading the prerequisites.''

: - P. Halmos

==Carefully define the following terms, then give one example and one non-example of each:==

# Cartesian product (of two sets).
# Power set (of a set).
# Equinumerous (sets).
# Countable set.
# Uncountable set.
# Cardinality of the continuum.
# Partial order.
# Maximal element (of a partially ordered set).
# Largest element (of a partially ordered set).
# Chain (in a partially ordered set).

==Carefully state the following theorems (you need not prove them):==

# Cantor-Bernstein Theorem.
# Cantor's Theorem.
# Continuum Hypothesis (of course this is not a theorem, though it is sometimes taken as an axiom).
# Axiom of Choice (see above).
# Zorn's Lemma.

==Solve the following problems:==

# Prove Cantor's Theorem (exercise 1I.1 contains many hints).
# Problems 1E and 1H (you will use the results of 1H incessantly for the rest of the semester).

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
#Cartesian Product of Two Sets:Let $A$ and $B$ be sets. The cartesian product of $A$ and $B$, $A \times B$, is the set:$$
A \times B = \{(a,b)| a\in A , b \in B\}
$$Example:The cartesian product of $\{1,2\}$ and $\{a,b\}$ is $\{(1,a), (1,b), (2,a), (2,b)\}$Non-Example:Remember that the Cartesian product is a set of tuples. Don't confuse it with the union of two sets, which would be $\{1,2,a,b\}$ for the above example.

#Power Set of a Set:Let $A$ be a set. The power set of $A$ is the set of all subsets of $A$: $$
\mathscr{P}(A) = \{B | B \subseteq A\}
$$Example:The power set of $\{1,2\}$ is $\{\emptyset,\{1\}, \{2\}, \{1,2\}\}$Non-Example:
#Equinumerous Sets:Two sets $A$ and $B$ are equinumerous if there is a bijection between the two sets.Example:$\mathbb{Z}$ and $\mathbb{Q}$ are equinumerous.Non-Example:$\mathbb{Z}$ and $\mathbb{R}$ are not equinumerous.
#Countable Set:A set $S$ is countable if there is an injection from $S$ to $\mathbb{Z}$ (or any other countable set.)Example:$\{a,b,c\}$ is countable - it can easily be injected to $\mathbb{Z}$.Non-Example:$\mathbb{R}$
#Uncountable Set:A set is uncountable if there is no injection from it to $\mathbb{Z}$.Example:$\mathbb{R}$.Non-Example:
#Cardinality of the Continuum:The cardinality of the continuum, $\mathbb{c}$, is the cardinality of the real numbers.Example:$\mathbb{R}$.Non-Example:$\mathbb{Z}$
#Partial Order:A partial ordering on a set $A$ is a relation that is reflexive, transitive, and anti-symmetric. A set with a partial order is also called a poset.Example:$(\mathbb{R},\leq)$ is a partial ordering of the reals.Non-Example:$(\mathbb{R},\lt)$ is not a partial ordering - it is not reflexive.
#Maximal Element of a Poset:If $(A,\leq)$ is a poset, then an element $b \in A$ is maximal if there is no $c \in A$ such that $b \leq c$.Example:a is maximal in the alphabet, by alphabetical order.Non-Example:There is no maximal natural number.
#Largest Element of a Poset:If $(A,\leq)$ is a poset, then an element $b \in A$ is the largest element of $A$ if for all $c\in A$ $b \geq c$.Example:3 is the largest element of the set $\{1,2,3\}$.Non-Example:The set $(0,1)$ has no largest element (within itself).
#Chain in a Poset:Given a poset $(A,\leq)$, a chain in $A$ is a totally ordered subset $B$. Meaning for every two elements $a,b \in B$, either $(a,b) \in \leq$ or $(b,a) \in \leq$.Example:The integers are a chain in the real numbers with respect to $\leq$.Non-Example:

====Theorems:====
#Cantor-Bernstein TheoremLet $A$ and $B$ be sets. If there are injections $f:A \rightarrow B$ and $g:B \rightarrow A$, then $|A| = |B|$. (Does not require the axiom of choice).
#Cantor's TheoremFor all sets $A$, $\mathscr{P}(A)$ has strictly greater cardinality than $A$.
#Continuum HypothesisThere does not exist a set $A$ such that $|\mathbb{Z}| \lt |A| \lt \mathbb{C}$.
#Axiom of ChoiceLet $X$ be a set of sets. Then there exists a function $f$ such that $f(x) \in x$ for all $x \in X$. Less formally, if we have a collection of sets, we have a way to choose one element from each of those sets.
#Zorn's LemmaLet $A$ be a poset. If every chain in $A$ has a maximal element, then $A$ has a maximal element.

====Book Problems:====

Math 440, Fall 2014, Assignment 1

2014-09-06T00:31:02Z

Vincent.Luczkow: /* Solutions: */

Math 440, Fall 2014, Assignment 1

2014-09-05T21:42:56Z

Vincent.Luczkow: /* Solutions: */

Math 440, Fall 2014, Assignment 1

2014-09-05T21:42:17Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 1

2014-09-05T18:15:22Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 1

2014-09-05T16:44:42Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 1

2014-09-05T16:42:09Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 1

2014-09-05T15:57:37Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 1

2014-09-05T15:57:03Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 1

2014-09-05T15:49:48Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 1

2014-09-05T15:49:20Z

Vincent.Luczkow: /* Definitions: */

Math 440, Fall 2014, Assignment 1

2014-09-05T15:48:42Z

Vincent.Luczkow: /* Solutions: */

Math 361, Spring 2014, Assignment 14

2014-05-12T14:47:30Z

Vincent.Luczkow: /* Theorems */

__NOTOC__
==Carefully define the following terms, then give one example and one non-example of each:==
# Splitting field (of a nonconstant polynomial $p\in F[x]$).
# Galois group (of a nonconstant $p\in F[x]$).

==Carefully state the following theorems (you need not prove them):==
# Theorem on existence and uniqueness of splitting fields.
# Theorem concerning finiteness of the Galois group.

==Solve the following problems:==
# Give an example of a domain which is ''not'' a UFD.
# Consider the polynomials $p=x^2-2, q=x^3-2,$ and $r=x^4-2$ in $\mathbb{Q}[x]$. Let $E_1, E_2,$ and $E_3$ be the splitting fields of $p, q,$ and $r$, respectively, over $\mathbb{Q}$. Compute the dimensions $[E_1:\mathbb{Q}], [E_2:\mathbb{Q}],$ and $[E_3:\mathbb{Q}]$. ''(Hint: follow the construction used to prove existence of splitting fields, and use the Dimension Formula. To compute the dimension of $E_3$, you may find the following fact helpful: $x^4-2 = (x^2 + \sqrt{2})(x^2-\sqrt{2}) = (x^2 + \sqrt{2})(x + \sqrt[4]{2})(x - \sqrt[4]{2})$.)''
# Compute the Galois group of $x^2+1\in\mathbb{R}[x]$.

==Questions:==
# I just want to make sure my definition for Galois Group is right. When it is a non-constant polynomial and a field, I believe it is the splitting field of that non-constant polynomial over the field. But there was also an important Galois Group definition with two splitting fields where the group contains all isomophisms making the commutative diagram.
==Solutions:==
===Definitions===
'''Definition:''' Splitting field (of a nonconstant polynomial $p\in F[x]$). 
Fix a polynomial of strictly positive degree $p \in F[x]$. A splitting field for $p$ is a field extension $F \to E$ for which both of the following are true:
# p splits over E
# $E=F(\alpha_1,\alpha_2,\dots,\alpha_n)$,and every $\alpha_i$ is a root of $p$.
'''Example:''' Working over $\mathbb{R}$, $\mathbb{C}$ is a splitting field for $x^2+1$.

'''Definition:''' Galois Group (of a nonconstant $p\in F[x]$). 
Fix a polynomial of strictly positive degree $p \in F[x]$. Define the Galois Group of $p$, denoted $Gal_F(p)$, to be the splitting field of $p$ over $F$. 
'''Example:''' $Gal_{\mathbb{R}}(x^2+1)=\mathbb{C}$

===Theorems===
#Existence and Uniqueness of Splitting FieldsLEt $p$ be a polynomial in $F[x]$, where $F$ is a field. A splitting field of $p$ over $F$ exists, and it is isomorphic to all other splitting fields of $p$ over $F$.
#Finiteness of Galois GroupThe Galois group of $p$ over $F$ is finite, and is a subgroup of $S_n$, where $n$ is the number of distinct roots of $p$.

===Problems===

Math 361, Spring 2014, Assignment 14

2014-05-12T14:45:26Z

Vincent.Luczkow: /* Theorems */

Math 361, Spring 2014, Assignment 14

2014-05-12T14:41:58Z

Vincent.Luczkow: /* Solutions: */

Midterm 2 Notes:Math 361, Spring 2014

2014-04-13T23:25:32Z

Vincent.Luczkow: /* Algebraic Closure of a Field */

=Math 361 Midterm 2 Notes=
==Linear Algebra==
===Vector Space===

Let $F$ be a field. A vector space over $F$ is a set $V$ with two functions, $+$ and $\cdot$. $+$ is vector addition, $\cdot$ is scalar multiplication.
 
$V$ is an abelian group with respect to vector addition. $+$ is a function from $V\times V \rightarrow V$, and it's associative, commutative, $V$ contains an additive identity (the zero vector, or $\vec{0}$), and $V$ contains an additive inverse for every element. The additive inverse of $\vec{v}$ is $-\vec{v}$.
 
Scalar multiplication is more complicated. Scalar multiplication is an operator from $V \times F \rightarrow V$. So it multiplies a field element by a vector. Scalar multiplication has some properties that kind of look like associativity and distributivity, but they're not quite the same. Anyway:
$$
(c +d) \cdot \vec{v} = c \cdot \vec{v} + d \cdot \vec{v}\\
c \cdot (\vec{v} + \vec{w}) = c \cdot \vec{v} + c \cdot \vec{w}\\
cd \cdot \vec{v} = c \cdot (c \cdot \vec{v}) \\
1 \cdot \vec{v} = \vec{v}
$$
Okay, some explanation. For the purposes of the midterm, there are only three types of vector space that we care about.
 
First - the $n$ tuples of $F$. So $F \times F \times \ldots \times F$. This is a vector space. Vector addition and scalar multiplication is just component wise. $\mathbb{R}^3$ is an example of such a vector space. $(1,2,3)$ and $(2,3,4)$ are both elements of $\mathbb{R}^3$. We add them component-wise:
$$
(1,2,3) + (2,3,4) = (1+2, 2+3, 3+4) = (3,5,7)
$$
$3$ is a scalar (it's an element of the base field, $\mathbb{R}$, and we do scalar multiplication component-wise:
$$
3\cdot (1,2,3) = (3*1, 3*2, 3*3) = (3,6,9)
$$
 
The next example is the polynomials over a field. Here a polynomial is a vector, and vector addition is just polynomial addition. Scalar multiplication is not polynomial multiplication - the scalars are elements of the base field, i.e. they're constant polynomials.
 
Consider $\mathbb{R}[x]$ - the real polynomials. A scalar here is $5$. To do scalar multiplication, you turn this into a constant polynomial (getting, surprise, $5$), and then you multiply this constant polynomial by your actual polynomial. So:
$$
5 \cdot 5 + 3x + 4x^2 = 5(5 + 3x + 4x^2) = 25 + 15x + 20x^2
$$
You can also consider polynomials of limited degree - so all the polynomials up to degree 5, for instance. This is actually the same as 6-tuples of the base field. $5 + 4x + 3x^2 + 2x^3 + x^4 + x^5$ can just be represented as $(5,4,3,2,1,1)$, which is an element of $\mathbb{R}^6$.
 
The third example is the rational functions over a field. Vector addition is just rational function addition, and scalar multiplication is (like for polynomials) just multiplication by a constant rational function.


===Linear Combinations and Other Things===

Let $F$ be a field and $V$ a vector space over $F$. Let $\vec{v}_1, \ldots, \vec{v}_n$ be vectors in $V$. A linear combination of these vectors is:
$$
c_1v_1 + c_2v_2 + \ldots + c_nv_n
$$
where $c_i \in F$.
 
So let $F$ be $\mathbb{R}$ and $V$ be $\mathbb{R}^3$. Let our set of vectors be:
$$
(1,2,3)\\
(2,3,4)\\
(3,4,5)
$$
One possible linear combination of these vectors is:
$$
3(1,2,3) + 25(2,3,4) - \frac{1}{4}(3,4,5)
$$
We can replace 3, 25, and $-\frac{1}{4}$ with any elements of $\mathbb{R}$ we want, and still get a linear combination of these three vectors.
====The Span of a Set====

Let's say we have a set of vectors $V = \{\vec{v}_1, \ldots, \vec{v}_n\}$. The span of this set is the set of all vectors that are linear combinations of these vectors. So the span of $\{(1,0,0),(0,1,0)\}$ is $\{(x,y,0) | x,y \in \mathbb{R}\}$, because:
$$
x(1,0,0) + y(0,1,0) = (x,0,0) + (0,y,0) = (x,y,0)
$$


====A Spanning Set====

Let $T$ be a subset of a vector space $V$. A third set, $S$ is a spanning set of $T$ if every vector in $T$ is a linear combination of vectors in $S$. So we can multiply vectors in $S$ by scalars, add them together, and get anything in $T$.
 
We also say that $S$ spans $T$.
 
So let $T = \{(x,0,0)\in \mathbb{R}^3| x \in \mathbb{R}\}$. This is the set of points on the $x$-axis. One spanning set of $T$ is the set $S = \{(1,0,0)\}$. We can scale this vector by $x$ to get $(x,0,0)$, so we can get all of $T$. Another spanning set is all of $\mathbb{R}^3$. Obviously everything in $T$ is a linear combination of everything in $\mathbb{R}^3$, because everything in $T$ is already in $\mathbb{R}^3$. This illustrates an important point: $S$ doesn't have to only span $T$. The span of $S$ can be much larger than $T$, so long as $T$ is in the span of $S$.


====Linear Independence====

Let $V$ be a vector space and $S$ be a set of vectors. $S$ is linearly independent (well, really, the vectors in $S$ are linearly independent) if none of them is a linear combination of any of the others.
 
Let $S = \{\vec{v}_1,\ldots,\vec{v}_n$. These vectors are linearly independent if and only if:
$$
c_1\vec{v}_1 + \cdots + c_n\vec{v}_n = \vec{0}
$$
implies that $c_i = 0$ for all $i$. So the only way to linearly combine the vectors and get $\vec{0}$ is for the scalars to all be 0.
 
Consider $S = \{(1,0,0),(0,1,0)\}$. These vectors are linearly independent. If you add them up, you get:
$$
c(1,0,0) + d(0,1,0) = (c,d,0)
$$
This only equals 0 if $c$ and $d$ are 0, so the vectors are linearly independent.


====Basis for a Vector Space====

If $V$ is a vector space, than a basis for $V$ is a set $B$ of linearly independent vectors whose span is $V$. So none of the vectors in $B$ can be linearly combined to get another vector in $B$, but if we use all the vectors in $B$ we get every vector in the vector space.
 
$\{(1,0,0),(0,1,0),(0,0,1)\}$ is a basis for $\mathbb{R}^3$. Obviously two of them can't be combined to get the third (the proof is basically the same as for the last section), but if we combine all of them we get:
$$
c(1,0,0) + d(0,1,0) + e(0,0,1) = (c,d,e)
$$
which is everything in $\mathbb{R}^3$.
 
Another example. $\{1,x,x^2\}$ is a basis for the real polynomials of degree less than or equal to 2. Any polynomial of this degree can be written:
$$
p = a*1 + bx + cx^2 = a\cdot1 + b\cdot x + c\cdot x^2
$$


====Dimension of a Vector Space====

There is a very important theorem that says that all bases of a vector space contain the same number of elements. We call this number of elements the dimension of the vector space.
 
The dimension of $\mathbb{R}^3$ is 3, because we created a basis with three elements earlier.
 
When you consider vector spaces of polynomials, you need one basis element for every variable, and one basis element for every power of every variable, and one basis element for every way to multiply variables together plus one for the constant polynomials.
 
So $\mathbb{R}[x]$ is infinite dimensional. There's only one variable $x$, but we can have arbitrarily high powers of $x$. If we limit ourselves to, say, polynomials of degree 6 or less, we get a 7-dimensional vector space.
 
For $\mathbb{R}[x,y]$, limited to polynomials of total degree two or less, we get:
$$
\{1,x,x^2,xy,y,y^2\}
$$
which is 6-dimensional.


==More Field Theory==

So, time to apply some of that stuff to field theory.


===Extension Fields are Vector Spaces===

If we have a field extension $F,E,\iota$, then $E$ is a vector space over $F$.
 
This is pretty obvious - we already know, from the classification of simple extensions that $E$ is either a set of polynomials over $F$, or a set of rational expressions over $F$, both of which are vector spaces over $F$. (Scalar multiplication is just multiplication by a constant.)
 
I should say I'm not sure if this is true of all field extensions. It's definitely true of all finitely generated field extensions (field extensions created by adjoining a finite number of elements to the base field), but it might not be true for all field extensions.


====Extensions of Extensions====

If $E$ is an extension of $K$, and $K$ is an extension of $F$, then $E$ is an extension of $F$.


====Dimension of an Extension Field====

If $E$ is a (let's just say finitely-generated) extension field over $F$, then the dimension of $E$ over $F$, written $[E:F]$, is the dimension of $E$ as a vector space over $F$.


====Dimension Formula====

If $E$ is an extension of $K$ and $K$ is an extension of $F$, then $[E:F] = [E:K][K:F]$.


===Algebraic Extension===

Let $F,E,\iota$ be a field extension. This extension is algebraic if every element of $E$ is algebraic over $F$.
 
For example, $E$ is algebraic over $F$ if $E$ was created by adjoining only algebraic elements.
 
$\mathbb{Q}[x]/\langle irr(\sqrt{2})\rangle$ is algebraic over $\mathbb{Q}$.


====Finite Extensions====

An extension $E$ of $F$ is finite dimensional if and only if it is algebraic and finitely-generated.
 
This makes sense. If it's not algebraic, then it's isomorphic to some type of rational expressions over $F$. The rational expressions over a field are infinite dimensional, so $E$ couldn't be finite dimensional if it was the rational expressions.
 
What if it's algebraic? Well, then it's the set of polynomials over a lot of variables. Furthermore, the degree of these polynomials is limited to some finite value. So the dimension of this vector space will be finite if and only if the number of variables is finite. The number of variables is finite if and only if the extension is finitely generated (each extension adds a single variable).


===Algebraically Closed Fields===

$F$ is a field. $F$ is algebraically closed under the following (equivalent) conditions:
 
Every polynomial in $F[x]$ has a root in $F$.
 
Every polynomial in $F[x]$ splits into linear factors.
 
The only algebraic extension of $F$ is the trivial extension.


====Algebraic Closure of a Field====

Let $F$ be a field. An algebraic closure of $F$ is an extension


==Constructibility==

Constructibility deals with constructing shapes with a compass and straightedge.
 
Geometric constructions work like this:
 
You start out with two points. You are allowed to draw lines between any points you find. Also, if there is a line of length $n$ between two points, you can construct a circle of radius $n$ centered at any point you've found.
 
So really you're trying to create a sequence of points.
 
We say a number $x$ is constructible if we can find a line segment of length $x$.
 
I'm not going to explain how to do all the constructions, mainly because it's a graphical problem and this is a text only interface. But we can construct any integer, and any rational number, and some other things to be explained later.

===The Constructible Numbers are a Field===

So if we have two constructible numbers, we can add, multiply, subtract, and divide. Geometrically this means if we have to lines of length $x$ and $y$, we can find a third line of length $x+y$, and another of length $x-y$, and $xy$ and $x/y$.
 
Proving this is, unfortunately, a graphical proof. Try google.


===Constructible Numbers are Closed on Square Roots===

So we know the constructible numbers are a field. This means they must at least be the rational numbers. But given any constructible number, we can find its square root, meaning the constructible numbers are not just the rational numbers.
 
So $\sqrt{2}$ is in $C$, and $\sqrt{2} + \sqrt{3}$ is too, and so is $\sqrt{\sqrt{2} + \sqrt{3}}$.
 
Also, square roots are the only thing we add. This makes sense - getting a new constructible number means we need a new line, so we need a new point. We can only create new points by intersecting lines, intersecting a line and a circle, or intersecting circles. All of these end of being linear or quadratic operations, so there can't be any way to extract, say, a cube root.

====Doubling the Cube is Impossible====

We have a cube of side length $s$. We want to find a cube with twice the volume. Let the volume of the first cube be $V = s^3$, and the second cube be $W = t^3$. We also have $W = 2V$. So $t^3 = 2s^3$, so $t = \sqrt[3]{2}s$. Solving this would require finding the cube root of 2, which we can't do.


====Trisecting the Angle is Impossible====

The proof is to show that one angle (60 degrees) can't be trisected.


==Advanced Group Theory==
===New Group Operations===
====Meet====

Let $H$ and $K$ be subgroups of $G$. The meet of $H$ and $K$, written $H \wedge K$, is the intersection of $H$ and $K$. This is also the largest subgroup that is contained by both $H$ and $K$.


====Join====

Let $H$ and $K$ be subgroups of $G$. The join of $H$ and $K$, written $H \vee K$, is the smallest subgroup that contains both $H$ and $K$. This is not necessarily the union of $H$ and $K$.


====Product====

Let $H$ and $K$ be subgroups of $G$. The product of $H$ and $K$, written $HK$, is the set:
$$
HK = \{hk | h \in H, k\in K\}
$$
This is only a group if $H$ or $K$ is normal in $G$.
 
If $H$ or $K$ is normal, then $HK = H \vee K$.


===Isomorphism Theorems===

The three isomorphism theorems establish isomorphisms between groups formed from the new group operations.


====First Isomorphism Theorem====

This is exactly the same as the fundamental theorem of group homomorphisms, which we covered in the fall.
 
If we have two groups $G$ and $K$, and a homomorphism $\phi: G \rightarrow K$, then there is a monomorphism $\psi: G/ker \phi \rightarrow K$, and $Im(\psi) = Im(\phi)$. So $\psi$ could also be thought of as an isomorphism from $G/ ker\phi$ to $Im(\phi)$.
 
Also, if $\pi$ is the projection of $G$ onto $G/ ker\phi$, then $\phi(x) = \psi(\pi(x))$.


====Second Isomorphism Theorem====

If $G$ is a group, $H$ is a subgroup of $G$, and $N$ is a normal subgroup of $G$, then:
$$
HN/N \cong H/(H\cap N)
$$
There are a couple of lemmas for this. One, $H \cap N$ is a normal subgroup of $H$, or the right side of the equation would be meaningless. This is relatively easy to prove, so I'm not going to explain why.
 
Another lemma is that $N$ is normal in $HN$. This is similarly easy to prove.
 
Anyway, this is almost a cancellation law. When we mod $HN$ by $N$, we send all the elements of $N$ to 1. This means that everything in $H \cap N$ also gets sent to 1. Remember that the elements of $HN$ are products $hn$. But $n$ gets sent to 1, so really this is $h$, unless $h$ also gets sent to 1, which only happens when $h \in H \cap N$.


====Third Isomorphism Theorem====

Let $G$ be a group, and let $H$ and $K$ be normal subgroups of $G$. Also, let $K$ be a subgroup of $H$. Then:
$$
\frac{G/K}{H/K} \cong \frac{G}{H}
$$
So here we kind of do get a cancellation property.
 
When we mod $G$ by $K$, we send all the elements of $G$ to 1. When we mod $H$ by $K$, we also send all the elements of $K$ to 1. So the elements of $H/K$ are the elements of $H$ that aren't in $K$. When we mod $G/K$ by $H/K$, we set the elements of $H$ that aren't in $K$ to 1. But we already set the elements in $K$ to 1 when we did $G/K$. So now we've sent all of $H$ to 1.

===Producing Normal Subgroups===

There are quite a few little theorems about creating


===Subnormal Series===
====Factors====

====Saturated Subnormal Series====

===Some More Stuff===




====Review of $G$-Sets====

Midterm 2 Notes:Math 361, Spring 2014

2014-04-13T19:59:08Z

Vincent.Luczkow: /* Algebraic Extension */

=Math 361 Midterm 2 Notes=
==Linear Algebra==
===Vector Space===

Let $F$ be a field. A vector space over $F$ is a set $V$ with two functions, $+$ and $\cdot$. $+$ is vector addition, $\cdot$ is scalar multiplication.
 
$V$ is an abelian group with respect to vector addition. $+$ is a function from $V\times V \rightarrow V$, and it's associative, commutative, $V$ contains an additive identity (the zero vector, or $\vec{0}$), and $V$ contains an additive inverse for every element. The additive inverse of $\vec{v}$ is $-\vec{v}$.
 
Scalar multiplication is more complicated. Scalar multiplication is an operator from $V \times F \rightarrow V$. So it multiplies a field element by a vector. Scalar multiplication has some properties that kind of look like associativity and distributivity, but they're not quite the same. Anyway:
$$
(c +d) \cdot \vec{v} = c \cdot \vec{v} + d \cdot \vec{v}\\
c \cdot (\vec{v} + \vec{w}) = c \cdot \vec{v} + c \cdot \vec{w}\\
cd \cdot \vec{v} = c \cdot (c \cdot \vec{v}) \\
1 \cdot \vec{v} = \vec{v}
$$
Okay, some explanation. For the purposes of the midterm, there are only three types of vector space that we care about.
 
First - the $n$ tuples of $F$. So $F \times F \times \ldots \times F$. This is a vector space. Vector addition and scalar multiplication is just component wise. $\mathbb{R}^3$ is an example of such a vector space. $(1,2,3)$ and $(2,3,4)$ are both elements of $\mathbb{R}^3$. We add them component-wise:
$$
(1,2,3) + (2,3,4) = (1+2, 2+3, 3+4) = (3,5,7)
$$
$3$ is a scalar (it's an element of the base field, $\mathbb{R}$, and we do scalar multiplication component-wise:
$$
3\cdot (1,2,3) = (3*1, 3*2, 3*3) = (3,6,9)
$$
 
The next example is the polynomials over a field. Here a polynomial is a vector, and vector addition is just polynomial addition. Scalar multiplication is not polynomial multiplication - the scalars are elements of the base field, i.e. they're constant polynomials.
 
Consider $\mathbb{R}[x]$ - the real polynomials. A scalar here is $5$. To do scalar multiplication, you turn this into a constant polynomial (getting, surprise, $5$), and then you multiply this constant polynomial by your actual polynomial. So:
$$
5 \cdot 5 + 3x + 4x^2 = 5(5 + 3x + 4x^2) = 25 + 15x + 20x^2
$$
You can also consider polynomials of limited degree - so all the polynomials up to degree 5, for instance. This is actually the same as 6-tuples of the base field. $5 + 4x + 3x^2 + 2x^3 + x^4 + x^5$ can just be represented as $(5,4,3,2,1,1)$, which is an element of $\mathbb{R}^6$.
 
The third example is the rational functions over a field. Vector addition is just rational function addition, and scalar multiplication is (like for polynomials) just multiplication by a constant rational function.


===Linear Combinations and Other Things===

Let $F$ be a field and $V$ a vector space over $F$. Let $\vec{v}_1, \ldots, \vec{v}_n$ be vectors in $V$. A linear combination of these vectors is:
$$
c_1v_1 + c_2v_2 + \ldots + c_nv_n
$$
where $c_i \in F$.
 
So let $F$ be $\mathbb{R}$ and $V$ be $\mathbb{R}^3$. Let our set of vectors be:
$$
(1,2,3)\\
(2,3,4)\\
(3,4,5)
$$
One possible linear combination of these vectors is:
$$
3(1,2,3) + 25(2,3,4) - \frac{1}{4}(3,4,5)
$$
We can replace 3, 25, and $-\frac{1}{4}$ with any elements of $\mathbb{R}$ we want, and still get a linear combination of these three vectors.
====The Span of a Set====

Let's say we have a set of vectors $V = \{\vec{v}_1, \ldots, \vec{v}_n\}$. The span of this set is the set of all vectors that are linear combinations of these vectors. So the span of $\{(1,0,0),(0,1,0)\}$ is $\{(x,y,0) | x,y \in \mathbb{R}\}$, because:
$$
x(1,0,0) + y(0,1,0) = (x,0,0) + (0,y,0) = (x,y,0)
$$


====A Spanning Set====

Let $T$ be a subset of a vector space $V$. A third set, $S$ is a spanning set of $T$ if every vector in $T$ is a linear combination of vectors in $S$. So we can multiply vectors in $S$ by scalars, add them together, and get anything in $T$.
 
We also say that $S$ spans $T$.
 
So let $T = \{(x,0,0)\in \mathbb{R}^3| x \in \mathbb{R}\}$. This is the set of points on the $x$-axis. One spanning set of $T$ is the set $S = \{(1,0,0)\}$. We can scale this vector by $x$ to get $(x,0,0)$, so we can get all of $T$. Another spanning set is all of $\mathbb{R}^3$. Obviously everything in $T$ is a linear combination of everything in $\mathbb{R}^3$, because everything in $T$ is already in $\mathbb{R}^3$. This illustrates an important point: $S$ doesn't have to only span $T$. The span of $S$ can be much larger than $T$, so long as $T$ is in the span of $S$.


====Linear Independence====

Let $V$ be a vector space and $S$ be a set of vectors. $S$ is linearly independent (well, really, the vectors in $S$ are linearly independent) if none of them is a linear combination of any of the others.
 
Let $S = \{\vec{v}_1,\ldots,\vec{v}_n$. These vectors are linearly independent if and only if:
$$
c_1\vec{v}_1 + \cdots + c_n\vec{v}_n = \vec{0}
$$
implies that $c_i = 0$ for all $i$. So the only way to linearly combine the vectors and get $\vec{0}$ is for the scalars to all be 0.
 
Consider $S = \{(1,0,0),(0,1,0)\}$. These vectors are linearly independent. If you add them up, you get:
$$
c(1,0,0) + d(0,1,0) = (c,d,0)
$$
This only equals 0 if $c$ and $d$ are 0, so the vectors are linearly independent.


====Basis for a Vector Space====

If $V$ is a vector space, than a basis for $V$ is a set $B$ of linearly independent vectors whose span is $V$. So none of the vectors in $B$ can be linearly combined to get another vector in $B$, but if we use all the vectors in $B$ we get every vector in the vector space.
 
$\{(1,0,0),(0,1,0),(0,0,1)\}$ is a basis for $\mathbb{R}^3$. Obviously two of them can't be combined to get the third (the proof is basically the same as for the last section), but if we combine all of them we get:
$$
c(1,0,0) + d(0,1,0) + e(0,0,1) = (c,d,e)
$$
which is everything in $\mathbb{R}^3$.
 
Another example. $\{1,x,x^2\}$ is a basis for the real polynomials of degree less than or equal to 2. Any polynomial of this degree can be written:
$$
p = a*1 + bx + cx^2 = a\cdot1 + b\cdot x + c\cdot x^2
$$


====Dimension of a Vector Space====

There is a very important theorem that says that all bases of a vector space contain the same number of elements. We call this number of elements the dimension of the vector space.
 
The dimension of $\mathbb{R}^3$ is 3, because we created a basis with three elements earlier.
 
When you consider vector spaces of polynomials, you need one basis element for every variable, and one basis element for every power of every variable, and one basis element for every way to multiply variables together plus one for the constant polynomials.
 
So $\mathbb{R}[x]$ is infinite dimensional. There's only one variable $x$, but we can have arbitrarily high powers of $x$. If we limit ourselves to, say, polynomials of degree 6 or less, we get a 7-dimensional vector space.
 
For $\mathbb{R}[x,y]$, limited to polynomials of total degree two or less, we get:
$$
\{1,x,x^2,xy,y,y^2\}
$$
which is 6-dimensional.


==More Field Theory==

So, time to apply some of that stuff to field theory.


===Extension Fields are Vector Spaces===

If we have a field extension $F,E,\iota$, then $E$ is a vector space over $F$.
 
This is pretty obvious - we already know, from the classification of simple extensions that $E$ is either a set of polynomials over $F$, or a set of rational expressions over $F$, both of which are vector spaces over $F$. (Scalar multiplication is just multiplication by a constant.)
 
I should say I'm not sure if this is true of all field extensions. It's definitely true of all finitely generated field extensions (field extensions created by adjoining a finite number of elements to the base field), but it might not be true for all field extensions.


====Extensions of Extensions====

If $E$ is an extension of $K$, and $K$ is an extension of $F$, then $E$ is an extension of $F$.


====Dimension of an Extension Field====

If $E$ is a (let's just say finitely-generated) extension field over $F$, then the dimension of $E$ over $F$, written $[E:F]$, is the dimension of $E$ as a vector space over $F$.


====Dimension Formula====

If $E$ is an extension of $K$ and $K$ is an extension of $F$, then $[E:F] = [E:K][K:F]$.


===Algebraic Extension===

Let $F,E,\iota$ be a field extension. This extension is algebraic if every element of $E$ is algebraic over $F$.
 
For example, $E$ is algebraic over $F$ if $E$ was created by adjoining only algebraic elements.
 
$\mathbb{Q}[x]/\langle irr(\sqrt{2})\rangle$ is algebraic over $\mathbb{Q}$.


====Finite Extensions====

An extension $E$ of $F$ is finite dimensional if and only if it is algebraic and finitely-generated.
 
This makes sense. If it's not algebraic, then it's isomorphic to some type of rational expressions over $F$. The rational expressions over a field are infinite dimensional, so $E$ couldn't be finite dimensional if it was the rational expressions.
 
What if it's algebraic? Well, then it's the set of polynomials over a lot of variables. Furthermore, the degree of these polynomials is limited to some finite value. So the dimension of this vector space will be finite if and only if the number of variables is finite. The number of variables is finite if and only if the extension is finitely generated (each extension adds a single variable).


===Algebraic Closure of a Field===




==Constructibility==

Constructibility deals with constructing shapes with a compass and straightedge.
 
Geometric constructions work like this:
 
You start out with two points. You are allowed to draw lines between any points you find. Also, if there is a line of length $n$ between two points, you can construct a circle of radius $n$ centered at any point you've found.
 
So really you're trying to create a sequence of points.
 
We say a number $x$ is constructible if we can find a line segment of length $x$.
 
I'm not going to explain how to do all the constructions, mainly because it's a graphical problem and this is a text only interface. But we can construct any integer, and any rational number, and some other things to be explained later.

===The Constructible Numbers are a Field===

So if we have two constructible numbers, we can add, multiply, subtract, and divide. Geometrically this means if we have to lines of length $x$ and $y$, we can find a third line of length $x+y$, and another of length $x-y$, and $xy$ and $x/y$.
 
Proving this is, unfortunately, a graphical proof. Try google.


===Constructible Numbers are Closed on Square Roots===

So we know the constructible numbers are a field. This means they must at least be the rational numbers. But given any constructible number, we can find its square root, meaning the constructible numbers are not just the rational numbers.
 
So $\sqrt{2}$ is in $C$, and $\sqrt{2} + \sqrt{3}$ is too, and so is $\sqrt{\sqrt{2} + \sqrt{3}}$.
 
Also, square roots are the only thing we add. This makes sense - getting a new constructible number means we need a new line, so we need a new point. We can only create new points by intersecting lines, intersecting a line and a circle, or intersecting circles. All of these end of being linear or quadratic operations, so there can't be any way to extract, say, a cube root.

====Doubling the Cube is Impossible====

We have a cube of side length $s$. We want to find a cube with twice the volume. Let the volume of the first cube be $V = s^3$, and the second cube be $W = t^3$. We also have $W = 2V$. So $t^3 = 2s^3$, so $t = \sqrt[3]{2}s$. Solving this would require finding the cube root of 2, which we can't do.


====Trisecting the Angle is Impossible====

The proof is to show that one angle (60 degrees) can't be trisected.


==Advanced Group Theory==
===New Group Operations===
====Meet====

Let $H$ and $K$ be subgroups of $G$. The meet of $H$ and $K$, written $H \wedge K$, is the intersection of $H$ and $K$. This is also the largest subgroup that is contained by both $H$ and $K$.


====Join====

Let $H$ and $K$ be subgroups of $G$. The join of $H$ and $K$, written $H \vee K$, is the smallest subgroup that contains both $H$ and $K$. This is not necessarily the union of $H$ and $K$.


====Product====

Let $H$ and $K$ be subgroups of $G$. The product of $H$ and $K$, written $HK$, is the set:
$$
HK = \{hk | h \in H, k\in K\}
$$
This is only a group if $H$ or $K$ is normal in $G$.
 
If $H$ or $K$ is normal, then $HK = H \vee K$.


===Isomorphism Theorems===

The three isomorphism theorems establish isomorphisms between groups formed from the new group operations.


====First Isomorphism Theorem====

This is exactly the same as the fundamental theorem of group homomorphisms, which we covered in the fall.
 
If we have two groups $G$ and $K$, and a homomorphism $\phi: G \rightarrow K$, then there is a monomorphism $\psi: G/ker \phi \rightarrow K$, and $Im(\psi) = Im(\phi)$. So $\psi$ could also be thought of as an isomorphism from $G/ ker\phi$ to $Im(\phi)$.
 
Also, if $\pi$ is the projection of $G$ onto $G/ ker\phi$, then $\phi(x) = \psi(\pi(x))$.


====Second Isomorphism Theorem====

If $G$ is a group, $H$ is a subgroup of $G$, and $N$ is a normal subgroup of $G$, then:
$$
HN/N \cong H/(H\cap N)
$$
There are a couple of lemmas for this. One, $H \cap N$ is a normal subgroup of $H$, or the right side of the equation would be meaningless. This is relatively easy to prove, so I'm not going to explain why.
 
Another lemma is that $N$ is normal in $HN$. This is similarly easy to prove.
 
Anyway, this is almost a cancellation law. When we mod $HN$ by $N$, we send all the elements of $N$ to 1. This means that everything in $H \cap N$ also gets sent to 1. Remember that the elements of $HN$ are products $hn$. But $n$ gets sent to 1, so really this is $h$, unless $h$ also gets sent to 1, which only happens when $h \in H \cap N$.


====Third Isomorphism Theorem====

Let $G$ be a group, and let $H$ and $K$ be normal subgroups of $G$. Also, let $K$ be a subgroup of $H$. Then:
$$
\frac{G/K}{H/K} \cong \frac{G}{H}
$$
So here we kind of do get a cancellation property.
 
When we mod $G$ by $K$, we send all the elements of $G$ to 1. When we mod $H$ by $K$, we also send all the elements of $K$ to 1. So the elements of $H/K$ are the elements of $H$ that aren't in $K$. When we mod $G/K$ by $H/K$, we set the elements of $H$ that aren't in $K$ to 1. But we already set the elements in $K$ to 1 when we did $G/K$. So now we've sent all of $H$ to 1.

===Producing Normal Subgroups===

There are quite a few little theorems about creating


===Subnormal Series===
====Factors====

====Saturated Subnormal Series====

===Some More Stuff===




====Review of $G$-Sets====

Midterm 2 Notes:Math 361, Spring 2014

2014-04-13T19:56:57Z

Vincent.Luczkow: /* Dimension of a Vector Space */

=Math 361 Midterm 2 Notes=
==Linear Algebra==
===Vector Space===

Let $F$ be a field. A vector space over $F$ is a set $V$ with two functions, $+$ and $\cdot$. $+$ is vector addition, $\cdot$ is scalar multiplication.
 
$V$ is an abelian group with respect to vector addition. $+$ is a function from $V\times V \rightarrow V$, and it's associative, commutative, $V$ contains an additive identity (the zero vector, or $\vec{0}$), and $V$ contains an additive inverse for every element. The additive inverse of $\vec{v}$ is $-\vec{v}$.
 
Scalar multiplication is more complicated. Scalar multiplication is an operator from $V \times F \rightarrow V$. So it multiplies a field element by a vector. Scalar multiplication has some properties that kind of look like associativity and distributivity, but they're not quite the same. Anyway:
$$
(c +d) \cdot \vec{v} = c \cdot \vec{v} + d \cdot \vec{v}\\
c \cdot (\vec{v} + \vec{w}) = c \cdot \vec{v} + c \cdot \vec{w}\\
cd \cdot \vec{v} = c \cdot (c \cdot \vec{v}) \\
1 \cdot \vec{v} = \vec{v}
$$
Okay, some explanation. For the purposes of the midterm, there are only three types of vector space that we care about.
 
First - the $n$ tuples of $F$. So $F \times F \times \ldots \times F$. This is a vector space. Vector addition and scalar multiplication is just component wise. $\mathbb{R}^3$ is an example of such a vector space. $(1,2,3)$ and $(2,3,4)$ are both elements of $\mathbb{R}^3$. We add them component-wise:
$$
(1,2,3) + (2,3,4) = (1+2, 2+3, 3+4) = (3,5,7)
$$
$3$ is a scalar (it's an element of the base field, $\mathbb{R}$, and we do scalar multiplication component-wise:
$$
3\cdot (1,2,3) = (3*1, 3*2, 3*3) = (3,6,9)
$$
 
The next example is the polynomials over a field. Here a polynomial is a vector, and vector addition is just polynomial addition. Scalar multiplication is not polynomial multiplication - the scalars are elements of the base field, i.e. they're constant polynomials.
 
Consider $\mathbb{R}[x]$ - the real polynomials. A scalar here is $5$. To do scalar multiplication, you turn this into a constant polynomial (getting, surprise, $5$), and then you multiply this constant polynomial by your actual polynomial. So:
$$
5 \cdot 5 + 3x + 4x^2 = 5(5 + 3x + 4x^2) = 25 + 15x + 20x^2
$$
You can also consider polynomials of limited degree - so all the polynomials up to degree 5, for instance. This is actually the same as 6-tuples of the base field. $5 + 4x + 3x^2 + 2x^3 + x^4 + x^5$ can just be represented as $(5,4,3,2,1,1)$, which is an element of $\mathbb{R}^6$.
 
The third example is the rational functions over a field. Vector addition is just rational function addition, and scalar multiplication is (like for polynomials) just multiplication by a constant rational function.


===Linear Combinations and Other Things===

Let $F$ be a field and $V$ a vector space over $F$. Let $\vec{v}_1, \ldots, \vec{v}_n$ be vectors in $V$. A linear combination of these vectors is:
$$
c_1v_1 + c_2v_2 + \ldots + c_nv_n
$$
where $c_i \in F$.
 
So let $F$ be $\mathbb{R}$ and $V$ be $\mathbb{R}^3$. Let our set of vectors be:
$$
(1,2,3)\\
(2,3,4)\\
(3,4,5)
$$
One possible linear combination of these vectors is:
$$
3(1,2,3) + 25(2,3,4) - \frac{1}{4}(3,4,5)
$$
We can replace 3, 25, and $-\frac{1}{4}$ with any elements of $\mathbb{R}$ we want, and still get a linear combination of these three vectors.
====The Span of a Set====

Let's say we have a set of vectors $V = \{\vec{v}_1, \ldots, \vec{v}_n\}$. The span of this set is the set of all vectors that are linear combinations of these vectors. So the span of $\{(1,0,0),(0,1,0)\}$ is $\{(x,y,0) | x,y \in \mathbb{R}\}$, because:
$$
x(1,0,0) + y(0,1,0) = (x,0,0) + (0,y,0) = (x,y,0)
$$


====A Spanning Set====

Let $T$ be a subset of a vector space $V$. A third set, $S$ is a spanning set of $T$ if every vector in $T$ is a linear combination of vectors in $S$. So we can multiply vectors in $S$ by scalars, add them together, and get anything in $T$.
 
We also say that $S$ spans $T$.
 
So let $T = \{(x,0,0)\in \mathbb{R}^3| x \in \mathbb{R}\}$. This is the set of points on the $x$-axis. One spanning set of $T$ is the set $S = \{(1,0,0)\}$. We can scale this vector by $x$ to get $(x,0,0)$, so we can get all of $T$. Another spanning set is all of $\mathbb{R}^3$. Obviously everything in $T$ is a linear combination of everything in $\mathbb{R}^3$, because everything in $T$ is already in $\mathbb{R}^3$. This illustrates an important point: $S$ doesn't have to only span $T$. The span of $S$ can be much larger than $T$, so long as $T$ is in the span of $S$.


====Linear Independence====

Let $V$ be a vector space and $S$ be a set of vectors. $S$ is linearly independent (well, really, the vectors in $S$ are linearly independent) if none of them is a linear combination of any of the others.
 
Let $S = \{\vec{v}_1,\ldots,\vec{v}_n$. These vectors are linearly independent if and only if:
$$
c_1\vec{v}_1 + \cdots + c_n\vec{v}_n = \vec{0}
$$
implies that $c_i = 0$ for all $i$. So the only way to linearly combine the vectors and get $\vec{0}$ is for the scalars to all be 0.
 
Consider $S = \{(1,0,0),(0,1,0)\}$. These vectors are linearly independent. If you add them up, you get:
$$
c(1,0,0) + d(0,1,0) = (c,d,0)
$$
This only equals 0 if $c$ and $d$ are 0, so the vectors are linearly independent.


====Basis for a Vector Space====

If $V$ is a vector space, than a basis for $V$ is a set $B$ of linearly independent vectors whose span is $V$. So none of the vectors in $B$ can be linearly combined to get another vector in $B$, but if we use all the vectors in $B$ we get every vector in the vector space.
 
$\{(1,0,0),(0,1,0),(0,0,1)\}$ is a basis for $\mathbb{R}^3$. Obviously two of them can't be combined to get the third (the proof is basically the same as for the last section), but if we combine all of them we get:
$$
c(1,0,0) + d(0,1,0) + e(0,0,1) = (c,d,e)
$$
which is everything in $\mathbb{R}^3$.
 
Another example. $\{1,x,x^2\}$ is a basis for the real polynomials of degree less than or equal to 2. Any polynomial of this degree can be written:
$$
p = a*1 + bx + cx^2 = a\cdot1 + b\cdot x + c\cdot x^2
$$


====Dimension of a Vector Space====

There is a very important theorem that says that all bases of a vector space contain the same number of elements. We call this number of elements the dimension of the vector space.
 
The dimension of $\mathbb{R}^3$ is 3, because we created a basis with three elements earlier.
 
When you consider vector spaces of polynomials, you need one basis element for every variable, and one basis element for every power of every variable, and one basis element for every way to multiply variables together plus one for the constant polynomials.
 
So $\mathbb{R}[x]$ is infinite dimensional. There's only one variable $x$, but we can have arbitrarily high powers of $x$. If we limit ourselves to, say, polynomials of degree 6 or less, we get a 7-dimensional vector space.
 
For $\mathbb{R}[x,y]$, limited to polynomials of total degree two or less, we get:
$$
\{1,x,x^2,xy,y,y^2\}
$$
which is 6-dimensional.


==More Field Theory==

So, time to apply some of that stuff to field theory.


===Extension Fields are Vector Spaces===

If we have a field extension $F,E,\iota$, then $E$ is a vector space over $F$.
 
This is pretty obvious - we already know, from the classification of simple extensions that $E$ is either a set of polynomials over $F$, or a set of rational expressions over $F$, both of which are vector spaces over $F$. (Scalar multiplication is just multiplication by a constant.)
 
I should say I'm not sure if this is true of all field extensions. It's definitely true of all finitely generated field extensions (field extensions created by adjoining a finite number of elements to the base field), but it might not be true for all field extensions.


====Extensions of Extensions====

If $E$ is an extension of $K$, and $K$ is an extension of $F$, then $E$ is an extension of $F$.


====Dimension of an Extension Field====

If $E$ is a (let's just say finitely-generated) extension field over $F$, then the dimension of $E$ over $F$, written $[E:F]$, is the dimension of $E$ as a vector space over $F$.


====Dimension Formula====

If $E$ is an extension of $K$ and $K$ is an extension of $F$, then $[E:F] = [E:K][K:F]$.


===Algebraic Extension===

Let $F,E,\iota$ be a field extension. This extension is algebraic if every element of $E$ is algebraic over $F$.
 
For example, $E$ is algebraic over $F$ if $E$ was created by adjoining only algebraic elements.
 
$\mathbb{Q}[x]/\langle irr(\sqrt{2})\rangle$ is algebraic over $\mathbb{Q}$.


====Finite Extensions====

An extension $E$ of $F$ is finite dimensional if and only if it is algebraic and finitely-generated.
 
This makes sense. If it's not algebraic, then it's isomorphic to some type of rational expressions over $F$. The rational expressions over a field are infinite dimensional, so $E$ couldn't be finite dimensional if it was the rational expressions.
 
What if it's algebraic? Well, then it's the set of polynomials over a lot of variables. Furthermore, the degree of these polynomials is limited


==Constructibility==

Constructibility deals with constructing shapes with a compass and straightedge.
 
Geometric constructions work like this:
 
You start out with two points. You are allowed to draw lines between any points you find. Also, if there is a line of length $n$ between two points, you can construct a circle of radius $n$ centered at any point you've found.
 
So really you're trying to create a sequence of points.
 
We say a number $x$ is constructible if we can find a line segment of length $x$.
 
I'm not going to explain how to do all the constructions, mainly because it's a graphical problem and this is a text only interface. But we can construct any integer, and any rational number, and some other things to be explained later.

===The Constructible Numbers are a Field===

So if we have two constructible numbers, we can add, multiply, subtract, and divide. Geometrically this means if we have to lines of length $x$ and $y$, we can find a third line of length $x+y$, and another of length $x-y$, and $xy$ and $x/y$.
 
Proving this is, unfortunately, a graphical proof. Try google.


===Constructible Numbers are Closed on Square Roots===

So we know the constructible numbers are a field. This means they must at least be the rational numbers. But given any constructible number, we can find its square root, meaning the constructible numbers are not just the rational numbers.
 
So $\sqrt{2}$ is in $C$, and $\sqrt{2} + \sqrt{3}$ is too, and so is $\sqrt{\sqrt{2} + \sqrt{3}}$.
 
Also, square roots are the only thing we add. This makes sense - getting a new constructible number means we need a new line, so we need a new point. We can only create new points by intersecting lines, intersecting a line and a circle, or intersecting circles. All of these end of being linear or quadratic operations, so there can't be any way to extract, say, a cube root.

====Doubling the Cube is Impossible====

We have a cube of side length $s$. We want to find a cube with twice the volume. Let the volume of the first cube be $V = s^3$, and the second cube be $W = t^3$. We also have $W = 2V$. So $t^3 = 2s^3$, so $t = \sqrt[3]{2}s$. Solving this would require finding the cube root of 2, which we can't do.


====Trisecting the Angle is Impossible====

The proof is to show that one angle (60 degrees) can't be trisected.


==Advanced Group Theory==
===New Group Operations===
====Meet====

Let $H$ and $K$ be subgroups of $G$. The meet of $H$ and $K$, written $H \wedge K$, is the intersection of $H$ and $K$. This is also the largest subgroup that is contained by both $H$ and $K$.


====Join====

Let $H$ and $K$ be subgroups of $G$. The join of $H$ and $K$, written $H \vee K$, is the smallest subgroup that contains both $H$ and $K$. This is not necessarily the union of $H$ and $K$.


====Product====

Let $H$ and $K$ be subgroups of $G$. The product of $H$ and $K$, written $HK$, is the set:
$$
HK = \{hk | h \in H, k\in K\}
$$
This is only a group if $H$ or $K$ is normal in $G$.
 
If $H$ or $K$ is normal, then $HK = H \vee K$.


===Isomorphism Theorems===

The three isomorphism theorems establish isomorphisms between groups formed from the new group operations.


====First Isomorphism Theorem====

This is exactly the same as the fundamental theorem of group homomorphisms, which we covered in the fall.
 
If we have two groups $G$ and $K$, and a homomorphism $\phi: G \rightarrow K$, then there is a monomorphism $\psi: G/ker \phi \rightarrow K$, and $Im(\psi) = Im(\phi)$. So $\psi$ could also be thought of as an isomorphism from $G/ ker\phi$ to $Im(\phi)$.
 
Also, if $\pi$ is the projection of $G$ onto $G/ ker\phi$, then $\phi(x) = \psi(\pi(x))$.


====Second Isomorphism Theorem====

If $G$ is a group, $H$ is a subgroup of $G$, and $N$ is a normal subgroup of $G$, then:
$$
HN/N \cong H/(H\cap N)
$$
There are a couple of lemmas for this. One, $H \cap N$ is a normal subgroup of $H$, or the right side of the equation would be meaningless. This is relatively easy to prove, so I'm not going to explain why.
 
Another lemma is that $N$ is normal in $HN$. This is similarly easy to prove.
 
Anyway, this is almost a cancellation law. When we mod $HN$ by $N$, we send all the elements of $N$ to 1. This means that everything in $H \cap N$ also gets sent to 1. Remember that the elements of $HN$ are products $hn$. But $n$ gets sent to 1, so really this is $h$, unless $h$ also gets sent to 1, which only happens when $h \in H \cap N$.


====Third Isomorphism Theorem====

Let $G$ be a group, and let $H$ and $K$ be normal subgroups of $G$. Also, let $K$ be a subgroup of $H$. Then:
$$
\frac{G/K}{H/K} \cong \frac{G}{H}
$$
So here we kind of do get a cancellation property.
 
When we mod $G$ by $K$, we send all the elements of $G$ to 1. When we mod $H$ by $K$, we also send all the elements of $K$ to 1. So the elements of $H/K$ are the elements of $H$ that aren't in $K$. When we mod $G/K$ by $H/K$, we set the elements of $H$ that aren't in $K$ to 1. But we already set the elements in $K$ to 1 when we did $G/K$. So now we've sent all of $H$ to 1.

===Producing Normal Subgroups===

There are quite a few little theorems about creating


===Subnormal Series===
====Factors====

====Saturated Subnormal Series====

===Some More Stuff===




====Review of $G$-Sets====

Midterm 2 Notes:Math 361, Spring 2014

2014-04-13T19:56:37Z

Vincent.Luczkow: /* Linear Independence */

=Math 361 Midterm 2 Notes=
==Linear Algebra==
===Vector Space===

Let $F$ be a field. A vector space over $F$ is a set $V$ with two functions, $+$ and $\cdot$. $+$ is vector addition, $\cdot$ is scalar multiplication.
 
$V$ is an abelian group with respect to vector addition. $+$ is a function from $V\times V \rightarrow V$, and it's associative, commutative, $V$ contains an additive identity (the zero vector, or $\vec{0}$), and $V$ contains an additive inverse for every element. The additive inverse of $\vec{v}$ is $-\vec{v}$.
 
Scalar multiplication is more complicated. Scalar multiplication is an operator from $V \times F \rightarrow V$. So it multiplies a field element by a vector. Scalar multiplication has some properties that kind of look like associativity and distributivity, but they're not quite the same. Anyway:
$$
(c +d) \cdot \vec{v} = c \cdot \vec{v} + d \cdot \vec{v}\\
c \cdot (\vec{v} + \vec{w}) = c \cdot \vec{v} + c \cdot \vec{w}\\
cd \cdot \vec{v} = c \cdot (c \cdot \vec{v}) \\
1 \cdot \vec{v} = \vec{v}
$$
Okay, some explanation. For the purposes of the midterm, there are only three types of vector space that we care about.
 
First - the $n$ tuples of $F$. So $F \times F \times \ldots \times F$. This is a vector space. Vector addition and scalar multiplication is just component wise. $\mathbb{R}^3$ is an example of such a vector space. $(1,2,3)$ and $(2,3,4)$ are both elements of $\mathbb{R}^3$. We add them component-wise:
$$
(1,2,3) + (2,3,4) = (1+2, 2+3, 3+4) = (3,5,7)
$$
$3$ is a scalar (it's an element of the base field, $\mathbb{R}$, and we do scalar multiplication component-wise:
$$
3\cdot (1,2,3) = (3*1, 3*2, 3*3) = (3,6,9)
$$
 
The next example is the polynomials over a field. Here a polynomial is a vector, and vector addition is just polynomial addition. Scalar multiplication is not polynomial multiplication - the scalars are elements of the base field, i.e. they're constant polynomials.
 
Consider $\mathbb{R}[x]$ - the real polynomials. A scalar here is $5$. To do scalar multiplication, you turn this into a constant polynomial (getting, surprise, $5$), and then you multiply this constant polynomial by your actual polynomial. So:
$$
5 \cdot 5 + 3x + 4x^2 = 5(5 + 3x + 4x^2) = 25 + 15x + 20x^2
$$
You can also consider polynomials of limited degree - so all the polynomials up to degree 5, for instance. This is actually the same as 6-tuples of the base field. $5 + 4x + 3x^2 + 2x^3 + x^4 + x^5$ can just be represented as $(5,4,3,2,1,1)$, which is an element of $\mathbb{R}^6$.
 
The third example is the rational functions over a field. Vector addition is just rational function addition, and scalar multiplication is (like for polynomials) just multiplication by a constant rational function.


===Linear Combinations and Other Things===

Let $F$ be a field and $V$ a vector space over $F$. Let $\vec{v}_1, \ldots, \vec{v}_n$ be vectors in $V$. A linear combination of these vectors is:
$$
c_1v_1 + c_2v_2 + \ldots + c_nv_n
$$
where $c_i \in F$.
 
So let $F$ be $\mathbb{R}$ and $V$ be $\mathbb{R}^3$. Let our set of vectors be:
$$
(1,2,3)\\
(2,3,4)\\
(3,4,5)
$$
One possible linear combination of these vectors is:
$$
3(1,2,3) + 25(2,3,4) - \frac{1}{4}(3,4,5)
$$
We can replace 3, 25, and $-\frac{1}{4}$ with any elements of $\mathbb{R}$ we want, and still get a linear combination of these three vectors.
====The Span of a Set====

Let's say we have a set of vectors $V = \{\vec{v}_1, \ldots, \vec{v}_n\}$. The span of this set is the set of all vectors that are linear combinations of these vectors. So the span of $\{(1,0,0),(0,1,0)\}$ is $\{(x,y,0) | x,y \in \mathbb{R}\}$, because:
$$
x(1,0,0) + y(0,1,0) = (x,0,0) + (0,y,0) = (x,y,0)
$$


====A Spanning Set====

Let $T$ be a subset of a vector space $V$. A third set, $S$ is a spanning set of $T$ if every vector in $T$ is a linear combination of vectors in $S$. So we can multiply vectors in $S$ by scalars, add them together, and get anything in $T$.
 
We also say that $S$ spans $T$.
 
So let $T = \{(x,0,0)\in \mathbb{R}^3| x \in \mathbb{R}\}$. This is the set of points on the $x$-axis. One spanning set of $T$ is the set $S = \{(1,0,0)\}$. We can scale this vector by $x$ to get $(x,0,0)$, so we can get all of $T$. Another spanning set is all of $\mathbb{R}^3$. Obviously everything in $T$ is a linear combination of everything in $\mathbb{R}^3$, because everything in $T$ is already in $\mathbb{R}^3$. This illustrates an important point: $S$ doesn't have to only span $T$. The span of $S$ can be much larger than $T$, so long as $T$ is in the span of $S$.


====Linear Independence====

Let $V$ be a vector space and $S$ be a set of vectors. $S$ is linearly independent (well, really, the vectors in $S$ are linearly independent) if none of them is a linear combination of any of the others.
 
Let $S = \{\vec{v}_1,\ldots,\vec{v}_n$. These vectors are linearly independent if and only if:
$$
c_1\vec{v}_1 + \cdots + c_n\vec{v}_n = \vec{0}
$$
implies that $c_i = 0$ for all $i$. So the only way to linearly combine the vectors and get $\vec{0}$ is for the scalars to all be 0.
 
Consider $S = \{(1,0,0),(0,1,0)\}$. These vectors are linearly independent. If you add them up, you get:
$$
c(1,0,0) + d(0,1,0) = (c,d,0)
$$
This only equals 0 if $c$ and $d$ are 0, so the vectors are linearly independent.


====Basis for a Vector Space====

If $V$ is a vector space, than a basis for $V$ is a set $B$ of linearly independent vectors whose span is $V$. So none of the vectors in $B$ can be linearly combined to get another vector in $B$, but if we use all the vectors in $B$ we get every vector in the vector space.
 
$\{(1,0,0),(0,1,0),(0,0,1)\}$ is a basis for $\mathbb{R}^3$. Obviously two of them can't be combined to get the third (the proof is basically the same as for the last section), but if we combine all of them we get:
$$
c(1,0,0) + d(0,1,0) + e(0,0,1) = (c,d,e)
$$
which is everything in $\mathbb{R}^3$.
 
Another example. $\{1,x,x^2\}$ is a basis for the real polynomials of degree less than or equal to 2. Any polynomial of this degree can be written:
$$
p = a*1 + bx + cx^2 = a\cdot1 + b\cdot x + c\cdot x^2
$$


====Dimension of a Vector Space====

There is a very important theorem that says that all bases of a vector space contain the same number of elements. We call this number of elements the dimension of the vector space.
 
The dimension of $\mathbb{R}^3$ is 3, because we created a basis with three elements earlier.
 
When you consider vector spaces of polynomials, you need one basis element for every variable, and one basis element for every power of every variable, and one basis element for every way to multiply variables together plus one for the constant polynomials.
 
So $\mathbb{R}[x]$ is infinite dimensional. There's only one variable $x$, but we can have arbitrarily high powers of $x$. If we limit ourselves to, say, polynomials of degree 6 or less, we get a 7-dimensional vector space.
 
For $\mathbb{R}[x,y]$, limited to polynomials of total degree two or less, we get:
$$
\{1,x,x^2,xy,y,y^2\}
$$
which is 6-dimensional.


==More Field Theory==

So, time to apply some of that stuff to field theory.

===Extension Fields are Vector Spaces===

If we have a field extension $F,E,\iota$, then $E$ is a vector space over $F$.
 
This is pretty obvious - we already know, from the classification of simple extensions that $E$ is either a set of polynomials over $F$, or a set of rational expressions over $F$, both of which are vector spaces over $F$. (Scalar multiplication is just multiplication by a constant.)
 
I should say I'm not sure if this is true of all field extensions. It's definitely true of all finitely generated field extensions (field extensions created by adjoining a finite number of elements to the base field), but it might not be true for all field extensions.


====Extensions of Extensions====

If $E$ is an extension of $K$, and $K$ is an extension of $F$, then $E$ is an extension of $F$.


====Dimension of an Extension Field====

If $E$ is a (let's just say finitely-generated) extension field over $F$, then the dimension of $E$ over $F$, written $[E:F]$, is the dimension of $E$ as a vector space over $F$.


====Dimension Formula====

If $E$ is an extension of $K$ and $K$ is an extension of $F$, then $[E:F] = [E:K][K:F]$.


===Algebraic Extension===

Let $F,E,\iota$ be a field extension. This extension is algebraic if every element of $E$ is algebraic over $F$.
 
For example, $E$ is algebraic over $F$ if $E$ was created by adjoining only algebraic elements.
 
$\mathbb{Q}[x]/\langle irr(\sqrt{2})\rangle$ is algebraic over $\mathbb{Q}$.


====Finite Extensions====

An extension $E$ of $F$ is finite dimensional if and only if it is algebraic and finitely-generated.
 
This makes sense. If it's not algebraic, then it's isomorphic to some type of rational expressions over $F$. The rational expressions over a field are infinite dimensional, so $E$ couldn't be finite dimensional if it was the rational expressions.
 
What if it's algebraic? Well, then it's the set of polynomials over a lot of variables. Furthermore, the degree of these polynomials is limited


==Constructibility==

Constructibility deals with constructing shapes with a compass and straightedge.
 
Geometric constructions work like this:
 
You start out with two points. You are allowed to draw lines between any points you find. Also, if there is a line of length $n$ between two points, you can construct a circle of radius $n$ centered at any point you've found.
 
So really you're trying to create a sequence of points.
 
We say a number $x$ is constructible if we can find a line segment of length $x$.
 
I'm not going to explain how to do all the constructions, mainly because it's a graphical problem and this is a text only interface. But we can construct any integer, and any rational number, and some other things to be explained later.

===The Constructible Numbers are a Field===

So if we have two constructible numbers, we can add, multiply, subtract, and divide. Geometrically this means if we have to lines of length $x$ and $y$, we can find a third line of length $x+y$, and another of length $x-y$, and $xy$ and $x/y$.
 
Proving this is, unfortunately, a graphical proof. Try google.


===Constructible Numbers are Closed on Square Roots===

So we know the constructible numbers are a field. This means they must at least be the rational numbers. But given any constructible number, we can find its square root, meaning the constructible numbers are not just the rational numbers.
 
So $\sqrt{2}$ is in $C$, and $\sqrt{2} + \sqrt{3}$ is too, and so is $\sqrt{\sqrt{2} + \sqrt{3}}$.
 
Also, square roots are the only thing we add. This makes sense - getting a new constructible number means we need a new line, so we need a new point. We can only create new points by intersecting lines, intersecting a line and a circle, or intersecting circles. All of these end of being linear or quadratic operations, so there can't be any way to extract, say, a cube root.

====Doubling the Cube is Impossible====

We have a cube of side length $s$. We want to find a cube with twice the volume. Let the volume of the first cube be $V = s^3$, and the second cube be $W = t^3$. We also have $W = 2V$. So $t^3 = 2s^3$, so $t = \sqrt[3]{2}s$. Solving this would require finding the cube root of 2, which we can't do.


====Trisecting the Angle is Impossible====

The proof is to show that one angle (60 degrees) can't be trisected.


==Advanced Group Theory==
===New Group Operations===
====Meet====

Let $H$ and $K$ be subgroups of $G$. The meet of $H$ and $K$, written $H \wedge K$, is the intersection of $H$ and $K$. This is also the largest subgroup that is contained by both $H$ and $K$.


====Join====

Let $H$ and $K$ be subgroups of $G$. The join of $H$ and $K$, written $H \vee K$, is the smallest subgroup that contains both $H$ and $K$. This is not necessarily the union of $H$ and $K$.


====Product====

Let $H$ and $K$ be subgroups of $G$. The product of $H$ and $K$, written $HK$, is the set:
$$
HK = \{hk | h \in H, k\in K\}
$$
This is only a group if $H$ or $K$ is normal in $G$.
 
If $H$ or $K$ is normal, then $HK = H \vee K$.


===Isomorphism Theorems===

The three isomorphism theorems establish isomorphisms between groups formed from the new group operations.


====First Isomorphism Theorem====

This is exactly the same as the fundamental theorem of group homomorphisms, which we covered in the fall.
 
If we have two groups $G$ and $K$, and a homomorphism $\phi: G \rightarrow K$, then there is a monomorphism $\psi: G/ker \phi \rightarrow K$, and $Im(\psi) = Im(\phi)$. So $\psi$ could also be thought of as an isomorphism from $G/ ker\phi$ to $Im(\phi)$.
 
Also, if $\pi$ is the projection of $G$ onto $G/ ker\phi$, then $\phi(x) = \psi(\pi(x))$.


====Second Isomorphism Theorem====

If $G$ is a group, $H$ is a subgroup of $G$, and $N$ is a normal subgroup of $G$, then:
$$
HN/N \cong H/(H\cap N)
$$
There are a couple of lemmas for this. One, $H \cap N$ is a normal subgroup of $H$, or the right side of the equation would be meaningless. This is relatively easy to prove, so I'm not going to explain why.
 
Another lemma is that $N$ is normal in $HN$. This is similarly easy to prove.
 
Anyway, this is almost a cancellation law. When we mod $HN$ by $N$, we send all the elements of $N$ to 1. This means that everything in $H \cap N$ also gets sent to 1. Remember that the elements of $HN$ are products $hn$. But $n$ gets sent to 1, so really this is $h$, unless $h$ also gets sent to 1, which only happens when $h \in H \cap N$.


====Third Isomorphism Theorem====

Let $G$ be a group, and let $H$ and $K$ be normal subgroups of $G$. Also, let $K$ be a subgroup of $H$. Then:
$$
\frac{G/K}{H/K} \cong \frac{G}{H}
$$
So here we kind of do get a cancellation property.
 
When we mod $G$ by $K$, we send all the elements of $G$ to 1. When we mod $H$ by $K$, we also send all the elements of $K$ to 1. So the elements of $H/K$ are the elements of $H$ that aren't in $K$. When we mod $G/K$ by $H/K$, we set the elements of $H$ that aren't in $K$ to 1. But we already set the elements in $K$ to 1 when we did $G/K$. So now we've sent all of $H$ to 1.

===Producing Normal Subgroups===

There are quite a few little theorems about creating


===Subnormal Series===
====Factors====

====Saturated Subnormal Series====

===Some More Stuff===




====Review of $G$-Sets====

Midterm 2 Notes:Math 361, Spring 2014

2014-04-13T19:56:19Z

Vincent.Luczkow: /* A Spanning Set */

=Math 361 Midterm 2 Notes=
==Linear Algebra==
===Vector Space===

Let $F$ be a field. A vector space over $F$ is a set $V$ with two functions, $+$ and $\cdot$. $+$ is vector addition, $\cdot$ is scalar multiplication.
 
$V$ is an abelian group with respect to vector addition. $+$ is a function from $V\times V \rightarrow V$, and it's associative, commutative, $V$ contains an additive identity (the zero vector, or $\vec{0}$), and $V$ contains an additive inverse for every element. The additive inverse of $\vec{v}$ is $-\vec{v}$.
 
Scalar multiplication is more complicated. Scalar multiplication is an operator from $V \times F \rightarrow V$. So it multiplies a field element by a vector. Scalar multiplication has some properties that kind of look like associativity and distributivity, but they're not quite the same. Anyway:
$$
(c +d) \cdot \vec{v} = c \cdot \vec{v} + d \cdot \vec{v}\\
c \cdot (\vec{v} + \vec{w}) = c \cdot \vec{v} + c \cdot \vec{w}\\
cd \cdot \vec{v} = c \cdot (c \cdot \vec{v}) \\
1 \cdot \vec{v} = \vec{v}
$$
Okay, some explanation. For the purposes of the midterm, there are only three types of vector space that we care about.
 
First - the $n$ tuples of $F$. So $F \times F \times \ldots \times F$. This is a vector space. Vector addition and scalar multiplication is just component wise. $\mathbb{R}^3$ is an example of such a vector space. $(1,2,3)$ and $(2,3,4)$ are both elements of $\mathbb{R}^3$. We add them component-wise:
$$
(1,2,3) + (2,3,4) = (1+2, 2+3, 3+4) = (3,5,7)
$$
$3$ is a scalar (it's an element of the base field, $\mathbb{R}$, and we do scalar multiplication component-wise:
$$
3\cdot (1,2,3) = (3*1, 3*2, 3*3) = (3,6,9)
$$
 
The next example is the polynomials over a field. Here a polynomial is a vector, and vector addition is just polynomial addition. Scalar multiplication is not polynomial multiplication - the scalars are elements of the base field, i.e. they're constant polynomials.
 
Consider $\mathbb{R}[x]$ - the real polynomials. A scalar here is $5$. To do scalar multiplication, you turn this into a constant polynomial (getting, surprise, $5$), and then you multiply this constant polynomial by your actual polynomial. So:
$$
5 \cdot 5 + 3x + 4x^2 = 5(5 + 3x + 4x^2) = 25 + 15x + 20x^2
$$
You can also consider polynomials of limited degree - so all the polynomials up to degree 5, for instance. This is actually the same as 6-tuples of the base field. $5 + 4x + 3x^2 + 2x^3 + x^4 + x^5$ can just be represented as $(5,4,3,2,1,1)$, which is an element of $\mathbb{R}^6$.
 
The third example is the rational functions over a field. Vector addition is just rational function addition, and scalar multiplication is (like for polynomials) just multiplication by a constant rational function.


===Linear Combinations and Other Things===

Let $F$ be a field and $V$ a vector space over $F$. Let $\vec{v}_1, \ldots, \vec{v}_n$ be vectors in $V$. A linear combination of these vectors is:
$$
c_1v_1 + c_2v_2 + \ldots + c_nv_n
$$
where $c_i \in F$.
 
So let $F$ be $\mathbb{R}$ and $V$ be $\mathbb{R}^3$. Let our set of vectors be:
$$
(1,2,3)\\
(2,3,4)\\
(3,4,5)
$$
One possible linear combination of these vectors is:
$$
3(1,2,3) + 25(2,3,4) - \frac{1}{4}(3,4,5)
$$
We can replace 3, 25, and $-\frac{1}{4}$ with any elements of $\mathbb{R}$ we want, and still get a linear combination of these three vectors.
====The Span of a Set====

Let's say we have a set of vectors $V = \{\vec{v}_1, \ldots, \vec{v}_n\}$. The span of this set is the set of all vectors that are linear combinations of these vectors. So the span of $\{(1,0,0),(0,1,0)\}$ is $\{(x,y,0) | x,y \in \mathbb{R}\}$, because:
$$
x(1,0,0) + y(0,1,0) = (x,0,0) + (0,y,0) = (x,y,0)
$$


====A Spanning Set====

Let $T$ be a subset of a vector space $V$. A third set, $S$ is a spanning set of $T$ if every vector in $T$ is a linear combination of vectors in $S$. So we can multiply vectors in $S$ by scalars, add them together, and get anything in $T$.
 
We also say that $S$ spans $T$.
 
So let $T = \{(x,0,0)\in \mathbb{R}^3| x \in \mathbb{R}\}$. This is the set of points on the $x$-axis. One spanning set of $T$ is the set $S = \{(1,0,0)\}$. We can scale this vector by $x$ to get $(x,0,0)$, so we can get all of $T$. Another spanning set is all of $\mathbb{R}^3$. Obviously everything in $T$ is a linear combination of everything in $\mathbb{R}^3$, because everything in $T$ is already in $\mathbb{R}^3$. This illustrates an important point: $S$ doesn't have to only span $T$. The span of $S$ can be much larger than $T$, so long as $T$ is in the span of $S$.


====Linear Independence====

Let $V$ be a vector space and $S$ be a set of vectors. $S$ is linearly independent (well, really, the vectors in $S$ are linearly independent) if none of them is a linear combination of any of the others.
 
Let $S = \{\vec{v}_1,\ldots,\vec{v}_n$. These vectors are linearly independent if and only if:
$$
c_1\vec{v}_1 + \cdots + c_n\vec{v}_n = \vec{0}
$$
implies that $c_i = 0$ for all $i$. So the only way to linearly combine the vectors and get $\vec{0}$ is for the scalars to all be 0.
 
Consider $S = \{(1,0,0),(0,1,0)\}. These vectors are linearly independent. If you add them up, you get:
$$
c(1,0,0) + d(0,1,0) = (c,d,0)
$$
This only equals 0 if \(c$ and $d$ are 0, so the vectors are linearly independent.


====Basis for a Vector Space====

If $V$ is a vector space, than a basis for $V$ is a set $B$ of linearly independent vectors whose span is $V$. So none of the vectors in $B$ can be linearly combined to get another vector in $B$, but if we use all the vectors in $B$ we get every vector in the vector space.
 
$\{(1,0,0),(0,1,0),(0,0,1)\}$ is a basis for $\mathbb{R}^3$. Obviously two of them can't be combined to get the third (the proof is basically the same as for the last section), but if we combine all of them we get:
$$
c(1,0,0) + d(0,1,0) + e(0,0,1) = (c,d,e)
$$
which is everything in $\mathbb{R}^3$.
 
Another example. $\{1,x,x^2\}$ is a basis for the real polynomials of degree less than or equal to 2. Any polynomial of this degree can be written:
$$
p = a*1 + bx + cx^2 = a\cdot1 + b\cdot x + c\cdot x^2
$$


====Dimension of a Vector Space====

There is a very important theorem that says that all bases of a vector space contain the same number of elements. We call this number of elements the dimension of the vector space.
 
The dimension of $\mathbb{R}^3$ is 3, because we created a basis with three elements earlier.
 
When you consider vector spaces of polynomials, you need one basis element for every variable, and one basis element for every power of every variable, and one basis element for every way to multiply variables together plus one for the constant polynomials.
 
So $\mathbb{R}[x]$ is infinite dimensional. There's only one variable $x$, but we can have arbitrarily high powers of $x$. If we limit ourselves to, say, polynomials of degree 6 or less, we get a 7-dimensional vector space.
 
For $\mathbb{R}[x,y]$, limited to polynomials of total degree two or less, we get:
$$
\{1,x,x^2,xy,y,y^2\}
$$
which is 6-dimensional.


==More Field Theory==

So, time to apply some of that stuff to field theory.

===Extension Fields are Vector Spaces===

If we have a field extension $F,E,\iota$, then $E$ is a vector space over $F$.
 
This is pretty obvious - we already know, from the classification of simple extensions that $E$ is either a set of polynomials over $F$, or a set of rational expressions over $F$, both of which are vector spaces over $F$. (Scalar multiplication is just multiplication by a constant.)
 
I should say I'm not sure if this is true of all field extensions. It's definitely true of all finitely generated field extensions (field extensions created by adjoining a finite number of elements to the base field), but it might not be true for all field extensions.


====Extensions of Extensions====

If $E$ is an extension of $K$, and $K$ is an extension of $F$, then $E$ is an extension of $F$.


====Dimension of an Extension Field====

If $E$ is a (let's just say finitely-generated) extension field over $F$, then the dimension of $E$ over $F$, written $[E:F]$, is the dimension of $E$ as a vector space over $F$.


====Dimension Formula====

If $E$ is an extension of $K$ and $K$ is an extension of $F$, then $[E:F] = [E:K][K:F]$.


===Algebraic Extension===

Let $F,E,\iota$ be a field extension. This extension is algebraic if every element of $E$ is algebraic over $F$.
 
For example, $E$ is algebraic over $F$ if $E$ was created by adjoining only algebraic elements.
 
$\mathbb{Q}[x]/\langle irr(\sqrt{2})\rangle$ is algebraic over $\mathbb{Q}$.


====Finite Extensions====

An extension $E$ of $F$ is finite dimensional if and only if it is algebraic and finitely-generated.
 
This makes sense. If it's not algebraic, then it's isomorphic to some type of rational expressions over $F$. The rational expressions over a field are infinite dimensional, so $E$ couldn't be finite dimensional if it was the rational expressions.
 
What if it's algebraic? Well, then it's the set of polynomials over a lot of variables. Furthermore, the degree of these polynomials is limited


==Constructibility==

Constructibility deals with constructing shapes with a compass and straightedge.
 
Geometric constructions work like this:
 
You start out with two points. You are allowed to draw lines between any points you find. Also, if there is a line of length $n$ between two points, you can construct a circle of radius $n$ centered at any point you've found.
 
So really you're trying to create a sequence of points.
 
We say a number $x$ is constructible if we can find a line segment of length $x$.
 
I'm not going to explain how to do all the constructions, mainly because it's a graphical problem and this is a text only interface. But we can construct any integer, and any rational number, and some other things to be explained later.

===The Constructible Numbers are a Field===

So if we have two constructible numbers, we can add, multiply, subtract, and divide. Geometrically this means if we have to lines of length $x$ and $y$, we can find a third line of length $x+y$, and another of length $x-y$, and $xy$ and $x/y$.
 
Proving this is, unfortunately, a graphical proof. Try google.


===Constructible Numbers are Closed on Square Roots===

So we know the constructible numbers are a field. This means they must at least be the rational numbers. But given any constructible number, we can find its square root, meaning the constructible numbers are not just the rational numbers.
 
So $\sqrt{2}$ is in $C$, and $\sqrt{2} + \sqrt{3}$ is too, and so is $\sqrt{\sqrt{2} + \sqrt{3}}$.
 
Also, square roots are the only thing we add. This makes sense - getting a new constructible number means we need a new line, so we need a new point. We can only create new points by intersecting lines, intersecting a line and a circle, or intersecting circles. All of these end of being linear or quadratic operations, so there can't be any way to extract, say, a cube root.

====Doubling the Cube is Impossible====

We have a cube of side length $s$. We want to find a cube with twice the volume. Let the volume of the first cube be $V = s^3$, and the second cube be $W = t^3$. We also have $W = 2V$. So $t^3 = 2s^3$, so $t = \sqrt[3]{2}s$. Solving this would require finding the cube root of 2, which we can't do.


====Trisecting the Angle is Impossible====

The proof is to show that one angle (60 degrees) can't be trisected.


==Advanced Group Theory==
===New Group Operations===
====Meet====

Let $H$ and $K$ be subgroups of $G$. The meet of $H$ and $K$, written $H \wedge K$, is the intersection of $H$ and $K$. This is also the largest subgroup that is contained by both $H$ and $K$.


====Join====

Let $H$ and $K$ be subgroups of $G$. The join of $H$ and $K$, written $H \vee K$, is the smallest subgroup that contains both $H$ and $K$. This is not necessarily the union of $H$ and $K$.


====Product====

Let $H$ and $K$ be subgroups of $G$. The product of $H$ and $K$, written $HK$, is the set:
$$
HK = \{hk | h \in H, k\in K\}
$$
This is only a group if $H$ or $K$ is normal in $G$.
 
If $H$ or $K$ is normal, then $HK = H \vee K$.


===Isomorphism Theorems===

The three isomorphism theorems establish isomorphisms between groups formed from the new group operations.


====First Isomorphism Theorem====

This is exactly the same as the fundamental theorem of group homomorphisms, which we covered in the fall.
 
If we have two groups $G$ and $K$, and a homomorphism $\phi: G \rightarrow K$, then there is a monomorphism $\psi: G/ker \phi \rightarrow K$, and $Im(\psi) = Im(\phi)$. So $\psi$ could also be thought of as an isomorphism from $G/ ker\phi$ to $Im(\phi)$.
 
Also, if $\pi$ is the projection of $G$ onto $G/ ker\phi$, then $\phi(x) = \psi(\pi(x))$.


====Second Isomorphism Theorem====

If $G$ is a group, $H$ is a subgroup of $G$, and $N$ is a normal subgroup of $G$, then:
$$
HN/N \cong H/(H\cap N)
$$
There are a couple of lemmas for this. One, $H \cap N$ is a normal subgroup of $H$, or the right side of the equation would be meaningless. This is relatively easy to prove, so I'm not going to explain why.
 
Another lemma is that $N$ is normal in $HN$. This is similarly easy to prove.
 
Anyway, this is almost a cancellation law. When we mod $HN$ by $N$, we send all the elements of $N$ to 1. This means that everything in $H \cap N$ also gets sent to 1. Remember that the elements of $HN$ are products $hn$. But $n$ gets sent to 1, so really this is $h$, unless $h$ also gets sent to 1, which only happens when $h \in H \cap N$.


====Third Isomorphism Theorem====

Let $G$ be a group, and let $H$ and $K$ be normal subgroups of $G$. Also, let $K$ be a subgroup of $H$. Then:
$$
\frac{G/K}{H/K} \cong \frac{G}{H}
$$
So here we kind of do get a cancellation property.
 
When we mod $G$ by $K$, we send all the elements of $G$ to 1. When we mod $H$ by $K$, we also send all the elements of $K$ to 1. So the elements of $H/K$ are the elements of $H$ that aren't in $K$. When we mod $G/K$ by $H/K$, we set the elements of $H$ that aren't in $K$ to 1. But we already set the elements in $K$ to 1 when we did $G/K$. So now we've sent all of $H$ to 1.

===Producing Normal Subgroups===

There are quite a few little theorems about creating


===Subnormal Series===
====Factors====

====Saturated Subnormal Series====

===Some More Stuff===




====Review of $G$-Sets====

Midterm 2 Notes:Math 361, Spring 2014

2014-04-13T19:55:58Z

Vincent.Luczkow: /* Vector Space */

=Math 361 Midterm 2 Notes=
==Linear Algebra==
===Vector Space===

Let $F$ be a field. A vector space over $F$ is a set $V$ with two functions, $+$ and $\cdot$. $+$ is vector addition, $\cdot$ is scalar multiplication.
 
$V$ is an abelian group with respect to vector addition. $+$ is a function from $V\times V \rightarrow V$, and it's associative, commutative, $V$ contains an additive identity (the zero vector, or $\vec{0}$), and $V$ contains an additive inverse for every element. The additive inverse of $\vec{v}$ is $-\vec{v}$.
 
Scalar multiplication is more complicated. Scalar multiplication is an operator from $V \times F \rightarrow V$. So it multiplies a field element by a vector. Scalar multiplication has some properties that kind of look like associativity and distributivity, but they're not quite the same. Anyway:
$$
(c +d) \cdot \vec{v} = c \cdot \vec{v} + d \cdot \vec{v}\\
c \cdot (\vec{v} + \vec{w}) = c \cdot \vec{v} + c \cdot \vec{w}\\
cd \cdot \vec{v} = c \cdot (c \cdot \vec{v}) \\
1 \cdot \vec{v} = \vec{v}
$$
Okay, some explanation. For the purposes of the midterm, there are only three types of vector space that we care about.
 
First - the $n$ tuples of $F$. So $F \times F \times \ldots \times F$. This is a vector space. Vector addition and scalar multiplication is just component wise. $\mathbb{R}^3$ is an example of such a vector space. $(1,2,3)$ and $(2,3,4)$ are both elements of $\mathbb{R}^3$. We add them component-wise:
$$
(1,2,3) + (2,3,4) = (1+2, 2+3, 3+4) = (3,5,7)
$$
$3$ is a scalar (it's an element of the base field, $\mathbb{R}$, and we do scalar multiplication component-wise:
$$
3\cdot (1,2,3) = (3*1, 3*2, 3*3) = (3,6,9)
$$
 
The next example is the polynomials over a field. Here a polynomial is a vector, and vector addition is just polynomial addition. Scalar multiplication is not polynomial multiplication - the scalars are elements of the base field, i.e. they're constant polynomials.
 
Consider $\mathbb{R}[x]$ - the real polynomials. A scalar here is $5$. To do scalar multiplication, you turn this into a constant polynomial (getting, surprise, $5$), and then you multiply this constant polynomial by your actual polynomial. So:
$$
5 \cdot 5 + 3x + 4x^2 = 5(5 + 3x + 4x^2) = 25 + 15x + 20x^2
$$
You can also consider polynomials of limited degree - so all the polynomials up to degree 5, for instance. This is actually the same as 6-tuples of the base field. $5 + 4x + 3x^2 + 2x^3 + x^4 + x^5$ can just be represented as $(5,4,3,2,1,1)$, which is an element of $\mathbb{R}^6$.
 
The third example is the rational functions over a field. Vector addition is just rational function addition, and scalar multiplication is (like for polynomials) just multiplication by a constant rational function.


===Linear Combinations and Other Things===

Let $F$ be a field and $V$ a vector space over $F$. Let $\vec{v}_1, \ldots, \vec{v}_n$ be vectors in $V$. A linear combination of these vectors is:
$$
c_1v_1 + c_2v_2 + \ldots + c_nv_n
$$
where $c_i \in F$.
 
So let $F$ be $\mathbb{R}$ and $V$ be $\mathbb{R}^3$. Let our set of vectors be:
$$
(1,2,3)\\
(2,3,4)\\
(3,4,5)
$$
One possible linear combination of these vectors is:
$$
3(1,2,3) + 25(2,3,4) - \frac{1}{4}(3,4,5)
$$
We can replace 3, 25, and $-\frac{1}{4}$ with any elements of $\mathbb{R}$ we want, and still get a linear combination of these three vectors.
====The Span of a Set====

Let's say we have a set of vectors $V = \{\vec{v}_1, \ldots, \vec{v}_n\}$. The span of this set is the set of all vectors that are linear combinations of these vectors. So the span of $\{(1,0,0),(0,1,0)\}$ is $\{(x,y,0) | x,y \in \mathbb{R}\}$, because:
$$
x(1,0,0) + y(0,1,0) = (x,0,0) + (0,y,0) = (x,y,0)
$$


====A Spanning Set====

Let $T$ be a subset of a vector space $V$. A third set, $S$ is a spanning set of $T$ if every vector in $T$ is a linear combination of vectors in $S$. So we can multiply vectors in $S$ by scalars, add them together, and get anything in $T$.
 
We also say that $S$ spans $T$.
 
So let $T = \{(x,0,0)\in \mathbb{R}^3| x \in \mathbb{R}\}. This is the set of points on the \(x$-axis. One spanning set of $T$ is the set $S = \{(1,0,0)\}$. We can scale this vector by $x$ to get $(x,0,0)$, so we can get all of $T$. Another spanning set is all of $\mathbb{R}^3$. Obviously everything in $T$ is a linear combination of everything in $\mathbb{R}^3$, because everything in $T$ is already in $\mathbb{R}^3$. This illustrates an important point: $S$ doesn't have to only span $T$. The span of $S$ can be much larger than $T$, so long as $T$ is in the span of $S$.


====Linear Independence====

Let $V$ be a vector space and $S$ be a set of vectors. $S$ is linearly independent (well, really, the vectors in $S$ are linearly independent) if none of them is a linear combination of any of the others.
 
Let $S = \{\vec{v}_1,\ldots,\vec{v}_n$. These vectors are linearly independent if and only if:
$$
c_1\vec{v}_1 + \cdots + c_n\vec{v}_n = \vec{0}
$$
implies that $c_i = 0$ for all $i$. So the only way to linearly combine the vectors and get $\vec{0}$ is for the scalars to all be 0.
 
Consider $S = \{(1,0,0),(0,1,0)\}. These vectors are linearly independent. If you add them up, you get:
$$
c(1,0,0) + d(0,1,0) = (c,d,0)
$$
This only equals 0 if \(c$ and $d$ are 0, so the vectors are linearly independent.


====Basis for a Vector Space====

If $V$ is a vector space, than a basis for $V$ is a set $B$ of linearly independent vectors whose span is $V$. So none of the vectors in $B$ can be linearly combined to get another vector in $B$, but if we use all the vectors in $B$ we get every vector in the vector space.
 
$\{(1,0,0),(0,1,0),(0,0,1)\}$ is a basis for $\mathbb{R}^3$. Obviously two of them can't be combined to get the third (the proof is basically the same as for the last section), but if we combine all of them we get:
$$
c(1,0,0) + d(0,1,0) + e(0,0,1) = (c,d,e)
$$
which is everything in $\mathbb{R}^3$.
 
Another example. $\{1,x,x^2\}$ is a basis for the real polynomials of degree less than or equal to 2. Any polynomial of this degree can be written:
$$
p = a*1 + bx + cx^2 = a\cdot1 + b\cdot x + c\cdot x^2
$$


====Dimension of a Vector Space====

There is a very important theorem that says that all bases of a vector space contain the same number of elements. We call this number of elements the dimension of the vector space.
 
The dimension of $\mathbb{R}^3$ is 3, because we created a basis with three elements earlier.
 
When you consider vector spaces of polynomials, you need one basis element for every variable, and one basis element for every power of every variable, and one basis element for every way to multiply variables together plus one for the constant polynomials.
 
So $\mathbb{R}[x]$ is infinite dimensional. There's only one variable $x$, but we can have arbitrarily high powers of $x$. If we limit ourselves to, say, polynomials of degree 6 or less, we get a 7-dimensional vector space.
 
For $\mathbb{R}[x,y]$, limited to polynomials of total degree two or less, we get:
$$
\{1,x,x^2,xy,y,y^2\}
$$
which is 6-dimensional.


==More Field Theory==

So, time to apply some of that stuff to field theory.

===Extension Fields are Vector Spaces===

If we have a field extension $F,E,\iota$, then $E$ is a vector space over $F$.
 
This is pretty obvious - we already know, from the classification of simple extensions that $E$ is either a set of polynomials over $F$, or a set of rational expressions over $F$, both of which are vector spaces over $F$. (Scalar multiplication is just multiplication by a constant.)
 
I should say I'm not sure if this is true of all field extensions. It's definitely true of all finitely generated field extensions (field extensions created by adjoining a finite number of elements to the base field), but it might not be true for all field extensions.


====Extensions of Extensions====

If $E$ is an extension of $K$, and $K$ is an extension of $F$, then $E$ is an extension of $F$.


====Dimension of an Extension Field====

If $E$ is a (let's just say finitely-generated) extension field over $F$, then the dimension of $E$ over $F$, written $[E:F]$, is the dimension of $E$ as a vector space over $F$.


====Dimension Formula====

If $E$ is an extension of $K$ and $K$ is an extension of $F$, then $[E:F] = [E:K][K:F]$.


===Algebraic Extension===

Let $F,E,\iota$ be a field extension. This extension is algebraic if every element of $E$ is algebraic over $F$.
 
For example, $E$ is algebraic over $F$ if $E$ was created by adjoining only algebraic elements.
 
$\mathbb{Q}[x]/\langle irr(\sqrt{2})\rangle$ is algebraic over $\mathbb{Q}$.


====Finite Extensions====

An extension $E$ of $F$ is finite dimensional if and only if it is algebraic and finitely-generated.
 
This makes sense. If it's not algebraic, then it's isomorphic to some type of rational expressions over $F$. The rational expressions over a field are infinite dimensional, so $E$ couldn't be finite dimensional if it was the rational expressions.
 
What if it's algebraic? Well, then it's the set of polynomials over a lot of variables. Furthermore, the degree of these polynomials is limited


==Constructibility==

Constructibility deals with constructing shapes with a compass and straightedge.
 
Geometric constructions work like this:
 
You start out with two points. You are allowed to draw lines between any points you find. Also, if there is a line of length $n$ between two points, you can construct a circle of radius $n$ centered at any point you've found.
 
So really you're trying to create a sequence of points.
 
We say a number $x$ is constructible if we can find a line segment of length $x$.
 
I'm not going to explain how to do all the constructions, mainly because it's a graphical problem and this is a text only interface. But we can construct any integer, and any rational number, and some other things to be explained later.

===The Constructible Numbers are a Field===

So if we have two constructible numbers, we can add, multiply, subtract, and divide. Geometrically this means if we have to lines of length $x$ and $y$, we can find a third line of length $x+y$, and another of length $x-y$, and $xy$ and $x/y$.
 
Proving this is, unfortunately, a graphical proof. Try google.


===Constructible Numbers are Closed on Square Roots===

So we know the constructible numbers are a field. This means they must at least be the rational numbers. But given any constructible number, we can find its square root, meaning the constructible numbers are not just the rational numbers.
 
So $\sqrt{2}$ is in $C$, and $\sqrt{2} + \sqrt{3}$ is too, and so is $\sqrt{\sqrt{2} + \sqrt{3}}$.
 
Also, square roots are the only thing we add. This makes sense - getting a new constructible number means we need a new line, so we need a new point. We can only create new points by intersecting lines, intersecting a line and a circle, or intersecting circles. All of these end of being linear or quadratic operations, so there can't be any way to extract, say, a cube root.

====Doubling the Cube is Impossible====

We have a cube of side length $s$. We want to find a cube with twice the volume. Let the volume of the first cube be $V = s^3$, and the second cube be $W = t^3$. We also have $W = 2V$. So $t^3 = 2s^3$, so $t = \sqrt[3]{2}s$. Solving this would require finding the cube root of 2, which we can't do.


====Trisecting the Angle is Impossible====

The proof is to show that one angle (60 degrees) can't be trisected.


==Advanced Group Theory==
===New Group Operations===
====Meet====

Let $H$ and $K$ be subgroups of $G$. The meet of $H$ and $K$, written $H \wedge K$, is the intersection of $H$ and $K$. This is also the largest subgroup that is contained by both $H$ and $K$.


====Join====

Let $H$ and $K$ be subgroups of $G$. The join of $H$ and $K$, written $H \vee K$, is the smallest subgroup that contains both $H$ and $K$. This is not necessarily the union of $H$ and $K$.


====Product====

Let $H$ and $K$ be subgroups of $G$. The product of $H$ and $K$, written $HK$, is the set:
$$
HK = \{hk | h \in H, k\in K\}
$$
This is only a group if $H$ or $K$ is normal in $G$.
 
If $H$ or $K$ is normal, then $HK = H \vee K$.


===Isomorphism Theorems===

The three isomorphism theorems establish isomorphisms between groups formed from the new group operations.


====First Isomorphism Theorem====

This is exactly the same as the fundamental theorem of group homomorphisms, which we covered in the fall.
 
If we have two groups $G$ and $K$, and a homomorphism $\phi: G \rightarrow K$, then there is a monomorphism $\psi: G/ker \phi \rightarrow K$, and $Im(\psi) = Im(\phi)$. So $\psi$ could also be thought of as an isomorphism from $G/ ker\phi$ to $Im(\phi)$.
 
Also, if $\pi$ is the projection of $G$ onto $G/ ker\phi$, then $\phi(x) = \psi(\pi(x))$.


====Second Isomorphism Theorem====

If $G$ is a group, $H$ is a subgroup of $G$, and $N$ is a normal subgroup of $G$, then:
$$
HN/N \cong H/(H\cap N)
$$
There are a couple of lemmas for this. One, $H \cap N$ is a normal subgroup of $H$, or the right side of the equation would be meaningless. This is relatively easy to prove, so I'm not going to explain why.
 
Another lemma is that $N$ is normal in $HN$. This is similarly easy to prove.
 
Anyway, this is almost a cancellation law. When we mod $HN$ by $N$, we send all the elements of $N$ to 1. This means that everything in $H \cap N$ also gets sent to 1. Remember that the elements of $HN$ are products $hn$. But $n$ gets sent to 1, so really this is $h$, unless $h$ also gets sent to 1, which only happens when $h \in H \cap N$.


====Third Isomorphism Theorem====

Let $G$ be a group, and let $H$ and $K$ be normal subgroups of $G$. Also, let $K$ be a subgroup of $H$. Then:
$$
\frac{G/K}{H/K} \cong \frac{G}{H}
$$
So here we kind of do get a cancellation property.
 
When we mod $G$ by $K$, we send all the elements of $G$ to 1. When we mod $H$ by $K$, we also send all the elements of $K$ to 1. So the elements of $H/K$ are the elements of $H$ that aren't in $K$. When we mod $G/K$ by $H/K$, we set the elements of $H$ that aren't in $K$ to 1. But we already set the elements in $K$ to 1 when we did $G/K$. So now we've sent all of $H$ to 1.

===Producing Normal Subgroups===

There are quite a few little theorems about creating


===Subnormal Series===
====Factors====

====Saturated Subnormal Series====

===Some More Stuff===




====Review of $G$-Sets====

Midterm 2 Notes:Math 361, Spring 2014

2014-04-13T19:55:37Z

Vincent.Luczkow: /* Math 361 Midterm 2 Notes */

=Math 361 Midterm 2 Notes=
==Linear Algebra==
===Vector Space===

Let $F$ be a field. A vector space over $F$ is a set $V$ with two functions, $+$ and $\cdot$. $+$ is vector addition, $\cdot$ is scalar multiplication.
 
$V$ is an abelian group with respect to vector addition. $+$ is a function from $V\times V \rightarrow V$, and it's associative, commutative, $V$ contains an additive identity (the zero vector, or $\vec{0}$), and $V$ contains an additive inverse for every element. The additive inverse of $\vec{v}$ is $-\vec{v}$.
 
Scalar multiplication is more complicated. Scalar multiplication is an operator from $V \times F \rightarrow V$. So it multiplies a field element by a vector. Scalar multiplication has some properties that kind of look like associativity and distributivity, but they're not quite the same. Anyway:
$$
(c +d) \cdot \vec{v} = c \cdot \vec{v} + d \cdot \vec{v}\\
c \cdot (\vec{v} + \vec{w}) = c \cdot \vec{v} + c \cdot \vec{w}\\
cd \cdot \vec{v} = c \cdot (c \cdot \vec{v})\) \\
1 \cdot \vec{v} = \vec{v}
$$
Okay, some explanation. For the purposes of the midterm, there are only three types of vector space that we care about.
 
First - the $n$ tuples of $F$. So $F \times F \times \ldots \times F$. This is a vector space. Vector addition and scalar multiplication is just component wise. $\mathbb{R}^3$ is an example of such a vector space. $(1,2,3)$ and $(2,3,4)$ are both elements of $\mathbb{R}^3$. We add them component-wise:
$$
(1,2,3) + (2,3,4) = (1+2, 2+3, 3+4) = (3,5,7)
$$
$3$ is a scalar (it's an element of the base field, $\mathbb{R}$, and we do scalar multiplication component-wise:
$$
3\cdot (1,2,3) = (3*1, 3*2, 3*3) = (3,6,9)
$$
 
The next example is the polynomials over a field. Here a polynomial is a vector, and vector addition is just polynomial addition. Scalar multiplication is not polynomial multiplication - the scalars are elements of the base field, i.e. they're constant polynomials.
 
Consider $\mathbb{R}[x]$ - the real polynomials. A scalar here is $5$. To do scalar multiplication, you turn this into a constant polynomial (getting, surprise, $5$), and then you multiply this constant polynomial by your actual polynomial. So:
$$
5 \cdot 5 + 3x + 4x^2 = 5(5 + 3x + 4x^2) = 25 + 15x + 20x^2
$$
You can also consider polynomials of limited degree - so all the polynomials up to degree 5, for instance. This is actually the same as 6-tuples of the base field. $5 + 4x + 3x^2 + 2x^3 + x^4 + x^5$ can just be represented as $(5,4,3,2,1,1)$, which is an element of $\mathbb{R}^6$.
 
The third example is the rational functions over a field. Vector addition is just rational function addition, and scalar multiplication is (like for polynomials) just multiplication by a constant rational function.


===Linear Combinations and Other Things===

Let $F$ be a field and $V$ a vector space over $F$. Let $\vec{v}_1, \ldots, \vec{v}_n$ be vectors in $V$. A linear combination of these vectors is:
$$
c_1v_1 + c_2v_2 + \ldots + c_nv_n
$$
where $c_i \in F$.
 
So let $F$ be $\mathbb{R}$ and $V$ be $\mathbb{R}^3$. Let our set of vectors be:
$$
(1,2,3)\\
(2,3,4)\\
(3,4,5)
$$
One possible linear combination of these vectors is:
$$
3(1,2,3) + 25(2,3,4) - \frac{1}{4}(3,4,5)
$$
We can replace 3, 25, and $-\frac{1}{4}$ with any elements of $\mathbb{R}$ we want, and still get a linear combination of these three vectors.
====The Span of a Set====

Let's say we have a set of vectors $V = \{\vec{v}_1, \ldots, \vec{v}_n\}$. The span of this set is the set of all vectors that are linear combinations of these vectors. So the span of $\{(1,0,0),(0,1,0)\}$ is $\{(x,y,0) | x,y \in \mathbb{R}\}$, because:
$$
x(1,0,0) + y(0,1,0) = (x,0,0) + (0,y,0) = (x,y,0)
$$


====A Spanning Set====

Let $T$ be a subset of a vector space $V$. A third set, $S$ is a spanning set of $T$ if every vector in $T$ is a linear combination of vectors in $S$. So we can multiply vectors in $S$ by scalars, add them together, and get anything in $T$.
 
We also say that $S$ spans $T$.
 
So let $T = \{(x,0,0)\in \mathbb{R}^3| x \in \mathbb{R}\}. This is the set of points on the \(x$-axis. One spanning set of $T$ is the set $S = \{(1,0,0)\}$. We can scale this vector by $x$ to get $(x,0,0)$, so we can get all of $T$. Another spanning set is all of $\mathbb{R}^3$. Obviously everything in $T$ is a linear combination of everything in $\mathbb{R}^3$, because everything in $T$ is already in $\mathbb{R}^3$. This illustrates an important point: $S$ doesn't have to only span $T$. The span of $S$ can be much larger than $T$, so long as $T$ is in the span of $S$.


====Linear Independence====

Let $V$ be a vector space and $S$ be a set of vectors. $S$ is linearly independent (well, really, the vectors in $S$ are linearly independent) if none of them is a linear combination of any of the others.
 
Let $S = \{\vec{v}_1,\ldots,\vec{v}_n$. These vectors are linearly independent if and only if:
$$
c_1\vec{v}_1 + \cdots + c_n\vec{v}_n = \vec{0}
$$
implies that $c_i = 0$ for all $i$. So the only way to linearly combine the vectors and get $\vec{0}$ is for the scalars to all be 0.
 
Consider $S = \{(1,0,0),(0,1,0)\}. These vectors are linearly independent. If you add them up, you get:
$$
c(1,0,0) + d(0,1,0) = (c,d,0)
$$
This only equals 0 if \(c$ and $d$ are 0, so the vectors are linearly independent.


====Basis for a Vector Space====

If $V$ is a vector space, than a basis for $V$ is a set $B$ of linearly independent vectors whose span is $V$. So none of the vectors in $B$ can be linearly combined to get another vector in $B$, but if we use all the vectors in $B$ we get every vector in the vector space.
 
$\{(1,0,0),(0,1,0),(0,0,1)\}$ is a basis for $\mathbb{R}^3$. Obviously two of them can't be combined to get the third (the proof is basically the same as for the last section), but if we combine all of them we get:
$$
c(1,0,0) + d(0,1,0) + e(0,0,1) = (c,d,e)
$$
which is everything in $\mathbb{R}^3$.
 
Another example. $\{1,x,x^2\}$ is a basis for the real polynomials of degree less than or equal to 2. Any polynomial of this degree can be written:
$$
p = a*1 + bx + cx^2 = a\cdot1 + b\cdot x + c\cdot x^2
$$


====Dimension of a Vector Space====

There is a very important theorem that says that all bases of a vector space contain the same number of elements. We call this number of elements the dimension of the vector space.
 
The dimension of $\mathbb{R}^3$ is 3, because we created a basis with three elements earlier.
 
When you consider vector spaces of polynomials, you need one basis element for every variable, and one basis element for every power of every variable, and one basis element for every way to multiply variables together plus one for the constant polynomials.
 
So $\mathbb{R}[x]$ is infinite dimensional. There's only one variable $x$, but we can have arbitrarily high powers of $x$. If we limit ourselves to, say, polynomials of degree 6 or less, we get a 7-dimensional vector space.
 
For $\mathbb{R}[x,y]$, limited to polynomials of total degree two or less, we get:
$$
\{1,x,x^2,xy,y,y^2\}
$$
which is 6-dimensional.


==More Field Theory==

So, time to apply some of that stuff to field theory.

===Extension Fields are Vector Spaces===

If we have a field extension $F,E,\iota$, then $E$ is a vector space over $F$.
 
This is pretty obvious - we already know, from the classification of simple extensions that $E$ is either a set of polynomials over $F$, or a set of rational expressions over $F$, both of which are vector spaces over $F$. (Scalar multiplication is just multiplication by a constant.)
 
I should say I'm not sure if this is true of all field extensions. It's definitely true of all finitely generated field extensions (field extensions created by adjoining a finite number of elements to the base field), but it might not be true for all field extensions.


====Extensions of Extensions====

If $E$ is an extension of $K$, and $K$ is an extension of $F$, then $E$ is an extension of $F$.


====Dimension of an Extension Field====

If $E$ is a (let's just say finitely-generated) extension field over $F$, then the dimension of $E$ over $F$, written $[E:F]$, is the dimension of $E$ as a vector space over $F$.


====Dimension Formula====

If $E$ is an extension of $K$ and $K$ is an extension of $F$, then $[E:F] = [E:K][K:F]$.


===Algebraic Extension===

Let $F,E,\iota$ be a field extension. This extension is algebraic if every element of $E$ is algebraic over $F$.
 
For example, $E$ is algebraic over $F$ if $E$ was created by adjoining only algebraic elements.
 
$\mathbb{Q}[x]/\langle irr(\sqrt{2})\rangle$ is algebraic over $\mathbb{Q}$.


====Finite Extensions====

An extension $E$ of $F$ is finite dimensional if and only if it is algebraic and finitely-generated.
 
This makes sense. If it's not algebraic, then it's isomorphic to some type of rational expressions over $F$. The rational expressions over a field are infinite dimensional, so $E$ couldn't be finite dimensional if it was the rational expressions.
 
What if it's algebraic? Well, then it's the set of polynomials over a lot of variables. Furthermore, the degree of these polynomials is limited


==Constructibility==

Constructibility deals with constructing shapes with a compass and straightedge.
 
Geometric constructions work like this:
 
You start out with two points. You are allowed to draw lines between any points you find. Also, if there is a line of length $n$ between two points, you can construct a circle of radius $n$ centered at any point you've found.
 
So really you're trying to create a sequence of points.
 
We say a number $x$ is constructible if we can find a line segment of length $x$.
 
I'm not going to explain how to do all the constructions, mainly because it's a graphical problem and this is a text only interface. But we can construct any integer, and any rational number, and some other things to be explained later.

===The Constructible Numbers are a Field===

So if we have two constructible numbers, we can add, multiply, subtract, and divide. Geometrically this means if we have to lines of length $x$ and $y$, we can find a third line of length $x+y$, and another of length $x-y$, and $xy$ and $x/y$.
 
Proving this is, unfortunately, a graphical proof. Try google.


===Constructible Numbers are Closed on Square Roots===

So we know the constructible numbers are a field. This means they must at least be the rational numbers. But given any constructible number, we can find its square root, meaning the constructible numbers are not just the rational numbers.
 
So $\sqrt{2}$ is in $C$, and $\sqrt{2} + \sqrt{3}$ is too, and so is $\sqrt{\sqrt{2} + \sqrt{3}}$.
 
Also, square roots are the only thing we add. This makes sense - getting a new constructible number means we need a new line, so we need a new point. We can only create new points by intersecting lines, intersecting a line and a circle, or intersecting circles. All of these end of being linear or quadratic operations, so there can't be any way to extract, say, a cube root.

====Doubling the Cube is Impossible====

We have a cube of side length $s$. We want to find a cube with twice the volume. Let the volume of the first cube be $V = s^3$, and the second cube be $W = t^3$. We also have $W = 2V$. So $t^3 = 2s^3$, so $t = \sqrt[3]{2}s$. Solving this would require finding the cube root of 2, which we can't do.


====Trisecting the Angle is Impossible====

The proof is to show that one angle (60 degrees) can't be trisected.


==Advanced Group Theory==
===New Group Operations===
====Meet====

Let $H$ and $K$ be subgroups of $G$. The meet of $H$ and $K$, written $H \wedge K$, is the intersection of $H$ and $K$. This is also the largest subgroup that is contained by both $H$ and $K$.


====Join====

Let $H$ and $K$ be subgroups of $G$. The join of $H$ and $K$, written $H \vee K$, is the smallest subgroup that contains both $H$ and $K$. This is not necessarily the union of $H$ and $K$.


====Product====

Let $H$ and $K$ be subgroups of $G$. The product of $H$ and $K$, written $HK$, is the set:
$$
HK = \{hk | h \in H, k\in K\}
$$
This is only a group if $H$ or $K$ is normal in $G$.
 
If $H$ or $K$ is normal, then $HK = H \vee K$.


===Isomorphism Theorems===

The three isomorphism theorems establish isomorphisms between groups formed from the new group operations.


====First Isomorphism Theorem====

This is exactly the same as the fundamental theorem of group homomorphisms, which we covered in the fall.
 
If we have two groups $G$ and $K$, and a homomorphism $\phi: G \rightarrow K$, then there is a monomorphism $\psi: G/ker \phi \rightarrow K$, and $Im(\psi) = Im(\phi)$. So $\psi$ could also be thought of as an isomorphism from $G/ ker\phi$ to $Im(\phi)$.
 
Also, if $\pi$ is the projection of $G$ onto $G/ ker\phi$, then $\phi(x) = \psi(\pi(x))$.


====Second Isomorphism Theorem====

If $G$ is a group, $H$ is a subgroup of $G$, and $N$ is a normal subgroup of $G$, then:
$$
HN/N \cong H/(H\cap N)
$$
There are a couple of lemmas for this. One, $H \cap N$ is a normal subgroup of $H$, or the right side of the equation would be meaningless. This is relatively easy to prove, so I'm not going to explain why.
 
Another lemma is that $N$ is normal in $HN$. This is similarly easy to prove.
 
Anyway, this is almost a cancellation law. When we mod $HN$ by $N$, we send all the elements of $N$ to 1. This means that everything in $H \cap N$ also gets sent to 1. Remember that the elements of $HN$ are products $hn$. But $n$ gets sent to 1, so really this is $h$, unless $h$ also gets sent to 1, which only happens when $h \in H \cap N$.


====Third Isomorphism Theorem====

Let $G$ be a group, and let $H$ and $K$ be normal subgroups of $G$. Also, let $K$ be a subgroup of $H$. Then:
$$
\frac{G/K}{H/K} \cong \frac{G}{H}
$$
So here we kind of do get a cancellation property.
 
When we mod $G$ by $K$, we send all the elements of $G$ to 1. When we mod $H$ by $K$, we also send all the elements of $K$ to 1. So the elements of $H/K$ are the elements of $H$ that aren't in $K$. When we mod $G/K$ by $H/K$, we set the elements of $H$ that aren't in $K$ to 1. But we already set the elements in $K$ to 1 when we did $G/K$. So now we've sent all of $H$ to 1.

===Producing Normal Subgroups===

There are quite a few little theorems about creating


===Subnormal Series===
====Factors====

====Saturated Subnormal Series====

===Some More Stuff===




====Review of $G$-Sets====

Midterm 2 Notes:Math 361, Spring 2014

2014-04-13T16:57:46Z

Vincent.Luczkow: Created page with "=Math 361 Midterm 2 Notes="

=Math 361 Midterm 2 Notes=

Math 361, Spring 2014

2014-04-13T16:57:07Z

Vincent.Luczkow: /* Notes */

==Course information==

* See the [http://cartan.math.umb.edu/classes/s14_ma361/s14_ma361_syllabus.pdf syllabus] for general information and the schedule of readings.
* Class meets Monday, Wednesday, and Friday, 11:00 a.m.-11:50 a.m., in W-1-57.
* Textbook: John Fraleigh, ''A First Course in Abstract Algebra,'' Seventh Edition.
* Instructor: [http://www.math.umb.edu/~jackson Steven Jackson].
* Office: [http://www.math.umb.edu/contact/office-maps/office-Jackson.html S-3-82]
* Office hours: by appointment.
* E-mail: [mailto:jackson@math.umb.edu jackson@math.umb.edu].
* Telephone: (617) 287-6469.

==Important dates==

* Weekly quizzes happen on Mondays during the last fifteen minutes of class. (In weeks when Monday is a holiday, the quiz happens on Wednesday.) The first quiz is on Monday, February 3.
* First midterm: Monday, March 3.
* Second midterm: Monday, April 14.
* Final exam: Monday, May 19 from 11:30 AM to 2:30 PM in W01 - 0057 (same classroom)

==Study Group==
* For those interested in the MATH 361 Spring 2014 weekly study group, please click [[Study Group | here]].

==Notes==
I have posted my midterm notes [[Midterm 1 Notes:Math_361,_Spring_2014 | here]]. They are incomplete - I've only covered the first three assignments. I will try to cover assignment four tomorrow.
The notes for the second midterm are [[Midterm 2 Notes:Math_361,_Spring_2014 | here]].

==How to use this page==

Below you will find links to the weekly assignment pages. Each of these pages
is editable by anyone in the class, so apart from telling you what problems to
work on they are excellent spaces in which to ask questions. (If you are very
shy you may ask your questions privately, either by [mailto:jackson@math.umb.edu email]
or in person. But we will all work more efficiently if you ask them on
the wiki, so that each question only needs to be answered once.) It is also
extremely helpful to try to answer questions posed by other students. I will
monitor these pages to ensure that no wrong answers go uncorrected.

If you are not already familiar with them, you may wish to read about
[http://en.wikipedia.org/wiki/Help:Wiki_markup wiki markup] and
[http://en.wikipedia.org/wiki/MOS:MATH#Typesetting_of_mathematical_formulae typesetting mathematics].
Also, you may wish to add this page and the assignment pages to your
[[Special:Watchlist|watchlist]]
using the link in the upper right corner of each page, then change your
[[Special:Preferences|preferences]]
to enable e-mail notifications; this way you will know about page activity
without constantly re-checking all the pages.

==Weekly assignments==

* [[Math 361, Spring 2014, Assignment 1|Assignment 1]], due Monday, February 3.
* [[Math 361, Spring 2014, Assignment 2|Assignment 2]], due Monday, February 10.
* [[Math 361, Spring 2014, Assignment 3|Assignment 3]], due Wednesday, February 19.
* [[Math 361, Spring 2014, Assignment 4|Assignment 4]], due Monday, February 24.
* [[Math 361, Spring 2014, Assignment 5|Assignment 5]], due Monday, March 3.
* [[Math 361, Spring 2014, Assignment 6|Assignment 6]], due Monday, March 10.
* [[Math 361, Spring 2014, Assignment 7|Assignment 7]], due Monday, March 24.
* [[Math 361, Spring 2014, Assignment 8|Assignment 8]], due Monday, March 31. ''(Note: we will have the quiz on Friday, April 4.)''
* [[Math 361, Spring 2014, Assignment 9|Assignment 9]], due Monday, April 7. ''(Note: we will have the quiz on Friday, April 11.)''
* [[Math 361, Spring 2014, Assignment 10|Assignment 10]], due Monday, April 14.
* [[Math 361, Spring 2014, Assignment 11|Assignment 11]], due Wednesday, April 23.
* [[Math 361, Spring 2014, Assignment 12|Assignment 12]], due Monday, April 28.
* [[Math 361, Spring 2014, Assignment 13|Assignment 13]], due Monday, May 5.
* [[Math 361, Spring 2014, Assignment 14|Assignment 14]], due Monday, May 12.
* [[Math 361, Spring 2014, Assignment 15|Assignment 15]], due Wednesday, May 14.

Math 361, Spring 2014, Assignment 7

2014-03-21T12:19:13Z

Vincent.Luczkow: /* Solve the following problems: */

__NOTOC__
==Carefully state the following theorems (you need not prove them):==

# Theorem characterizing constructible numbers in terms of certain towers of field extensions (we stated this only in class; it is not in the book).
# Theorem characterizing $\mathrm{deg}(\mathrm{irr}(\alpha,\mathbb{Q}))$ whenever $\alpha$ is a constructible number.

==Solve the following problems:==

# Section 32, problems 2(a-f) and 10.
# Prove the quadratic formula. Specifically: let $F$ be any field with $\mathrm{char}\ F\neq2$, and let $p = ax^2 + bx + c$ be any polynomial in $F[x]$ of degree 2. Show that, if there is an element $\delta\in F$ with $\delta^2=b^2-4ac$, then $p$ splits over $F$ and has roots $\frac{-b\pm\delta}{2a}$. Also show that if there is no such $\delta$, then the quotient ring $E=F[x]/\left\langle x^2 - (b^2 - 4ac)\right\rangle$ is a field with $[E:F]=2$. Letting $\delta$ denote the coset of $x$, show that $p$ splits over $E$, again with roots $\frac{-b\pm\delta}{2a}$.
# Prove that $\cos(20^{\circ})$ is not a constructible number. ''(Hint: use elementary trigonometric identities to prove that, for any angle $\theta$, one has $\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$. Conclude that $\cos(20^{\circ})$ is a root of the polynomial $p = 8x^3 - 6x - 1$. Next, use the Rational Root Theorem to show that $p$ is irreducible over $\mathbb{Q}$, and is hence a constant multiple of the minimal polynomial of $\cos(20^{\circ})$. The rest is up to you.)''
# Prove that the regular nonagon is not constructible with compass and straightedge.
# Prove that the regular pentagon and the regular decagon ''are'' constructible with compass and straightedge, using the following steps:
::(a) Draw a unit circle diagram showing that $\sin(144^{\circ}) = \sin(36^{\circ})$.
::(b) Use trigonometric identities to show that, for any angle $\theta$, we have $\sin(4\theta) = 4\sin(\theta)\cos(\theta)(2\cos^2(\theta)-1)$.
::(c) Show that $\cos(36^{\circ})$ is a root of the polynomial $p=8x^3-4x-1$.
::(d) Show that $p$ admits the factorization $p=(2x+1)(4x^2-2x-1)$.
::(e) Show that $\cos(36^{\circ})$ is a root of the polynomial $4x^2-2x-1$.
::(f) Use the quadratic formula together with some geometric reasoning to show that $\cos(36^{\circ}) = \frac{1+\sqrt{5}}{4}$.
::(g) Use more trigonometric identities and some elementary algebra to show that $\cos(72^{\circ}) = \frac{\sqrt{5}-1}{4}$.
::(h) Locate an actual compass and straightedge, and construct a regular pentagon and a regular decagon. ''(Obviously this is not the quiz question, but it is extremely fun. For bonus points, take sticks and rope into a muddy field and lay out a pentagonal fence.)''

==Theorems==

==Problems==

Midterm 1 Notes:Math 361, Spring 2014

2014-03-03T01:25:07Z

Vincent.Luczkow: /* More on Extension Fields */

=Math 361 Midterm 1 Notes=



==Rings and Fields==

Brief review: a ring is a set $R$ with two operations, $+$ and $*$. $R$ is an abelian group with respect to $+$, and $*$ is associative. Lots of useful things are rings - integers, rational numbers, real numbers, complex numbers, matrices, and polynomials.

Anyway, this is algebra, so we need to be able to compare structures with homomorphisms. Since a ring is an abelian group with an extra operation, it makes sense for the ring homomorphisms to be group homomorphisms with some extra structure. So, if $R$ and $S$ are rings, and we have $\phi:R \rightarrow S$, then we want $\phi$ to have:
$$
\phi(0_R) = 0_S\\
\phi(a+b) = \phi(a)+\phi(b)
$$
These are just the requirements for a group homomorphism. Things on the left hand side are in $R$, things on the right hand side are in $S$. To extend $\phi$ to be a ring homomorphism, we're just going to make it preserve $\cdot$ as well:
$$
\phi(a * b) = \phi(a) * \phi(b)
$$
The definitions of epimorphism, monomorphism, and isomorphism extend to the rings as you would expect (surjection, injection, bijection respectively).
When we get homomorphisms, we always want to ask what kinds of things can be kernels. When we did this for groups, the answer was normal subgroups, and we ended up using them to build quotient groups. We're going to do something similar with rings.
Let's say we have a ring homomorphism $\phi:R \rightarrow S$. Let's say that the kernel of $\phi$ is the set $I$. What can we say about $I$.
Well, $\phi$ is a group homomorphism, so we know that $0\in I$, and that $I$ is a normal subgroup under addition. But $R$ is commutative - every subgroup is normal, so $I$ is a subgroup under addition. Namely, $I$ is closed under addition, i.e.
$$
a,b \in I \rightarrow a+b \ in I
$$
This makes sense. If $\phi(a) = 0$ and $\phi(b) = 0$ then the $\phi(a+b) = \phi(a) + \phi(b) = 0 + 0 = 0$.
What about multiplication. Well, obviously $I$ is going to be closed under multiplication - if $\phi(a) = 0$ and $\phi(b)=0$, then $\phi(ab) = \phi(a)\phi(b) = 0*0 = 0$. So now we can say that $I$ is at least a subring of $R$.

But we can say more. In rings, anything times 0 is 0. So let's take $c \in R$ - any element of the entire ring - and $a \in I$ - an element in the kernel of $\phi$. Where is $ca$? Well:
$$
\phi(ca) = \phi(c)\phi(a) = \phi(c) * 0 = 0
$$

So not only is $I$ a subring, but it absorbs products. If we take something in the kernel, multiply it by anything else, we get something in the kernel.

These are actually the only properties kernels need - contain 0, closure under addition, and absorb multiplication. Any subset of $R$ that has these properties is a ideal of $R$.

So ideals are the equivalent of normal subgroups for rings. With normal subgroups, we created quotient groups. We're going to use ideals to create quotient rings, and in basically the same way.

We're going to mod out by an ideal. Ideals are normal subgroups, though, so really we just need to extend our definition of quotient groups.

So we're going to create cosets. I'll work with $\mathbb{Z}$ as an example ring, and $\langle 5 \rangle$ (the multiples of 5) as an ideal (I promise this is an ideal.) We already know how to create a quotient group using $\mathbb{Z}$ and $\langle 5 \rangle$, so we do that and get things of the form:
$$
n + \langle 5 \rangle
$$
We know how to add cosets of ideals. We also know how to tell if two ideals are equal:
$$
n + \langle 5 \rangle = m + \langle 5 \rangle \Longleftrightarrow n-m \in \langle 5 \rangle
$$
So $4 + \langle 5 \rangle = 14 + \langle 5 \rangle$ because $14 - 4 = 10 = 2*5 \in \langle 5 \rangle$.
 
But this is still just a quotient group. To get a quotient ring, we need some way to multiply. Fortunately, multiplying is easy:
$$
n + \langle 5 \rangle * m + \langle 5 \rangle = nm + \langle 5 \rangle
$$
Just make sure to reduce $mn$ mod $\langle 5 \rangle$ at the end.
 
To reiterate: We get a ring $R$. $I$ is an ideal of $R$. That means that $I$ is a subring of $R$ that absorbs products, or that $I$ is the kernel of a ring homomorphism $\phi:R \rightarrow S$.
 
Given such an $R$ and $I$, we can form the quotient ring $R/I$, which consists of the cosets $n + I$, where $n$ is any element in $R$. We need to define equality of elements of $R/I$, addition in $R/I$, and multiplication in $R/I$.
$$
a + I = b + I \Longleftrightarrow a-b \in I\\
(a+I) + (b+I) = (a+b + I)\\
(a+I)(b+I) = (ab + I)
$$
Remember how we had the fundamental theorem on group homomorphisms? The image of a group homomorphism is isomorphic to the domain modulo the kernel? Well turns out the same thing is true for rings.
To give a little more detail: we have a ring homomorphism $\phi:R \rightarrow S$. The kernel of $\phi$ is some ideal $I$. We can create a monomorphism $\psi:R/I \rightarrow S$. Furthermore, the image of $\psi$ is equal to the image of $\phi$.
When we send $R$ into $S$, the image of $R$ is some ring. Maybe it's $R$, maybe it's smaller, but in some way we compress $R$ into a subring of $S$. What this theorem says is that the new ring is equal to $R/I$, to the original ring modulo the kernel of the homomorphism.


===Some Useful Theorems===

The image of a ring homomorphism is a ring.

Imagine that we have two ideals, $I$ and $J$, and that $I \subset J$. Then $\pi[J]$ is an ideal of $R/I$. So when we mod $J$ by $I$, we still get an ideal.

The opposite is also true. If $K$ is an ideal of $R/I$, then $\pi^{-1}[K]$ is an ideal of $R$, and $I \subset J$.

To explain those last two - let $J$ be an ideal of $R$ that contains $I$. Then when we compress $J$ using $\pi: R \rightarrow R/I$, the result is still an ideal. $\pi[J]$ will contain $0+I$, because it contained all of $I$, and $I$ is what gets compressed to $0+I$. Also, since $\pi$ is a homomorphism, it will preserve addition and multiplication, so $\pi[J]$ will still be closed under addition and absorb multiplication. I don't feel like explaining why it works the other way, but it should be pretty clear.

If an ideal $I$ contains a unit, it is all of $R$. Let's say $u \in R$ is a unit, and $u$ is in $I$. By the definition of $u$, there is an element $v \in R$ such that $uv =1$. But $I$ absorbs multiplication, so $uv = 1 \in I$. We can extend this to say that everything is in $I$. If $x$ is any element of $R$, then $1x$ must be in $I$, because 1 is in $I$ and $I$ absorbs multiplication. But $1x = x$, so every $x$ is in $I$, and $I$ must be all of $R$.


All fields have exactly two ideals - the improper ideal and the zero ideal. Every element of a field,except zero, is a unit. If an ideal contains a unit, it is the improper ideal. So the only way to construct an ideal in a field is by taking just zero (the zero ideal), or by taking everything (the whole field/ the improper ideal).


As an extension of this - for any field there are really only two homomorphisms. This is because there are only two possible kernels, because there are only two possible ideals. If we take the kernel to be the zero ideal, then we get a monomorphism, and we inject $F$ into some other field. If we take the kernel to be the improper ideal, then we compress $F$ completely, and we get the zero ideal.


==Types of Ideals==

Once we know how to create rings, we want to know how to create specific types of rings. Specifically, how do we create integral domains and fields?
 
Imagine that $R/I$ was an integral domain. What does that imply?
 
Well, we can't have any zero divisors. So given $a + I \neq 0 + I$ and $b + I \neq 0 + I), then \(ab + I \neq 0 + I$. But what it really means, if $a +I \neq 0+ I$, is that $a-0 \not \in I$ (by the definition of equality). So $a \not \in I$. And $b \not \in I$ and $ab \not \in $.
 
So what we've shown, really, is that if we want to get an integral domain from $R/I$, we need $ab\in I$ to imply either $a\in I$ or $b \in I$.
 
I'm not going to prove this next part, but this is an if and only if relationship. So if $I$ is any ideal with this property, then $R/I$ is an integral domain, and if $R/I$ is an integral domain, then $I$ has this property. We call $I$ a prime ideal.
 
Fields are more complicated, and honestly hard to explain so I'm just going to say how to get them and not try to explain why it works.
 
If $I$ is an ideal, then$I$ is a maximal ideal if:
 
-$I$ is not all of $R$
 
-and $I$ isn't contained by any other ideal except $R$. Basically there's no way to make $I$ bigger without either making it all of $R$ or not an ideal.
 
If $I$ is a maximal ideal, then $R/I$ is a field. If $R/I$ is a field, then $I$ is maximal.


===More Theorems===

In $F[x]$, the prime ideals and the maximal ideals are the same.
 
In $F[x]$, the maximal ideals/ prime ideals are the ideals generated by irreducible polynomials.
 
So if we have $\mathbb{Q}[x]$ and $x^2-2$, we get a field from $\mathbb{Q}[x]/\langle x^2-2\rangle$.


==Why This All Matters==

We want to be able to solve any polynomial, i.e. we want to be able to find all the roots of a polynomial. If we get a polynomial in $F[x]$, those roots don't necessarily exist in $F$. So our goal is to make a new field that does contain the roots.
 
To be more specific, given a polynomial $p\ \in F[x]$, we want to find all $a \in F$ such that $p(a) = 0$. Alternatively, we want to factor $p$ as a product of linear polynomials $x-a$.
 
This is not necessarily possible in $F$. We want to find another field $E$ where it is.
 
First we need to define what it means to extend a field. Hence, field extensions.
 
A field extension is a triple $(F,E,\iota)$. $F$ and $E$ are fields, and $\iota:F\rightarrow E$ is a monomorphism. So there is a copy of $F$ in $E$.
 
This also means that $F[x]$ exists in $E[x]$. So we can take our original polynomial $p\in F[x]$ and make a new polynomial in $q \in E[x]$. We want to make sure these polynomials are equivalent, in the sense that all the roots of $p$ are roots of $q$. Turns out they are. Because $\iota$ is a monomorphism.
 
Also, $F$ is called the base field and $E$ is called the extension field.


==Finding the Right Extension Field==

The first question is whether or not a useful extension field even exists. The answer is yes. If we have a polynomial $p \in F[x]$, we can always find a extension field $E$ such that $p$ has a root in $E$. This is Kronecker's Theorem.
 
You might be thinking that Kronecker's Theorem only guarantees the existence of one root. Can we find a field where all the roots exist?
 
Again, yes. Remember that if $p(a) = 0$, then $x-a$ divides $p$. So let's say we found a field $E$, and we found a root $a \in E$. That means that we can divide $p$ by $x-a$. When we do this, we'll get a new polynomial. But by Kronecker's theorem, some field has to exist that contains a root of this new polynomial too. And obviously any root of the new polynomial is a root of $p$, because the new polynomial is a factor of $p$. So when we find a root for the new polynomial, we also find a root for $p$. And of course we can iterate this process. Eventually we will find a field $K$ where we can express $p$ as a product of linear polynomials, i.e. $x-a_1*x-a_2*\ldots*x-a_n$.
 
The next question is: can we actually find an extension field? Kronecker's Theorem says that a useful extension field exists, so can we find it? Yes. Yes we can.
 
Specifically, if we have an irreducible polynomial $p \in F[x]$, then $\langle p\rangle$ is maximal over $F[x]$. That means that $F[x]/\langle p \rangle$ is a field. Turns out that $p$ has a root over this field. Specifically, the element $x + \langle p \rangle$ is a root of $p$ over this field.
 
There is a very easy way to get confused here. $F[x]/\langle p \rangle$ is a field. Its elements are cosets of $\langle p \rangle$. So we take a polynomial $q \in F[x]$, and we create $q + \langle p \rangle$.
 
For example, $x^2-2$ is irreducible over $\mathbb{Q}$. So we can create the field $\mathbb{Q}[x]/\langle x^2-2 \rangle$. One element of this field is $x-4 + \langle x^2-2 \rangle$.
 
Now, that element looks an awful lot like a polynomial. But it isn't. It's a coset.
 
Computation in this field is tedious. You have to do polynomial arithmetic, and then mod out by the ideal $\langle x^2-2\rangle$. I will give a very brief example.
 
Two polynomials, $x-4 +\langle x^2-2 \rangle) and \(x+5 + \langle x^2-2 \rangle$. We're going to multiply them.
 
The rule for coset multiplication is $(a+I)(b+I) = ab + I$. In our example, $a = x+4$ and $b = x+5$ and $I = \langle x^2-2 \rangle$. So we need to multiply $x+4$ by $x+5$. This is just normal polynomial arithmetic:
$$
(x+4)(x+5) = x^2 + 9x + 20
$$
Now we have to reduce modulo $x^2 -2$. This is exactly the same as reducing modulo an integer - divide by the integer and take the remainder. So we divide $x^2 + 9x + 20$ by $x^2 -2$. This gives us $9x + 22$. So our final answer is:
$$
(x+4 + \langle x^2-2 \rangle)(x+5 + \langle x^2-2 \rangle) = 9x + 22 + \langle x^2-2 \rangle
$$
There's a trick for doing this. What exactly is $x^2 -2 + \langle x^2-2 \rangle$? Well, it's $0 + \langle x^2-2 \rangle$. But $x^2 - 2 + \langle x^2-2 \rangle$ is really just $(x^2 + \langle x^2-2 \rangle$ - (2 + \langle x^2-2 \rangle)\). So, in this field, $x^2-2 = 0$, meaning $x^2 = 2$. So instead of actually doing out the division and finding the remainder, we can just look at $x^2 + 9x + 20$, and replace every $x^2$ by a 2. This gives us $2 + 9x + 20 = 22$. (Note that I'm dropping the $+\langle x^2-2 \rangle$ for the sake of my laziness.)
 
Here's where it gets confusing. We can take this field, and we can create polynomials from the elements of this field. So we can take $\mathbb{Q}[x]/\langle x^2-2\rangle$, and we can create the polynomial ring:
$$
\frac{\mathbb{Q}[x]}{\langle x^2-2 \rangle}[x]
$$
So now our cosets become coefficients of polynomials. For example:
$$
(x-4 + \langle x^2-2\rangle) + (3x-10 + \langle x^2-2 \rangle)x + (x-5 + \langle x^2-2 \rangle)x^2
$$
is an element of this field. To make this horrifying monstrosity slightly more comprehensible, we're going to rename the coset $x + \langle x^2-2 \rangle$ to be $\alpha + \langle x^2 -2 \rangle$. So at least now we only have to deal with one set of 'x's.
 
I am not going to do a computation in this polynomial ring. But I will give a possible problem. Calculate:
$$
((\alpha -1 + \langle x^2-2 \rangle) + (\alpha+5 + \langle x^2-2 \rangle)x^2)) * ((\alpha + 2 + \langle x^2-2 \rangle)x
$$
This is just multiplying a quadratic polynomial by a linear polynomial. If you know how to do this, I would suggest you don't bother. If it is not immediately obvious how to calculate this, you should try it.


===Theorems===

If $p$ is linear, then $F[x] / \langle p \rangle \cong F$. Modding out by a linear polynomial does nothing.
 
Given a field $F[x]/\langle p \rangle$, where $p$ is irreducible. the element $\alpha = x+ \langle p \rangle$ is a root of $p$. This also means that $x - \alpha$ is a divisor of $p$.
 
To elaborate on that last point - if we have an irreducible polynomial $p \in F[x]$, we can find a root of $p$ in the field $F[x]/\langle p \rangle$.


==An Example==

Let's say we have a polynomial $p =x^5-9x^4+20x^3-2x^2+18x-40$, over $\mathbb{Q}$. We want to find all of its roots. First we find its roots over $\mathbb{Q}$. These are 5 and 4. So we didn't find all of the roots, meaning we're going to need to extend our field. In order to extend our field, we're going to need an irreducible polynomial. Specifically, we need the irreducible factors of $p$. $p$ itself has two linear factors, $x-4$ and $x-5$.(Assuming Wolfram Alpha didn't lie to me.) (Assuming Wolfram Alpha didn't lie to me.) When we $p$ by both of those factors, we get the polynomial $x^3 - 2$. This new polynomial is irreducible over $\mathbb{Q}$, meaning we can use it to create an extension field of $\mathbb{Q}$. So we get, as an extension field, $\mathbb{Q}/\langle x^3 -2\rangle$.
 
But we're not done. We know this new field contains at least one root of $p$. Does it contain all of them?
 
To find out, we need to do more. We've been given a polynomial over a field, we need to find a root. This is the same problem we started with, except instead of working with $p$, we're working with a factor of $p$. We can just do what we did before, only we have a smaller polynomial. We'll get an even smaller polynomial as a result, and polynomials can only get so small, so eventually we'll have found all the roots.
 
We know that $x - \alpha$ is a factor of $x^3 -2$ in $\mathbb{Q}/\langle x^3 -2 \rangle$. So, like we did before, we divide our starting polynomial by its linear factors. This gives us $x^2 + \alpha x + \alpha^2$. Does this have any roots in $\mathbb{Q}[x]/\langle x^3 -2 \rangle$? No, it doesn't. I can't prove it, but it doesn't. So we have to extend our field again, by modding out by $\langle x^3 + \alpha x + \alpha^2$. I am not going to demonstrate this - the notation is just too crazy.

===Theorems===

Not a theorem, a definition. A monic polynomial is a polynomial whose leading coefficient is 1. $x^2 + 4$ is monic. $5x^2 + 20$ is not.
 
Given a polynomial $p$, we can find a polynomial $q$ that is monic and generates $\langle p \rangle$. So we have $5x^2 + 20$. Well, if we divide by 5 we get $x^2 + 4$, and $\langle x^2 + 4\rangle = \langle 5x^2 + 20 \rangle$.


==More on Extension Fields==

First some definitions.
 
If $E$ is an extension field of $F$, and $a\in E$, and $a$ is a root of a (non-zero) polynomial $p\in F[x]$, then $a$ is algebraic over $F$. So $\sqrt{2}$ is algebraic over $\mathbb{Q}$, because it is a root of $x^2 -2$, but $\pi$ is not algebraic, because there is no rational polynomial that kills $\pi$. Note that $\pi$ is algebraic over $\mathbb{R}$, because it is a root of $x-\pi$.
 
Let's say $a\in E$ is algebraic over $F$. It must be the root of some non-zero polynomial $p \in F[x]$. But then it also has to be a root of $p^2$, and $p^3$, and $p+p$. Actually, it has to be a root of every multiple of $p$. (This should be fairly clear. Let $p = x^2 - 2$. If we multiply $p$ by $q$, we can always factor $pq$ to be $(x^2 -2)q$, and $\sqrt{2}$ is obviously a root of this polynomial.) Now $p$ might not generate all of the polynomials that kill $a$. For instance, if we took $p = (x^2-2)^2$\), we would still get an ideal of polynomials that are zero on $a$, but not every polynomial.
 
So let's say we take the set of all polynomials that are zero on $a$. This is still an ideal. We call this ideal the annihilators of $a$ over $F$: $ann(a,F)$. But this is an ideal of polynomials, i.e. an ideal in $F[x]$. But $F[x]$ is a principal ideal domain, meaning $ann(a,F)$ is principal, meaning it is generated by a single element. Actually, we can go farther than that. We can find a monic polynomial that generates $ann(a,F)$. We call this monic generator of $ann(a,F)$ the minimal polynomial of $a$ over $F$. It is written $irr(a,F)$. So $irr(\sqrt{2},\mathbb{Q})$ is $x^2-2$.
 
Another definition:

===Theorems===

If $p$ is any annihilator of $a$ over $F$, then $irr(a,F)$ must be a factor of $p$. So $(x^2-2)^2$ is an annihilator of $\sqrt{2}$, and the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $x^2-2$, which is obviously a factor of $(x^2-2)^2$.