Math 361, Spring 2016, Assignment 8

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

1. ($F$-)linear combination (of elements of a vector space $V$ with scalars from $F$).
2. ($F$-)span (of a subset of a vector space).
3. ($F$-)linear relation (among elements of $V$).
4. ($F$-)linearly independent set.
5. ($F$-)basis.
6. Dimension (of a vector space over $F$).
7. Dimension (of a field extension $F\rightarrow E$).
8. Degree (of a field extension).
9. Degree (of an algebraic element).

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

1. Theorem concerning extension of linearly independent sets to bases.
2. Theorem concerning refinement of spanning sets to bases.
3. Theorem relating the cardinalities of two bases for the same vector space.
4. Theorem concerning the orders of finite fields.
5. Theorem relating algebraic elements to finite-dimensional subextensions.
6. Dimension theorem.

Solve the following problems:

1. Section 30, problems 1, 3, 5, 7, and 9.
2. Let $F$ denote the field $\mathbb{Q}(\sqrt[3]{2})$, the smallest subfield of $\mathbb{R}$ containing $\mathbb{Q}$ and $\sqrt[3]{2}$.

(a) Compute the dimension of $F$ when regarded as a vector space over $\mathbb{Q}$.

(b) Show that the polynomial $x^2-2$ has no roots in $F$. (Hint: use the Dimension Theorem.)

(c) Compute $[\mathbb{Q}(\sqrt{2},\sqrt[3]{2}):\mathbb{Q}]$.

(d) Find an explicit basis for $\mathbb{Q}(\sqrt{2},\sqrt[3]{2})$, regarded as a vector space over $\mathbb{Q}$. Then give several sample calculations in this field.

--------------------End of assignment--------------------

Questions:

• Under definitions, #7 seems to be an elaboration or special case of #6, and #8 seems to be just another name for the phenomenon resulting from the two. Am I mistaken, or is consolidation of the definitions acceptable, or would you rather have specific examples illustrating the situations in which each term would arise? *cough* -IB (talk) 14:03, 28 March 2016 (EDT)

You're correct. #7 is a special case of #6, and #8 is identical to #7. No need for separate examples in #7 and #8 (though it would be nice to give an example for #6 that shows it's a more general concept than #7). -Steven.Jackson (talk) 14:19, 28 March 2016 (EDT)

Solutions:

Vocab

1. Suppose $S\subseteq V$. A linear combination of elements of $S$ is a vector that can be written as the sum of scalar multiples of elements of $S$. That is, for vectors $v_i\in S$, vector $w$ can be written

$c_1\cdot v_1+\ldots + c_i\cdot v_i$ for scalars $c_i$ if and only if $w$ is a linear combination of those elements in $S$. ::* The zero vector is a linear combination of any nonempty subset of any vector space. Set all $c_i$ to the zero element of the field, and the sum becomes a sum of zero vectors, which is the zero vector. ::* If $S$ is the singleton corresponding to the zero vector, any nonzero vector in $V$ is not a linear combination of elements of $S$ (Prove these - they seem a bit too obvious (or find better examples)) :2. The ''span'' of a subset $S\subseteq V$ is the set of all possible linear combinations of elements of $S$. ::* ::* :3. A ''linear relation'' of a subset $S\subseteq V$ is an ordered set of scalars which, when multiplied with their respective vectors in $S$ and those resulting vectors are added, the sum is the zero vector. ::<small> N.B. The relation where all the scalars are the zero element is the ''trivial relation''</small> ::* ::* :4. If there are no possible linear relations in a subset of vectors of $V$, the members of that set are said to be linearly independent, and the set is a ''linearly independent set''. In other words, a set is ::linearly independent if there is no way to get the zero vector with any sums of scalar products of the elements of the set. ::* ::* :5. Keep the definitions of ''span'' and ''linear independent set'' in mind: a ''basis'' for $V$ is a set which spans $V$ and is a linearly independent set. ::* ::* :6. The cardinality of any basis of a vector space $V$ is the ''dimension'' of $V$. The reason it is meaningful is because any two bases of a vector space have the same cardinality (see #3 in Theorems). ::* The dimension of $\mathbb{R}^2$ over $\mathbb{R}$ is $2$, because a basis for $\mathbb{R}^2$ over $\mathbb{R}$ is $\{(0, 1), (1, 0)\}$, and the cardinality of that set is $2$. ::* :7. If the vector space $E$ is a field extension of some field $F$, then the ''dimension of the field extension'' $F\rightarrow E$ is the dimension of vector space $E$ over $F$. This is often denoted $[E:F]$ :8. That last term ("dimension of the field extension $F\rightarrow E$") can be interchanged with the term ''the degree of the extension'' $F\rightarrow E$. ::* Examples here would be redundant and are left as further exercise for the reader. :9. If $\alpha$ is algebraic over field $F$, then $F(\alpha)$ is a field extension of $F$ and $[F(\alpha):F]\equiv$deg($\alpha , F$) is the ''degree of'' $\alpha$. ::* $\mathbb{Q}$ is a field, and $\mathbb{Q}(\sqrt[3]{2})$ is isomorphic to $\mathbb{Q}[x]/<x^3-2>$, so $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$, so the degree of $\sqrt[3]{2}$ is $3$. ::* ==='"`UNIQ--h-6--QINU`"'Theorems=== Given a vector space $V$ and $S\subseteq V$'''...''' :1. If $S$ is a linearly independent set, it can be "grown" into a basis by adding linearly independent vectors until $S$ spans. :2. If $S$ is a spanning set, it can be "trimmed" into a basis by removing linearly dependent vectors from $S$ until $S$ is linearly independent, without removing the spanning property. :3. Given any two bases $B$ and $\beta$ of $V$, by theorems 1 and 2, $B$ and $\beta$ have the same cardinality. (This is important, because it means that the dimension of a vector space, which depends on this ::value, can only possibly be one number. You can't get conflicting dimensions for the same vector space.) <br> Given a finite field $F$'''...''' :4. The order of $F$ must be a prime power. (''i.e.'' $F$ must have $q$ elements, where $q=p^n$ for some prime number $p$ and natural number $n$) :5. If $E$ is a field extension of $F$, and $[E:F]$ <small>(see definition 7)</small> is finite, every element of $E$ is algebraic over $F$. <br> :6. Given field $F$ with extension $E$, and field $K$ which is a subextension of $E$ ''s.t.'' $F\leq K\leq E$'''...''' *$E$ is a vector space over $F$, :$K$ is a vector space over $F$, and :$E$ is a vector space over $K$. *$[E:F]=[E:K]\cdot [K:F]$, where $\cdot$ is standard integer multiplication. (example: if $F$, $K$ and $E$ are isomorphic to each other, $[E:F]=[E:K]\cdot [K:F] = 1\cdot 1 = 1$ (the trivial case holds))