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Math 380, Spring 2018, Assignment 14

2018-05-14T17:46:30Z

John.DeBrota: /* Solutions: */

__NOTOC__
''I must study politics and war that my sons may have liberty to study mathematics and philosophy.''

: - John Adams, letter to Abigail Adams, May 12, 1780

==Solve the following problems:==

:1. Working in $\mathbb{R}^2$, consider the variety $V=\mathbb{V}(x^2y^2-1)$.
::(a) Make a sketch of $V$. What do you think its "dimension" ought to be?
::(b) Compute a Grőbner basis for the ideal $I=\left\langle x^2y^2-1\right\rangle$.
::(c) Make a diagram of $\left\langle\mathrm{LT}(I)\right\rangle$, showing the monomials of this ideal as integer points in the plane.
::(d) For each non-negative integer $d$, let $H(d)$ denote the number of monomials of total degree $d$ or less that lie ''outside'' $\left\langle\mathrm{LT}(I)\right\rangle$. Compute $H(d)$ for $0\leq d\leq 10$. (The function $H(d)$ is called the ''Hilbert function'' associated with $I$.)
::(e) Make a graph of $H(d)$ as a function of the integer $d$. Show that, although there is no line passing through all the points of this graph, there is a line that passes through all its points ''for all sufficiently large $d$''. That is, show that $H(d)$ is ''eventually linear.''
:2. Repeat the problem above, replacing $V$ by the variety $\mathbb{V}(x^2-1, y^2-1)$. Show in the end that in this case the Hilbert function is ''eventually constant.''

:3. (Optional) Devise an algorithm that loops through the leading terms of a Grőbner basis of an ideal, and decides whether its Hilbert function is eventually constant. In other words, devise an algorithm that decides whether a given variety is finite.

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

==Questions:==

==Solutions:==
'''1:''' (a) The sketch is the plot of the functions $y=1/x$ and $y=-1/x$, so there are lines in all four quadrants of the x-y plane. Intuitively this is a $1$-dimensional variety.

(b) There is one polynomial generator so it is already a Groebner basis: $G=\{x^2y^2-1\}$.

(c) By the definition of a GB, $\langle\text{LT}(I)\rangle=\langle x^2y^2\rangle$. Plotting this monomial ideal in the normal way, there is one corner at the grid position $(2,2)$.

(d) $H(0)=1$, $H(1)=3$, $H(2)=6$, $H(3)=10$, and then $H(4)$ through $H(10)$ result from adding $4$ to the previous number: $14,18,22,26,30,34$, and $38$.

(e) The asymptotic "adding $4$" implies the slope becomes constant, that is, that the $H$ function is asymptotically linear. From class, we associated this with defining what we mean by a $1$-dimensional variety.

'''2:''' (a) The variety is the four points $\{\pm 1, \pm 1\}$. A reasonable definition should designate this a $0$-dimensional ideal.

(b) Once again, we already have a Groebner basis which can be quickly verified by calculating the normal form of the only syzygy polynomial and finding it to be zero. $G=\{x^2-1,y^2-1\}$.

(c) By the definition of a GB, $\langle\text{LT}(I)\rangle=\langle x^2,y^2\rangle$. Plotting this monomial ideal in the normal way, there are two corners to the "boat": one at the grid position $(0,2)$ and the other at $(2,0)$.

(d) $H(0)=1$, $H(1)=3$, $H(2)=4$, $H(3)=4=H(4)=H(5)=\cdots$.

(e) From $d=3$ on, $H(d)=4$, that is, it is "asymptotically constant." This means the variety is $0$-dimensional.

Math 380, Spring 2018, Assignment 13

2018-05-01T16:14:01Z

John.DeBrota: /* Solutions: */

__NOTOC__
''"Reeling and Writhing, of course, to begin with,"'' ''the Mock Turtle replied;''
''"And then the different branches of Arithmetic'' - ''Ambition, Distraction, Uglification, and Derision."''

: - Lewis Carroll, ''Alice's Adventures in Wonderland''

==Read:==

# [http://cartan.math.umb.edu/classes/s18_ma380/agc.pdf Notes on abstract Galois correspondences].

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

# Poset.
# Homomorphism (of posets; a.k.a. ''order-preserving map'').
# Isomorphism (of posets).
# Anti-homomorphism (of posets; a.k.a. ''order-reversing map'').
# Anti-isomorphism (of posets).
# Inflationary pair (of maps betweeen posets).
# Abstract Galois correspondence.
# Closure (of an element of a poset occurring in an abstract Galois correspondence).
# Closed element (of a poset occurring in an abstract Galois correspondence).
# Radical ideal.

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

# Theorem relating three passes through an abstract Galois correspondence to one pass.
# Theorem concerning the restriction of an abstract Galois correspondence to the actual images of its mappings.
# Theorem characterizing the closure of an element ("The closure $\overline{s}$ of $s$ is the smallest...")
# Theorem characterizing the smallest variety containing a given set $A\subseteq\mathsf{k}^n$ (a.k.a. the ''Zariski closure'' of $A$).
# Theorem relating the output of $\mathbb{I}$ to radical ideals.

==Solve the following problems:==

# (Orthogonal complement as abstract Galois correspondence) Working in $\mathbb{R}^n$ with its usual dot product, define the ''orthogonal complement'' of a set $S$ to be the set of all vectors in $\mathbb{R}^n$ that are orthogonal to all elements of $S$, i.e. $$S^{\perp}=\{\vec{x}\in\mathbb{R}^n\,|\,\vec{x}\cdot\vec{s}=0\quad\forall\vec{s}\in S\}.$$
::(a) Take $n=2$ and $S=\{(1,1),(2,2)\}$. Draw pictures of $S$, of $S^{\perp}$, and of the double complement $\left(S^\perp\right)^\perp$.
::(b) Take $n=2$ and $S=\{(1,1),(2,0)\}$. Draw pictures of $S$, of $S^{\perp}$, and of the double complement $\left(S^\perp\right)^\perp$.
::(c) Take $n=2$ and $S=\{(0,0)\}$. Draw pictures of $S$, of $S^{\perp}$, and of the double complement $\left(S^\perp\right)^\perp$.
::(d) Now let both $\mathcal{S}$ and $\mathcal{T}$ denote the power set of $\mathbb{R}^n$, regarded as a poset under inclusion. Define $f:\mathcal{S}\rightarrow\mathcal{T}$ by $f(S)=S^\perp$ and define $g:\mathcal{T}\rightarrow\mathcal{S}$ by $g(T)=T^\perp$. (Of course $f$ and $g$ are really the same object with two different names, as are $\mathcal{S}$ and $\mathcal{T}$. We are about to set up a Galois correspondence from a certain poset to itself.) Show that the pair $(f,g)$ is an abstract Galois correspondence between $\mathcal{S}$ and $\mathcal{T}$.
::(e) Describe the closure operation in the Galois correspondence defined above. ''(Hint: you have already met this operation in elementary Linear Algebra).''
::(f) Describe the closed elements in the Galois correspondence defined above.
::(g) Show that the set of linear subspaces of $\mathbb{R}^n$, regarded as a poset under inclusion, is anti-isomorphic to itself. Describe the anti-isomorphism explicitly, with pictures for small $n$.

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

==Questions:==

==Solutions:==
'''(a)''' $\mathcal{S}$ is the two colinear vectors plotted in $\mathbb{R}^2$. $\mathcal{S}^\perp$ is the line passing through the origin at a 45 degree angle through the second and fourth quadrants. In other words, it makes an angle of $3\pi/4$ with the positive $x$ axis. $(\mathcal{S}^\perp)^\perp$ is the line containing the two vectors of $\mathcal{S}$, but extending infinitely in both directions.

'''(b)''' $\mathcal{S}$ is the two vectors plotted in $\mathbb{R}^2$. They are not colinear. $\mathcal{S}^\perp$ is just the zero vector. Nothing else in $\mathbb{R}^2$ is perpendicular to both vectors in $\mathcal{S}$. $(\mathcal{S}^\perp)^\perp$ is the entire plane because everything is perpendicular to the zero vector.

'''(c)''' $\mathcal{S}$ is the zero vector. $\mathcal{S}^\perp$ is the whole plane. $(\mathcal{S}^\perp)^\perp$ is the zero vector again.

'''(d)''' We need to show that it is order reversing and inflationary under inclusion. For two sets $s_1$ and $s_2$ such that $s_1\subset s_2$, $s_1^\perp$ is all the vectors orthogonal to every vector in $s_1$ and $s_2^\perp$ is all the vectors orthogonal to every vector in $s_2$. Since $s_1\subset s_2$, everything in $s_2^\perp$ is orthogonal to everything in $s_1$ because every vector in $s_1$ is a vector in $s_2$. Thus, $s_2^\perp\subset s_1^\perp$. $s_2$ may contain more vectors than $s_1$ so the vectors in $s_2^\perp$ may have to satisfy more restrictions than those in $s_1^\perp$, so the inclusion can be strict. Thus the orthogonal complement is order reversing. It is trivially inflationary because a vector is always orthogonal to a vector which is orthogonal to it, that is, $s_1\subset(s_1^\perp)^\perp$ because a vector in $s_1$ is always perpendicular to every vector in $s_1^\perp$.

'''(e)''' The closure operation is the span operation from linear algebra.

'''(f)''' It follows that the closed sets are the linear subspaces of $\mathcal{R}^n$.

'''(g)''' We showed in class that the maps restricted to the subposets defined by the image sets of each map in an abstract Galois correspondence are anti-isomorphisms. That's what's going on here: Restricting to the linear subspaces on $\mathbb{R}^n$ is restricting to the image sets because $f(g(f(s)))=f(s)$ and the closure operation is $f(g(s))$ (and both reasons reverse for the other map). Since this is true for any AGC, it is true for ours.

Math 380, Spring 2018, Assignment 9

2018-04-01T18:36:23Z

John.DeBrota: /* Questions: */

__NOTOC__
''The moving power of mathematical invention is not reasoning but the imagination.''

: - Augustus de Morgan

==Read:==

# Section 2.6.
# Section 2.7.

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

# $\overline{f}^{(g_1,\dots,g_s)}$ (the ''normal form'' of $f$ modulo the ordered set $(g_1,\dots,g_s)$).
# Least common multiple (of two monomials).
# $S(f,g)$ (the ''syzygy polynomial'' determined by $f$ and $g$).

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

# Bound on the multidegree of $S(f,g)$.
# Cancellation lemma (this is Lemma 2.6.5 in the text).
# Buchberger's $S$-pair criterion.

==Carefully describe the following algorithms:==

# Buchberger's algorithm (to compute a Grőbner basis for a given ideal $\left\langle f_1,\dots,f_s\right\rangle$).

==Solve the following problems:==

# Section 2.6, problems 2, 5, 6, and 9.
# Section 2.7, problem 2(a), using lex order only.
# (Parameter elimination again) Using lex order in which $t>x>y$, compute a Grőbner basis for the ideal $\left\langle t-x, ty-1\right\rangle$. What does this tell you about the image of the parametrization $x=t,\quad y=1/t$?

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

==Questions:==

I really don't know what to do for the last part of question $3$. What does a parameterization derived from a set of polynomials which form a Groebner basis have over a parameterization derived otherwise?

==Solutions:==

'''2.6.2:''' (a) Performing the division algorithm with divisors $g_1=x+z$ and $g_2=y-z$, we get $q_1=y$, $q_2=-z$, and $r=-z^2$.

(b) Performing the division algorithm with the reversed divisors, $g_1=y-z$ and $g_2=x+z$, we get $q_1=x$, $q_2=z$, and $r=-z^2$. The quotients are different, but the remainders are the same. That the remainders are the same is a general property when we have a Groebner basis.

'''2.6.5:''' We compute the syzygy polynomial with lex order for each pair.

(a)
\[
S(4x^2z-7y^2,xyz^2+3xz^4)=-3x^2z^4-\frac{7}{4}y^3z
\]
(b)
\[
S(x^4y-z^2,3xz^2-y)=\frac{1}{3}x^3y^2-z^4
\]
(c) Assuming $i$ is the imaginary unit and not a variable.
\[
S(x^7y^2z+2ixyz,2x^7y^2z+4)=2ixyz-2
\]
(d) Assuming $3^z$ is a typo in the second polynomial because otherwise it wouldn't be a polynomial.
\[
S(xy+z^3,z^2-3z)=3xyz+z^5
\]

'''2.6.6:''' Yes. Take the pair from part (d) of the previous problem. With lex order $S(f,g)=3xyz+z^5$. With grlex order,
\[
S(f,g)=\frac{z^3}{z^3}(z^3+xy)-\frac{z^3}{z^2}(z^2-3z)=xy+3z^2\;.
\]

'''2.6.9:''' Apply the S-pair criterion, that is, show that $\overline{S(-x^2+y,-x^3+z)}^{\{-x^2+y,-x^3+z\}}\neq 0$. First we compute $S(-x^2+y,-x^3+z)=-xy-z$. Then, using the division algorithm, we see that $r=-xy-z\neq 0$

'''2.7.2a:''' We use Buchberger's algorithm. It is tedious to write it all out, but it converges after two additions to the basis. The final Groebner basis is $\{x^y-1,xy^2-x,x^2-y,y^2-1\}$. With this, the normal form of all 6 syzygy polynomials is zero.

'''3:''' $\{t-x,ty-1\}$ is already a Groebner basis for the ideal it generates. We check this with the S-pair criterion. The syzygy polynomial is $S(t-x,ty-1)=ty-xy$ and the remainder upon dividing this by $\{t-x,ty-1\}$ is $0$. I don't know what this tells us about the parameterization, maybe that it doesn't miss any points in the variety?

Math 380, Spring 2018, Assignment 9

2018-04-01T18:34:22Z

John.DeBrota: /* Solutions: */

__NOTOC__
''The moving power of mathematical invention is not reasoning but the imagination.''

: - Augustus de Morgan

==Read:==

# Section 2.6.
# Section 2.7.

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

# $\overline{f}^{(g_1,\dots,g_s)}$ (the ''normal form'' of $f$ modulo the ordered set $(g_1,\dots,g_s)$).
# Least common multiple (of two monomials).
# $S(f,g)$ (the ''syzygy polynomial'' determined by $f$ and $g$).

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

# Bound on the multidegree of $S(f,g)$.
# Cancellation lemma (this is Lemma 2.6.5 in the text).
# Buchberger's $S$-pair criterion.

==Carefully describe the following algorithms:==

# Buchberger's algorithm (to compute a Grőbner basis for a given ideal $\left\langle f_1,\dots,f_s\right\rangle$).

==Solve the following problems:==

# Section 2.6, problems 2, 5, 6, and 9.
# Section 2.7, problem 2(a), using lex order only.
# (Parameter elimination again) Using lex order in which $t>x>y$, compute a Grőbner basis for the ideal $\left\langle t-x, ty-1\right\rangle$. What does this tell you about the image of the parametrization $x=t,\quad y=1/t$?

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

==Questions:==

==Solutions:==

'''2.6.2:''' (a) Performing the division algorithm with divisors $g_1=x+z$ and $g_2=y-z$, we get $q_1=y$, $q_2=-z$, and $r=-z^2$.

(b) Performing the division algorithm with the reversed divisors, $g_1=y-z$ and $g_2=x+z$, we get $q_1=x$, $q_2=z$, and $r=-z^2$. The quotients are different, but the remainders are the same. That the remainders are the same is a general property when we have a Groebner basis.

'''2.6.5:''' We compute the syzygy polynomial with lex order for each pair.

(a)
\[
S(4x^2z-7y^2,xyz^2+3xz^4)=-3x^2z^4-\frac{7}{4}y^3z
\]
(b)
\[
S(x^4y-z^2,3xz^2-y)=\frac{1}{3}x^3y^2-z^4
\]
(c) Assuming $i$ is the imaginary unit and not a variable.
\[
S(x^7y^2z+2ixyz,2x^7y^2z+4)=2ixyz-2
\]
(d) Assuming $3^z$ is a typo in the second polynomial because otherwise it wouldn't be a polynomial.
\[
S(xy+z^3,z^2-3z)=3xyz+z^5
\]

'''2.6.6:''' Yes. Take the pair from part (d) of the previous problem. With lex order $S(f,g)=3xyz+z^5$. With grlex order,
\[
S(f,g)=\frac{z^3}{z^3}(z^3+xy)-\frac{z^3}{z^2}(z^2-3z)=xy+3z^2\;.
\]

'''2.6.9:''' Apply the S-pair criterion, that is, show that $\overline{S(-x^2+y,-x^3+z)}^{\{-x^2+y,-x^3+z\}}\neq 0$. First we compute $S(-x^2+y,-x^3+z)=-xy-z$. Then, using the division algorithm, we see that $r=-xy-z\neq 0$

'''2.7.2a:''' We use Buchberger's algorithm. It is tedious to write it all out, but it converges after two additions to the basis. The final Groebner basis is $\{x^y-1,xy^2-x,x^2-y,y^2-1\}$. With this, the normal form of all 6 syzygy polynomials is zero.

'''3:''' $\{t-x,ty-1\}$ is already a Groebner basis for the ideal it generates. We check this with the S-pair criterion. The syzygy polynomial is $S(t-x,ty-1)=ty-xy$ and the remainder upon dividing this by $\{t-x,ty-1\}$ is $0$. I don't know what this tells us about the parameterization, maybe that it doesn't miss any points in the variety?

Math 380, Spring 2018, Assignment 8

2018-03-26T17:15:29Z

John.DeBrota: /* Solutions: */

__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''

==Read:==

# Section 2.4.
# Section 2.5.

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

# $\left\langle LT(I)\right\rangle$ (the ''leading term ideal'' of an ideal $I$).
# Grőbner basis (for an ideal $I$).

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

# Dickson's Lemma.
# Ascending chain condition for monomial ideals.
# Hilbert basis theorem.
# Theorem concerning the existence of Grőbner bases.
# Ascending chain condition for arbitrary ideals.

==Carefully describe the following algorithms:==

# Algorithm to decide whether $f\in\left\langle g_1,\dots,g_s\right\rangle$ ''when $g_1,\dots,g_s$ form a Grőbner basis for the ideal that they generate.''

==Solve the following problems:==

# Section 2.5, problems 1, 7, 8, 10, 17, and 18.

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

==Questions:==

==Solutions:==
'''2.5.1:''' $\langle\text{LT}(g_1),\text{LT}(g_2),\text{LT}(g_3)\rangle=\langle xy^2,xy,x\rangle$ with lex order. I need to find a polynomial $g\in I$ such that $\text{LT}(g)\notin\langle xy^2,xy,x\rangle$. A monomial is contained in a monomial ideal iff it is divisible by one of the generators of the monomial ideal. Note that every generator of $\langle xy^2,xy,x\rangle$ contains a factor of $x$. Thus, any $g\in I$ with a leading term not containing $x$ will work for our purposes. Consider $g=g_2-yg_3=y^2z^4-z^2$ which has leading term $\text{LT}(g)=y^2z^4$. This is not divisible by any of the generators of $\langle xy^2,xy,x\rangle$, so it is not in this ideal.

'''2.5.7:''' A Groebner basis is a set of polynomials $g_1,\ldots,g_s\in I$ such that $\langle \text{LT}(I)\rangle=\langle\text{LT}(g_1),\ldots,\text{LT}(g_s)\rangle$. We want to know if the polynomials generating $I=\langle x^4y^2-z^5,x^3y^3-1,x^2y^4-2z\rangle$ with grlex order form a Groebner basis for $I$. Taking the leading terms of each generator, the question is whether $\langle \text{LT}(I)\rangle=\langle x^4y^2,x^3y^3,x^2y^4\rangle$. Consider
\[
y^2(x^4y^2-z^5)-x^2(x^2y^4-2z)=-y^2z^5-2x^2z\in I \implies y^2z^5\in \langle \text{LT}(I)\rangle\;.
\]
But $y^2z^5\notin \langle x^4y^2,x^3y^3,x^2y^4\rangle$ because it is not divisible by any of the generators which implies that $\{x^4y^2-z^5,x^3y^3-1,x^2y^4-2z\}$ is not a Groebner basis for $I$.

'''2.5.8:''' We want to determine whether $\{x-z^2,y-z^3\}$ is a Groebner basis with lex order for the ideal $I$ they generate. We will show that $\langle\text{LT}(g_1),\text{LT}(g_2)\rangle=\langle x,y\rangle=\langle\text{LT}(I)\rangle$, and so the set is a Groebner basis. A general element of $I$ is of the form $a_1(x-z^2)+a_2(y-z^3)$ where $a_i\in\mathbb{R}[x,y,z]$. It is easy to see from this general form that the leading term of any element in $I$ has at least one power of $x$ or $y$. This will necessarily also be true for each term of any polynomial in $\langle \text{LT}(I)\rangle$. Since every term of any $q\in\langle\text{LT}(I)\rangle$ will be divisible by either $x$ or $y$ (or both), we have that $q\in\langle x,y\rangle$ which shows $\langle \text{LT}(I)\rangle\subseteq\langle x,y\rangle$. The reverse containment is trivial because $x=\text{LT}(g_1)\in\langle\text{LT}(I)\rangle$ and $y=\text{LT}(g_2)\in\langle\text{LT}(I)\rangle$.

'''2.5.10:''' Let $I=\langle g_1\rangle$ be a principal ideal. We want to show $\{g_1,\ldots,g_s\}$ is a Groebner basis for $I$. So we must show $\langle\text{LT}(I)\rangle=\langle\text{LT}(g_1),\ldots,\text{LT}(g_s)\rangle$. A general element of $I$ is the polynomial $f=hg_1$. The leading term is then $\text{LT}(f)=\text{LT}(hg_1)=\text{LT}(h)\text{LT}(g_1)$ where we used the result of previous homework problem showing $\text{LT}(ab)=\text{LT}(a)\text{LT}(b)$. Every term of any polynomial in the monomial ring $\langle \text{LT}(I)\rangle$ will be divisible by $\text{LT}(g_1)$ which shows $\text{LT}(I)\subseteq\langle\text{LT}(g1),\ldots,\text{LT}(g_s)\rangle$. For the other direction, $\text{LT}(g_1)$ is divisible by the generator of $\langle \text{LT}(I)\rangle$ formed when $h=1$ and $\text{LT}(g_i)$ is divisible by the generator formed when $h=g_i$.

'''2.5.17:'''
(a) The decompositions $y+x^2-4=-(x^2-y)+2(x^2-2)$ and $x^2-2=(1/2)(x^2-y)+(1/2)(y+x^2-4)$ demonstrate the nontrivial polynomial inclusions necessary for the reverse containment argument.

(b) $\mathbb{V}(I)=\mathbb{V}(x^2-y,x^2-2)$. The second equation is clearly solved by $x=\pm\sqrt{2}$. When these values are substituted into the first, we see that $y=2$. Thus the variety is the points $\{(\pm\sqrt{2},2)\}$.

'''2.5.18:''' (a) Let $a\in\mathbb{V}(f,g)$. Then $f(a)=0$ and $f(a)=0=g_1(a)g_2(a)$. So either $g_1(a)=0$ or $g_2(a)=0$ which shows $a\in \mathbb{V}(f,g_1)\cup\mathbb{f,g_2}$. OTOH, let $a\in\mathbb{V}(f,g_1)\cup\mathbb{f,g_2}$. Then either $a\in\mathbb{V}(f,g_1)$ or $a\in\mathbb{V}(f,g_2)$. Say $a\in\mathbb{V}(f,g_1)$. Then $f(a)=0$ and $g_1(a)=0$. So $g(a)=g_1(a)g_2(a)=0$ which implies $a\in\mathbb{V}(f,g)$. It is similar for the other option.

(b) The decompositions $xz-x^4=(xz-y^2)+(y+x^2)(y-x^2)$ and $xz-y^2=(xz-x^4)-(y+x^2)(y-x^2)$ demonstrate the nontrivial polynomial inclusions necessary for the reverse containment argument.

(c) $\mathbb{V}(y-x^2,xz-x^4)=\mathbb{V}(y-x^2,x)\cup\mathbb{V}(y-x^2,z-x^3)$. The first of these varieties is the z-axis because the solution set of the system of equations $\{x=0,y=x^2\}$ is $\{0,0,z\}$ for any $z\in\mathbb{R}$. The second variety is the twisted cubic which we have seen before. So the original variety is the union of the z-axis and the twisted cubic.

Math 380, Spring 2018, Assignment 4

2018-02-18T20:24:42Z

John.DeBrota: /* Solutions: */

__NOTOC__
''I was at the mathematical school, where the master taught his pupils after a method scarce imaginable to us in Europe. The proposition and demonstration were fairly written on a thin wafer, with ink composed of a cephalic tincture. This the student was to swallow upon a fasting stomach, and for three days following eat nothing but bread and water. As the wafer digested the tincture mounted to the brain, bearing the proposition along with it.''

: - Jonathan Swift, ''Gulliver's Travels''

==Read:==

# Section 1.5.
# Section 2.1.

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

# Monic polynomial (in $\mathsf{k}[x]$).
# $\mathrm{gcd}(f,g)$ (where $f,g\in\mathsf{k}[x]$).

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

# Theorem concerning the division algorithm.
# Classification of ideals in $\mathsf{k}[x]$ ("Every ideal in $\mathsf{k}[x]$ is generated by...").
# Theorem concerning monic generators of ideals in $\mathsf{k}[x]$.
# Theorem relating $\mathrm{gcd}(f,g)$ to common divisors of $f$ and $g$.
# Theorem relating $\left\langle f,g\right\rangle$ to $\left\langle g,r\right\rangle$ when $f=gq+r$.

==Carefully describe the following algorithms:==

# Division algorithm (to compute quotient and remainder when $f$ is divided by $g$).
# Euclid's algorithm (to compute $\mathrm{gcd}(f,g)$).
# Algorithm to replace any system of finitely many univariate equations by a single univariate equation.
# Algorithm to factor polynomials over $\mathbb{C}$. ''(Note: this algorithm is conceptually simple but should NOT be used in engineering applications; it is numerically unstable. See books on numerical analysis for more practical algorithms.)''

==Solve the following problems:==

# Section 1.5, problems 1, 3, 4, 11, and 12.
# Prove that for any $f_1,\dots,f_s\in\mathsf{k}[x]$, one has $\left\langle f_1,f_2,f_3,\dots,f_s\right\rangle = \left\langle \mathrm{gcd}(f_1,f_2),f_3,\dots,f_s\right\rangle$. ''(Hint: use the ideal containment criterion from the previous assignment.)''
# Optional (but interesting): section 1.5, problem 17. (You may need to use the results of problems 13, 14, and 15.)

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

==Questions:==

==Solutions:==

'''1.5.1:''' It is known that every nonconstant polynomial in $\mathbb{C}[x]$ has a root in $\mathbb{C}$. Let $a_1$ be such a root. By the factor theorem, $f$ is divisible by $x-a_1$ with remainder zero, that is $f=(x-a_1)q$ where $q$ is the quotient. If $q$ is a nonconstant polynomial, repeat this argument with $q$ instead of $f$, otherwise $q$ is the constant c in the desired form. Thus every univariate polynomial of degree $n>0$ over the complex numbers may be written $f=c(x-a_1)\cdots(x-a_n)$ where $a_1,\ldots,a_n,c\in\mathbb{C}$ and $c\neq0$.

'''1.5.3:''' We wish to prove that $I=\langle x,y\rangle\subseteq k[x,y]$ is not a principal ideal, that is, that it is not generated by one element. Suppose it were. Let $f$ be the single generator, i.e., $\langle x,y\rangle=\langle f\rangle$. Obviously $f$ cannot be a constant. Then every polynomial in $\langle x,y\rangle$ is equal to $fg$ for some $g\in k[x,y]$. Consider $x\in\langle x,y\rangle$. Then $x=fg$ for some $g$. Taking the degree of both sides we have $\text{deg}(x)=1=\text{deg}(fg)=\text{deg}(f)+\text{deg}(g)$. The only way this can be satisfied is if $f$ or $g$ is a constant and the other is degree 1. Since $f$ cannot be a constant, $g$ is. Say $g=c$, then $x=cf\implies f=\frac{x}{c}$. Since $f$ generates $\langle x,y\rangle$, it must also be that $y=fh$ for some $h\in k[x,y]$. Then $y=\frac{x}{c}h$, but this clearly cannot be satisfied, so we have a contradiction. Thus $\langle x,y\rangle$ is not a principal ideal.

'''1.5.4:''' $\text{gcd}(f,g)$ is the monic generator of the ideal $\langle f,g\rangle$, so, if $h=\text{gcd}(f,g)$, then $\langle h\rangle=\langle f,g\rangle$. By definition, this means that $h\in\langle f,g\rangle$, so there exists $A,B\in k[x]$ such that $Af+Bg=h$.

'''1.5.11:''' (a) As noted in class, every nonconstant polynomial has a root so $\mathbb{V}(f)\neq \emptyset$ iff $f$ is not a constant.

(b) If $\text{gcd}(f_1,\ldots,f_s)=1$, we have $\langle f_1,\ldots,f_s\rangle=\langle 1 \rangle$ and so $\mathbb{V}(f_1,\ldots,f_s)=\mathbb{V}(1)=\emptyset$. If $\mathbb{V}(f_1,\ldots f_s)=\emptyset$, then $\mathbb{V}(\text{gcd}(f_1,\ldots,f_s))=\emptyset$. This means that $\text{gcd}(f_1,\ldots,f_s)$ is a polynomial in $\mathbb{C}[x]$ which doesn't have a root in $\mathbb{C}$. Therefore it is a constant polynomial. By the definition of the gcd, it is monic, so $\text{gcd}(f_1,\ldots,f_s)=1$.

(c) Determine $\text{gcd}(f_1,\ldots,f_s)$. If this is not equal to $1$ then the affine variety $\mathbb{V}(f_1,\ldots,f_s)$ is not empty.

'''1.5.12:''' (a) By inspection of the factored form of $f$, the polynomial vanishes at the points $\{a_1,\ldots a_l\}$ and doesn't vanish at any others. Thus $\mathbb{V}(f)=\{a_1,\ldots, a_l\}$.

(b) We wish to show $\mathbb{I}(\mathbb{V}(f))=\langle f_\text{red}\rangle$. We proceed by mutual inclusion. $\mathbb{I}(\mathbb{V}(f))$ is the set of all polynomials which vanish at $\{a_1,\ldots,a_l\}$. We can clearly see that $f_\text{red}$, as defined, vanishes at all of these points, so $f_\text{red}\in\mathbb{I}(\mathbb{V}(f))\implies \langle f_\text{red}\rangle\subseteq\mathbb{I}(\mathbb{V}(f))$. Now let $g\in\mathbb{I}(\mathbb{V}(f))$. We want to show that $g\in\langle f_\text{red}\rangle$. Since $g\in\mathbb{I}(\mathbb{V}(f))$, $g$ has, among its roots, $\{a_1\ldots,a_l\}$. As it is a univariate polynomial over $\mathbb{C}$, we may write $g$ in the fully factored form
\[
g=c(x-b_1)^{n_1}\cdots(x-b_s)^{n_s}\;.
\]
But it must be that $l$ of the roots are the $a_i$, so we can arrange to write $g$ as
\[
g=c(x-a_1)^{n_1}\cdots(x-a_l)^{n_l}(x-b_{s-l})^{n_{s-l}}\cdots(x-b_s)^{n_s}\;.
\]
By inspection, then, we can see that $g=h f_\text{red}$ for an appropriately choosen $h\in\mathbb{C}[x]$, and so $g\in\langle f_\text{red}\rangle\implies \mathbb{I}(\mathbb{V}(f))\subseteq\langle f_\text{red}\rangle$.

'''2:''' $\text{gcd}(f_1,f_2)$ is the unique monic generator of $\langle f_1,f_2\rangle$. Consequently, $\langle \text{gcd}(f_1,f_2)\rangle=\langle f_1,f_2\rangle\implies\text{gcd}(f_1,f_2)\in\langle f_1,f_2\rangle\implies\text{gcd}(f_1,f_2)\in\langle f_1,\ldots,f_s\rangle\implies\langle \text{gcd}(f_1,f_2),f_3,\ldots,f_s\rangle\subseteq\langle f_1,\ldots,f_s\rangle$. Since $\text{gcd}(f_1,f_2)$ is a divisor of $f_1$, $f_1=h_1 \text{gcd}(f_1,f_2)$. Likewise $f_2=h_2 \text{gcd}(f_1,f_2)$. This shows that $f_1,f_2\in\langle \text{gcd}(f_1,f_2)\rangle\implies\langle f_1,f_2\rangle\subseteq\langle\text{gcd}(f_1,f_2)\rangle$. We can add the generators $f_3,\ldots,f_s$ to both ideals without changing the inclusion, so we have $\langle f_1,\ldots,f_s\rangle\subseteq\langle\text{gcd}(f_1,f_2),f_3,\ldots,f_s\rangle$.

'''1.5.17:''' First let's convert the variety into one with only one equation by finding the gcd of the two polynomials. Note that $x$ can be factored out of the first polynomial, but not out of the second. Clearly $x$ is not a common divisor, so we seek $\text{gcd}(x^4-2x^3+2x-1,x^5-x^4-2x^3+2x^2+x-1)$. Polynomial division quickly reveals that the first divides the second with remainder zero, so $x^4-2x^3+2x-1$ is the gcd. From problem (1.5.12), we learned that $\mathbb{I}(\mathbb{V}(f))=\langle f_\text{red}\rangle$ and the result of problem (1.5.15) tells us that
\[
f_\text{red}=\frac{f}{\text{gcd}(f,f')}\;,
\]
where $f'$ is the formal derivative of $f$. The formal derivative of $x^4-2x^3+2x-1$ is $4x^3-6x^2+2$. Running the gcd algorithm takes a couple steps, which I'm not going to typeset. We obtain $\text{gcd}(x^4-2x^3+2x-1,4x^3-6x^2+2)=x^2-2x+1$. Thus,
\[
f_\text{red}=\frac{x^4-2x^3+2x-1}{x^2-2x+1}=x^2-1\;.
\]
Thus we have $\mathbb{I}(\mathbb{V}(x^5-2x^4+2x^2-x,x^5-x^4-2x^3+2x^2+x-1))=\langle x^2-1\rangle$, that is, a basis for the ideal is $x^2-1$.

Math 380, Spring 2018, Assignment 4

2018-02-18T20:09:20Z

John.DeBrota: /* Solutions: */

Math 380, Spring 2018, Assignment 3

2018-02-13T17:53:43Z

John.DeBrota: /* 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''

==Read:==

# Section 1.4.

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

# Ideal (in $\mathsf{k}[x_1,\dots,x_n]$).
# $\left\langle S\right\rangle$ (the ideal generated by a set $S$ of polynomials).
# $\left\langle f_1,\dots,f_s\right\rangle$ (the ideal generated by the finite set $\{f_1,\dots,f_s\}$).
# $\mathbb{I}(T)$ (the ''ideal'' of the set of points $T\subseteq\mathsf{k}^n$).

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

# Theorem relating $\mathbb{V}(S)$ to $\mathbb{V}(\left\langle S\right\rangle)$.
# Theorem characterizing when $\left\langle f_1,\dots,f_s\right\rangle\subseteq\left\langle g_1,\dots,g_t\right\rangle$.
# Theorem characterizing when $\left\langle f_1,\dots,f_s\right\rangle=\left\langle g_1,\dots,g_t\right\rangle$.
# Theorem concerning the ''inclusion-reversing'' and ''inflationary'' character of the fundamental pairing $(\mathbb{V},\mathbb{I})$.

==Solve the following problems:==

# Section 4, problems 1, 3, 5, 6(a), 6(b), 7, and 8. ''(In problem 6, the word "basis" means "generating set.")''
# Prove that, for any set $T\subseteq\mathsf{k}^n$, the set $\mathbb{I}(T)$ is always an ''ideal'' of $\mathsf{k}[x_1,\dots,x_n]$.
# Working in $\mathbb{R}^1$, describe $\mathbb{V}(\mathbb{I}(\mathbb{Z}))$. ''(In topological language, this exercise shows that the "Zariski closure" of a set may be quite different from its ordinary closure. This is because the Zariski topology is extremely "coarse.")''

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

==Questions:==
This came up because of exercise 1.4.7 but is just a general question: Are we allowed to assume we're working in the smallest affine space that will fit around the polynomials concerned i.e: does the answer boil down to the same thing whether you do your computations in k[x,y] or k[x,y,z,..,w]?
:That is a very good question! For purposes of the class, yes, unless the problem explicitly states otherwise. So for problem 1.4.7 the varieties are subsets of $\mathsf{k}^2$. But: this particular problem makes sense, and its result is true, in $\mathsf{k}^n$ for any $n\geq2$.

:It is moderately enlightening to think about the problem for larger $n$. You will find the following, in general: suppose $f_1,\dots,f_s\in\mathsf{k}[x_1,\dots,x_n]$ and let $V=\mathbb{V}(f_1,\dots,f_s)$ (a subset of $\mathsf{k}^n$). But we could also think of $f_1,\dots,f_s$ as polynomials in $x_1,\dots,x_n,x_{n+1},\dots,x_{n+m}$ and then use them to define $W=\mathbb{V}(f_1,\dots,f_s)$ (a subset of $\mathsf{k}^{n+m}$). Then $W=V\times\mathsf{k}^m$. One says that $W$ is a ''cylinder over $V$''. (If you think about some cases where $n=2$ and $m=1$ you will understand why.)

:Cylinders are usually regarded as uninteresting; most geometric properties of $W$ are already carried by $V$, because most algebraic algorithms, carried out to answer geometric questions about $W$, will only involve $x_1,\dots,x_n$ and so they will automatically answer the same question about $V$.

::-[[User:Steven.Jackson|Steven.Jackson]] ([[User talk:Steven.Jackson|talk]]) 19:16, 12 February 2018 (EST)

We know every ideal of a variety is a radical ideal. Is the reverse true? That is, is every radical ideal the ideal of some variety?

==Solutions:==
'''1.4.1:'''

(a) From second equation, $y=1/x$. Plugging into the first, we have $x^2+\frac{1}{x^2}-1=0$ which we can write as $x^4-x^2+1=0$.

(b) Multiplying the second equation by $xy+1$, it becomes $x^2y^2-1=0$. If we multiply the first equation by $x^2$, we get $x^4+x^2y^2-x^2=0$ and we can now substitute $x^2y^2=1$ to obtain $x^4-x^2+1=0$, just as in part (a). Explicitly,
\[
x^4-x^2+1=x^2(x^2+y^2-1)-(xy+1)(xy-1)\;,
\]
where the coefficient polynomials are $x^2$ and $-(xy+1)$.

'''1.4.3:'''

In each case, to show equality we show containment in both directions.

(a) To show $\langle x+y, x-y\rangle\subseteq\langle x,y\rangle$, we need to show that each polynomial defining the ideal is an element of the other ideal. That is, we need $x+y\in\langle x,y\rangle$ and $x-y\in\langle x,y\rangle$. This is obvious: Since $x$ and $y$ are in $\langle x,y\rangle$, so is $x+y$ and $x-y$ because ideals are closed under consequences. Next, to show $\langle x+y, x-y\rangle\supseteq\langle x,y\rangle$, we need to show $x,y\in\langle x+y,x-y\rangle$. By inspection, we see
\[
x=\frac{1}{2}(x+y)+\frac{1}{2}(x-y)\quad \text{and}\quad y=\frac{1}{2}(x+y)-\frac{1}{2}(x-y)\;,
\]
so we are done.

(b) We trivially see $\langle x+xy,y+xy,x^2,y^2\rangle\subseteq\langle x,y\rangle$, the constituent polynomials of the left hand side are obvious consequences of the polynomials comprising the right. We will show $\langle x+xy,y+xy,x^2,y^2\rangle\supseteq\langle x,y\rangle$. After some algebraic annoyances we see

\[
x=(1-y)(x+xy)+y(y+xy)-y^2 \quad \text{and} \quad y=x(x+xy)+(1-x)(y+xy)-x^2\;.
\]

(c) $\langle 2x^2+3y^2-11,x^2-y^2-3\rangle\subseteq\langle x^2-4,y^2-1\rangle$:
\[
2x^2+3y^2-11=2(x^2-4)+3(y^2-1)\quad\text{and}\quad x^2-y^2-3=(x^2-4)-(y^2-1)\;.
\]

$\langle 2x^2+3y^2-11,x^2-y^2-3\rangle\supseteq\langle x^2-4,y^2-1\rangle$:
\[
x^2-4=\frac{1}{5}(2x^2+3y^2-11)+\frac{3}{5}(x^2-y^2-3)\quad \text{and}\quad y^2-1=\frac{1}{5}(2x^2+3y^2-11)-\frac{2}{5}(x^2-y^2-3)\;.
\]

'''1.4.5:''' As shown in class, the affine variety defined by a set of polynomials is equal to the affine variety defined by the ideal generated by its constituent polynomials. From problem 3b, we know the ideals are equal and so the varieties are also equal.

'''1.4.6a:''' If any vector space basis is infinite, ''every'' vector space basis is infinite. $I=\langle x\rangle\subseteq k[x]$ is the ideal generated by the polynomial $x$. The general form of any polynomial in this set is
\[
x(a_0+a_1x+a_2x^2+\cdots+a_nx^n)=a_0x+a_1x^2+\cdots+a_nx^{n+1}\;,
\]
where $n$ can be any finite number. We need a linearly independent spanning set for the ideal. The obvious choice is $\{x,x^2,x^3,\ldots\}$. This infinite set obviously spans the ideal. To show linear independence, we need to convince ourselves that $a_0x+a_1x^2+a_2x^3\cdots=0$ iff $a_i=0$ for all $i$. The forward direction is obvious. For the reverse, note that $x$ may be factored out of the left hand side. $x$ is a placeholder for an arbitrary field element and so may be divided out on both sides. This leaves $a_0+a_1x+a_2x^2+\cdots=0$. The constant term in this resulting polynomial must equal zero or the whole thing cannot equal zero. We are then left with the original situation with $a_i\to a_{i+1}$, so the argument repeats. Thus we have an infinite basis for the vectorspace $I$, proving that all bases for this set are infinite, even though the basis for the ideal is finite.

'''1.4.6b:''' $yx-xy=0$.

'''1.4.7:''' We are interested in the ideal $\mathbb{I}(\mathbb{V}(x^n,y^m))$ where $n$ and $m$ are arbitrary positive integers. Unpacking the definitions, $\mathbb{V}(x^n,y^m)$ is the set of points where the polynomials $x^n$ and $y^m$ vanish. This set is the singular point $\{0,0\}$. $\mathbb{I}(\{0,0\})$ is the set of all polynomials in k[x,y] which vanish at $\{0,0\}$. If the coefficient of the degree zero term vanishes, this is assured. A general term of the ideal may then be written
\[
a_{10}x+a_{01}y+a_{20}x^2+a_{11}xy+a_{02}y^2+a_{30}x^3+a_{21}x^2y+a_{12}xy^2+a_{03}y^3+\cdots
\]
We compare this with a general element of the ideal $\langle x,y\rangle=g_1x+g_2y$ where $g_1,g_2\in k[x,y]$. Let $g_1=b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\cdots+b_{ij}x^iy^j$ and $g_2=c_{00}+c_{10}x+c_{01}y+c_{20}x^2+c_{11}xy+c_{02}y^2+\cdots+c_{kl}x^ky^l$. Then
\[
\langle x,y\rangle = (b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\cdots+b_{ij}x^iy^j)x+(c_{00}+c_{10}x+c_{01}y+c_{20}x^2+c_{11}xy+c_{02}y^2+\cdots+c_{kl}x^ky^l)y\;.
\]
We can quite easily see that there is no constant term in the simplified result. It is clear that every other term is present, which could be demonstrated explicitly by associating to each $a_{ij}$ a sum of some $b_{kl}$ and $c_{mn}$. Thus we have the desired equality by explicit computation. (I didn't use the containment argument, but I hope this still demonstrates the fact.)

'''1.4.8:''' (a) We prove both directions. Suppose $f\in\mathbb{I}(\mathbb{V})$, then $f^m\in\mathbb{I}(\mathbb{V})$ because ideals are closed under consequences, that is, we can just multiply $f$ by the polynomial $f^{m-1}$ and it has to be in the ideal. Now suppose $f^m\in\mathbb{I}(\mathbb{V})$. $\mathbb{I}(\mathbb{V})$ is the set of polynomials which vanish on the variety $\mathbb{V}$. So $f^m=0$ at all points on $\mathbb{V}$. By simple algebra, it must be that $f=0$ also at these points, thus $f\in\mathbb{I}(\mathbb{V})$.

(b) By part (a) it suffices to show that $x\notin\langle x^2,y^2\rangle$. If it were, it must be that $x=g_1x^2+g_2y^2$ for $g_1,g_2\in k[x,y]$. This is clearly impossible. Thus $\langle x^2,y^2\rangle$ is not radical and is consequently not the ideal of any variety.

'''2:''' The set $\mathbb{I}(T)$ is the set of all polynomials which vanish on the set of points $T\subseteq k^n$. We need to check whether all three axioms of an ideal hold for this set. The zero polynomial is clearly present because it vanishes everywhere, therefore it also vanishes on $T$. If $f_1,f_2\in \mathbb{I}(T)$ then they each also vanish on $T$. In accordance with how we add polynomials and the functions they generate, it must be that for any $a\in T$, $(f_1+f_2)(a)=0$ as well. Therefore $f_1+f_2\in\mathbb{I}(T)$. Finally if $g$ is an arbitrary polynomial and $f_1\in\mathbb{I}(T)$ then $gf_1(a)=g(a)f_1(a)=g(a)0=0$ which means $gf_1\in\mathbb{I}(T)$. Thus $\mathbb{I}(T)$ is an ideal of $k[x_1,\ldots,x_n]$.

'''3:''' Unpacking the definitions, $\mathbb{I}(\mathbb{Z})$ is the set of polynomials which vanish on all integers. We know that nonconstant univariate polynomials may have at most as many roots as their degree. If these polynomials must vanish at ''all'' integer points, this is infinitely many roots which implies that $f=\text{constant}=0$. So we have $\mathbb{V}(0)$ which is the solution set to the equation $0=0$. This equation is satisfied by the entire affine space, that is all of $\mathbb{R}^1$, so $\mathbb{V}(\mathbb{I}(\mathbb{Z}))=\mathbb{R}^1$.

:Endorsing the solutions to 1.4.1-1.4.6 above. All correct, and beautifully written.

:$\quad$-[[User:Steven.Jackson|Steven.Jackson]] ([[User talk:Steven.Jackson|talk]]) 10:08, 13 February 2018 (EST)

Math 380, Spring 2018, Assignment 3

2018-02-13T17:47:18Z

John.DeBrota: /* 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''

==Read:==

# Section 1.4.

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

# Ideal (in $\mathsf{k}[x_1,\dots,x_n]$).
# $\left\langle S\right\rangle$ (the ideal generated by a set $S$ of polynomials).
# $\left\langle f_1,\dots,f_s\right\rangle$ (the ideal generated by the finite set $\{f_1,\dots,f_s\}$).
# $\mathbb{I}(T)$ (the ''ideal'' of the set of points $T\subseteq\mathsf{k}^n$).

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

# Theorem relating $\mathbb{V}(S)$ to $\mathbb{V}(\left\langle S\right\rangle)$.
# Theorem characterizing when $\left\langle f_1,\dots,f_s\right\rangle\subseteq\left\langle g_1,\dots,g_t\right\rangle$.
# Theorem characterizing when $\left\langle f_1,\dots,f_s\right\rangle=\left\langle g_1,\dots,g_t\right\rangle$.
# Theorem concerning the ''inclusion-reversing'' and ''inflationary'' character of the fundamental pairing $(\mathbb{V},\mathbb{I})$.

==Solve the following problems:==

# Section 4, problems 1, 3, 5, 6(a), 6(b), 7, and 8. ''(In problem 6, the word "basis" means "generating set.")''
# Prove that, for any set $T\subseteq\mathsf{k}^n$, the set $\mathbb{I}(T)$ is always an ''ideal'' of $\mathsf{k}[x_1,\dots,x_n]$.
# Working in $\mathbb{R}^1$, describe $\mathbb{V}(\mathbb{I}(\mathbb{Z}))$. ''(In topological language, this exercise shows that the "Zariski closure" of a set may be quite different from its ordinary closure. This is because the Zariski topology is extremely "coarse.")''

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

==Questions:==
This came up because of exercise 1.4.7 but is just a general question: Are we allowed to assume we're working in the smallest affine space that will fit around the polynomials concerned i.e: does the answer boil down to the same thing whether you do your computations in k[x,y] or k[x,y,z,..,w]?
:That is a very good question! For purposes of the class, yes, unless the problem explicitly states otherwise. So for problem 1.4.7 the varieties are subsets of $\mathsf{k}^2$. But: this particular problem makes sense, and its result is true, in $\mathsf{k}^n$ for any $n\geq2$.

:It is moderately enlightening to think about the problem for larger $n$. You will find the following, in general: suppose $f_1,\dots,f_s\in\mathsf{k}[x_1,\dots,x_n]$ and let $V=\mathbb{V}(f_1,\dots,f_s)$ (a subset of $\mathsf{k}^n$). But we could also think of $f_1,\dots,f_s$ as polynomials in $x_1,\dots,x_n,x_{n+1},\dots,x_{n+m}$ and then use them to define $W=\mathbb{V}(f_1,\dots,f_s)$ (a subset of $\mathsf{k}^{n+m}$). Then $W=V\times\mathsf{k}^m$. One says that $W$ is a ''cylinder over $V$''. (If you think about some cases where $n=2$ and $m=1$ you will understand why.)

:Cylinders are usually regarded as uninteresting; most geometric properties of $W$ are already carried by $V$, because most algebraic algorithms, carried out to answer geometric questions about $W$, will only involve $x_1,\dots,x_n$ and so they will automatically answer the same question about $V$.

::-[[User:Steven.Jackson|Steven.Jackson]] ([[User talk:Steven.Jackson|talk]]) 19:16, 12 February 2018 (EST)

We know every ideal of a variety is a radical ideal. Is the reverse true? That is, is every radical ideal the ideal of some variety?

==Solutions:==
'''1.4.1:'''

(a) From second equation, $y=1/x$. Plugging into the first, we have $x^2+\frac{1}{x^2}-1=0$ which we can write as $x^4-x^2+1=0$.

(b) Multiplying the second equation by $xy+1$, it becomes $x^2y^2-1=0$. If we multiply the first equation by $x^2$, we get $x^4+x^2y^2-x^2=0$ and we can now substitute $x^2y^2=1$ to obtain $x^4-x^2+1=0$, just as in part (a). Explicitly,
\[
x^4-x^2+1=x^2(x^2+y^2-1)-(xy+1)(xy-1)\;,
\]
where the coefficient polynomials are $x^2$ and $-(xy+1)$.

'''1.4.3:'''

In each case, to show equality we show containment in both directions.

(a) To show $\langle x+y, x-y\rangle\subseteq\langle x,y\rangle$, we need to show that each polynomial defining the ideal is an element of the other ideal. That is, we need $x+y\in\langle x,y\rangle$ and $x-y\in\langle x,y\rangle$. This is obvious: Since $x$ and $y$ are in $\langle x,y\rangle$, so is $x+y$ and $x-y$ because ideals are closed under consequences. Next, to show $\langle x+y, x-y\rangle\supseteq\langle x,y\rangle$, we need to show $x,y\in\langle x+y,x-y\rangle$. By inspection, we see
\[
x=\frac{1}{2}(x+y)+\frac{1}{2}(x-y)\quad \text{and}\quad y=\frac{1}{2}(x+y)-\frac{1}{2}(x-y)\;,
\]
so we are done.

(b) We trivially see $\langle x+xy,y+xy,x^2,y^2\rangle\subseteq\langle x,y\rangle$, the constituent polynomials of the left hand side are obvious consequences of the polynomials comprising the right. We will show $\langle x+xy,y+xy,x^2,y^2\rangle\supseteq\langle x,y\rangle$. After some algebraic annoyances we see

\[
x=(1-y)(x+xy)+y(y+xy)-y^2 \quad \text{and} \quad y=x(x+xy)+(1-x)(y+xy)-x^2\;.
\]

(c) $\langle 2x^2+3y^2-11,x^2-y^2-3\rangle\subseteq\langle x^2-4,y^2-1\rangle$:
\[
2x^2+3y^2-11=2(x^2-4)+3(y^2-1)\quad\text{and}\quad x^2-y^2-3=(x^2-4)-(y^2-1)\;.
\]

$\langle 2x^2+3y^2-11,x^2-y^2-3\rangle\supseteq\langle x^2-4,y^2-1\rangle$:
\[
x^2-4=\frac{1}{5}(2x^2+3y^2-11)+\frac{3}{5}(x^2-y^2-3)\quad \text{and}\quad y^2-1=\frac{1}{5}(2x^2+3y^2-11)-\frac{2}{5}(x^2-y^2-3)\;.
\]

'''1.4.5:''' As shown in class, the affine variety defined by a set of polynomials is equal to the affine variety defined by the ideal generated by its constituent polynomials. From problem 3b, we know the ideals are equal and so the varieties are also equal.

'''1.4.6a:''' If any vector space basis is infinite, ''every'' vector space basis is infinite. $I=\langle x\rangle\subseteq k[x]$ is the ideal generated by the polynomial $x$. The general form of any polynomial in this set is
\[
x(a_0+a_1x+a_2x^2+\cdots+a_nx^n)=a_0x+a_1x^2+\cdots+a_nx^{n+1}\;,
\]
where $n$ can be any finite number. We need a linearly independent spanning set for the ideal. The obvious choice is $\{x,x^2,x^3,\ldots\}$. This infinite set obviously spans the ideal. To show linear independence, we need to convince ourselves that $a_0x+a_1x^2+a_2x^3\cdots=0$ iff $a_i=0$ for all $i$. The forward direction is obvious. For the reverse, note that $x$ may be factored out of the left hand side. $x$ is a placeholder for an arbitrary field element and so may be divided out on both sides. This leaves $a_0+a_1x+a_2x^2+\cdots=0$. The constant term in this resulting polynomial must equal zero or the whole thing cannot equal zero. We are then left with the original situation with $a_i\to a_{i+1}$, so the argument repeats. Thus we have an infinite basis for the vectorspace $I$, proving that all bases for this set are infinite, even though the basis for the ideal is finite.

'''1.4.6b:''' $yx-xy=0$.

'''1.4.7:''' We are interested in the ideal $\mathbb{I}(\mathbb{V}(x^n,y^m))$ where $n$ and $m$ are arbitrary positive integers. Unpacking the definitions, $\mathbb{V}(x^n,y^m)$ is the set of points where the polynomials $x^n$ and $y^m$ vanish. This set is the singular point $\{0,0\}$. $\mathbb{I}(\{0,0\})$ is the set of all polynomials in k[x,y] which vanish at $\{0,0\}$. If the coefficient of the degree zero term vanishes, this is assured. A general term of the ideal may then be written
\[
a_{10}x+a_{01}y+a_{20}x^2+a_{11}xy+a_{02}y^2+a_{30}x^3+a_{21}x^2y+a_{12}xy^2+a_{03}y^3+\cdots
\]
We compare this with a general element of the ideal $\langle x,y\rangle=g_1x+g_2y$ where $g_1,g_2\in k[x,y]$. Let $g_1=b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\cdots+b_{ij}x^iy^j$ and $g_2=c_{00}+c_{10}x+c_{01}y+c_{20}x^2+c_{11}xy+c_{02}y^2+\cdots+c_{kl}x^ky^l$. Then
\[
\langle x,y\rangle = (b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\cdots+b_{ij}x^iy^j)x+(c_{00}+c_{10}x+c_{01}y+c_{20}x^2+c_{11}xy+c_{02}y^2+\cdots+c_{kl}x^ky^l)y\;.
\]
We can quite easily see that there is no constant term in the simplified result. It is clear that every other term is present, which could be demonstrated explicitly by associating to each $a_{ij}$ a sum of some $b_{kl}$ and $c_{mn}$. Thus we have the desired equality by explicit computation. (I didn't use the containment argument, but I hope this still demonstrates the fact.)

'''1.4.8:''' (a) We prove both directions. Suppose $f\in\mathbb{I}(\mathbb{V})$, then $f^m\in\mathbb{I}(\mathbb{V})$ because ideals are closed under consequences, that is, we can just multiply $f$ by the polynomial $f^{m-1}$ and it has to be in the ideal. Now suppose $f^m\in\mathbb{I}(\mathbb{V})$. $\mathbb{I}(\mathbb{V})$ is the set of polynomials which vanish on the variety $\mathbb{V}$. So $f^m=0$ at all points on $\mathbb{V}$. By simple algebra, it must be that $f=0$ also at these points, thus $f\in\mathbb{I}(\mathbb{V})$.

(b) By part (a) it suffices to show that $x\notin\langle x^2,y^2\rangle$. If it were, it must be that $x=g_1x^2+g_2y^2$ for $g_1,g_2\in k[x,y]$. This is clearly impossible. Thus $\langle x^2,y^2\rangle$ is not radical and is consequently not the ideal of any variety.

'''2:''' The set $\mathbb{I}(T)$ is the set of all polynomials which vanish on the set of points $T\subseteq k^n$. We need to check whether all three axioms of an ideal hold for this set. The zero polynomial is clearly present because it vanishes everywhere, therefore it also vanishes on $T$. If $f_1,f_2\in \mathbb{I}(T)$ then they each also vanish on $T$. In accordance with how we add polynomials and the functions they generate, it must be that for any $a\in T$, $(f_1+f_2)(a)=0$ as well. Therefore $f_1+f_2\in\mathbb{I}(T)$. Finally if $g$ is an arbitrary polynomial and $f_1\in\mathbb{I}(T)$ then $gf_1(a)=g(a)f_1(a)=g(a)0=0$ which means $gf_1\in\mathbb{I}(T)$. Thus $\mathbb{I}(T)$ is an ideal of $k[x_1,\ldots,x_n]$.

'''3:'''

:Endorsing the solutions to 1.4.1-1.4.6 above. All correct, and beautifully written.

:$\quad$-[[User:Steven.Jackson|Steven.Jackson]] ([[User talk:Steven.Jackson|talk]]) 10:08, 13 February 2018 (EST)

Math 380, Spring 2018, Assignment 3

2018-02-13T17:36:05Z

John.DeBrota: /* Questions: */

__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''

==Read:==

# Section 1.4.

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

# Ideal (in $\mathsf{k}[x_1,\dots,x_n]$).
# $\left\langle S\right\rangle$ (the ideal generated by a set $S$ of polynomials).
# $\left\langle f_1,\dots,f_s\right\rangle$ (the ideal generated by the finite set $\{f_1,\dots,f_s\}$).
# $\mathbb{I}(T)$ (the ''ideal'' of the set of points $T\subseteq\mathsf{k}^n$).

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

# Theorem relating $\mathbb{V}(S)$ to $\mathbb{V}(\left\langle S\right\rangle)$.
# Theorem characterizing when $\left\langle f_1,\dots,f_s\right\rangle\subseteq\left\langle g_1,\dots,g_t\right\rangle$.
# Theorem characterizing when $\left\langle f_1,\dots,f_s\right\rangle=\left\langle g_1,\dots,g_t\right\rangle$.
# Theorem concerning the ''inclusion-reversing'' and ''inflationary'' character of the fundamental pairing $(\mathbb{V},\mathbb{I})$.

==Solve the following problems:==

# Section 4, problems 1, 3, 5, 6(a), 6(b), 7, and 8. ''(In problem 6, the word "basis" means "generating set.")''
# Prove that, for any set $T\subseteq\mathsf{k}^n$, the set $\mathbb{I}(T)$ is always an ''ideal'' of $\mathsf{k}[x_1,\dots,x_n]$.
# Working in $\mathbb{R}^1$, describe $\mathbb{V}(\mathbb{I}(\mathbb{Z}))$. ''(In topological language, this exercise shows that the "Zariski closure" of a set may be quite different from its ordinary closure. This is because the Zariski topology is extremely "coarse.")''

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

==Questions:==
This came up because of exercise 1.4.7 but is just a general question: Are we allowed to assume we're working in the smallest affine space that will fit around the polynomials concerned i.e: does the answer boil down to the same thing whether you do your computations in k[x,y] or k[x,y,z,..,w]?
:That is a very good question! For purposes of the class, yes, unless the problem explicitly states otherwise. So for problem 1.4.7 the varieties are subsets of $\mathsf{k}^2$. But: this particular problem makes sense, and its result is true, in $\mathsf{k}^n$ for any $n\geq2$.

:It is moderately enlightening to think about the problem for larger $n$. You will find the following, in general: suppose $f_1,\dots,f_s\in\mathsf{k}[x_1,\dots,x_n]$ and let $V=\mathbb{V}(f_1,\dots,f_s)$ (a subset of $\mathsf{k}^n$). But we could also think of $f_1,\dots,f_s$ as polynomials in $x_1,\dots,x_n,x_{n+1},\dots,x_{n+m}$ and then use them to define $W=\mathbb{V}(f_1,\dots,f_s)$ (a subset of $\mathsf{k}^{n+m}$). Then $W=V\times\mathsf{k}^m$. One says that $W$ is a ''cylinder over $V$''. (If you think about some cases where $n=2$ and $m=1$ you will understand why.)

:Cylinders are usually regarded as uninteresting; most geometric properties of $W$ are already carried by $V$, because most algebraic algorithms, carried out to answer geometric questions about $W$, will only involve $x_1,\dots,x_n$ and so they will automatically answer the same question about $V$.

::-[[User:Steven.Jackson|Steven.Jackson]] ([[User talk:Steven.Jackson|talk]]) 19:16, 12 February 2018 (EST)

We know every ideal of a variety is a radical ideal. Is the reverse true? That is, is every radical ideal the ideal of some variety?

==Solutions:==
'''1.4.1:'''

(a) From second equation, $y=1/x$. Plugging into the first, we have $x^2+\frac{1}{x^2}-1=0$ which we can write as $x^4-x^2+1=0$.

(b) Multiplying the second equation by $xy+1$, it becomes $x^2y^2-1=0$. If we multiply the first equation by $x^2$, we get $x^4+x^2y^2-x^2=0$ and we can now substitute $x^2y^2=1$ to obtain $x^4-x^2+1=0$, just as in part (a). Explicitly,
\[
x^4-x^2+1=x^2(x^2+y^2-1)-(xy+1)(xy-1)\;,
\]
where the coefficient polynomials are $x^2$ and $-(xy+1)$.

'''1.4.3:'''

In each case, to show equality we show containment in both directions.

(a) To show $\langle x+y, x-y\rangle\subseteq\langle x,y\rangle$, we need to show that each polynomial defining the ideal is an element of the other ideal. That is, we need $x+y\in\langle x,y\rangle$ and $x-y\in\langle x,y\rangle$. This is obvious: Since $x$ and $y$ are in $\langle x,y\rangle$, so is $x+y$ and $x-y$ because ideals are closed under consequences. Next, to show $\langle x+y, x-y\rangle\supseteq\langle x,y\rangle$, we need to show $x,y\in\langle x+y,x-y\rangle$. By inspection, we see
\[
x=\frac{1}{2}(x+y)+\frac{1}{2}(x-y)\quad \text{and}\quad y=\frac{1}{2}(x+y)-\frac{1}{2}(x-y)\;,
\]
so we are done.

(b) We trivially see $\langle x+xy,y+xy,x^2,y^2\rangle\subseteq\langle x,y\rangle$, the constituent polynomials of the left hand side are obvious consequences of the polynomials comprising the right. We will show $\langle x+xy,y+xy,x^2,y^2\rangle\supseteq\langle x,y\rangle$. After some algebraic annoyances we see

\[
x=(1-y)(x+xy)+y(y+xy)-y^2 \quad \text{and} \quad y=x(x+xy)+(1-x)(y+xy)-x^2\;.
\]

(c) $\langle 2x^2+3y^2-11,x^2-y^2-3\rangle\subseteq\langle x^2-4,y^2-1\rangle$:
\[
2x^2+3y^2-11=2(x^2-4)+3(y^2-1)\quad\text{and}\quad x^2-y^2-3=(x^2-4)-(y^2-1)\;.
\]

$\langle 2x^2+3y^2-11,x^2-y^2-3\rangle\supseteq\langle x^2-4,y^2-1\rangle$:
\[
x^2-4=\frac{1}{5}(2x^2+3y^2-11)+\frac{3}{5}(x^2-y^2-3)\quad \text{and}\quad y^2-1=\frac{1}{5}(2x^2+3y^2-11)-\frac{2}{5}(x^2-y^2-3)\;.
\]

'''1.4.5:''' As shown in class, the affine variety defined by a set of polynomials is equal to the affine variety defined by the ideal generated by its constituent polynomials. From problem 3b, we know the ideals are equal and so the varieties are also equal.

'''1.4.6a:''' If any vector space basis is infinite, ''every'' vector space basis is infinite. $I=\langle x\rangle\subseteq k[x]$ is the ideal generated by the polynomial $x$. The general form of any polynomial in this set is
\[
x(a_0+a_1x+a_2x^2+\cdots+a_nx^n)=a_0x+a_1x^2+\cdots+a_nx^{n+1}\;,
\]
where $n$ can be any finite number. We need a linearly independent spanning set for the ideal. The obvious choice is $\{x,x^2,x^3,\ldots\}$. This infinite set obviously spans the ideal. To show linear independence, we need to convince ourselves that $a_0x+a_1x^2+a_2x^3\cdots=0$ iff $a_i=0$ for all $i$. The forward direction is obvious. For the reverse, note that $x$ may be factored out of the left hand side. $x$ is a placeholder for an arbitrary field element and so may be divided out on both sides. This leaves $a_0+a_1x+a_2x^2+\cdots=0$. The constant term in this resulting polynomial must equal zero or the whole thing cannot equal zero. We are then left with the original situation with $a_i\to a_{i+1}$, so the argument repeats. Thus we have an infinite basis for the vectorspace $I$, proving that all bases for this set are infinite, even though the basis for the ideal is finite.

'''1.4.6b:''' $yx-xy=0$.

'''1.4.7:''' We are interested in the ideal $\mathbb{I}(\mathbb{V}(x^n,y^m))$ where $n$ and $m$ are arbitrary positive integers. Unpacking the definitions, $\mathbb{V}(x^n,y^m)$ is the set of points where the polynomials $x^n$ and $y^m$ vanish. This set is the singular point $\{0,0\}$. $\mathbb{I}(\{0,0\})$ is the set of all polynomials in k[x,y] which vanish at $\{0,0\}$. If the coefficient of the degree zero term vanishes, this is assured. A general term of the ideal may then be written
\[
a_{10}x+a_{01}y+a_{20}x^2+a_{11}xy+a_{02}y^2+a_{30}x^3+a_{21}x^2y+a_{12}xy^2+a_{03}y^3+\cdots
\]
We compare this with a general element of the ideal $\langle x,y\rangle=g_1x+g_2y$ where $g_1,g_2\in k[x,y]$. Let $g_1=b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\cdots+b_{ij}x^iy^j$ and $g_2=c_{00}+c_{10}x+c_{01}y+c_{20}x^2+c_{11}xy+c_{02}y^2+\cdots+c_{kl}x^ky^l$. Then
\[
\langle x,y\rangle = (b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\cdots+b_{ij}x^iy^j)x+(c_{00}+c_{10}x+c_{01}y+c_{20}x^2+c_{11}xy+c_{02}y^2+\cdots+c_{kl}x^ky^l)y\;.
\]
We can quite easily see that there is no constant term in the simplified result. It is clear that every other term is present, which could be demonstrated explicitly by associating to each $a_{ij}$ a sum of some $b_{kl}$ and $c_{mn}$. Thus we have the desired equality by explicit computation. (I didn't use the containment argument, but I hope this still demonstrates the fact.)

'''1.4.8:''' (a) We prove both directions. Suppose $f\in\mathbb{I}(\mathbb{V})$, then $f^m\in\mathbb{I}(\mathbb{V})$ because ideals are closed under consequences, that is, we can just multiply $f$ by the polynomial $f^{m-1}$ and it has to be in the ideal. Now suppose $f^m\in\mathbb{I}(\mathbb{V})$. $\mathbb{I}(\mathbb{V})$ is the set of polynomials which vanish on the variety $\mathbb{V}$. So $f^m=0$ at all points on $\mathbb{V}$. By simple algebra, it must be that $f=0$ also at these points, thus $f\in\mathbb{I}(\mathbb{V})$.

(b) By part (a) it suffices to show that $x\notin\langle x^2,y^2\rangle$. If it were, it must be that $x=g_1x^2+g_2y^2$ for $g_1,g_2\in k[x,y]$. This is clearly impossible. Thus $\langle x^2,y^2\rangle$ is not radical and is consequently not the ideal of any variety.

:Endorsing the solutions to 1.4.1-1.4.6 above. All correct, and beautifully written.

:$\quad$-[[User:Steven.Jackson|Steven.Jackson]] ([[User talk:Steven.Jackson|talk]]) 10:08, 13 February 2018 (EST)

Math 380, Spring 2018, Assignment 3

2018-02-13T17:33:59Z

John.DeBrota: /* 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''

==Read:==

# Section 1.4.

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

# Ideal (in $\mathsf{k}[x_1,\dots,x_n]$).
# $\left\langle S\right\rangle$ (the ideal generated by a set $S$ of polynomials).
# $\left\langle f_1,\dots,f_s\right\rangle$ (the ideal generated by the finite set $\{f_1,\dots,f_s\}$).
# $\mathbb{I}(T)$ (the ''ideal'' of the set of points $T\subseteq\mathsf{k}^n$).

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

# Theorem relating $\mathbb{V}(S)$ to $\mathbb{V}(\left\langle S\right\rangle)$.
# Theorem characterizing when $\left\langle f_1,\dots,f_s\right\rangle\subseteq\left\langle g_1,\dots,g_t\right\rangle$.
# Theorem characterizing when $\left\langle f_1,\dots,f_s\right\rangle=\left\langle g_1,\dots,g_t\right\rangle$.
# Theorem concerning the ''inclusion-reversing'' and ''inflationary'' character of the fundamental pairing $(\mathbb{V},\mathbb{I})$.

==Solve the following problems:==

# Section 4, problems 1, 3, 5, 6(a), 6(b), 7, and 8. ''(In problem 6, the word "basis" means "generating set.")''
# Prove that, for any set $T\subseteq\mathsf{k}^n$, the set $\mathbb{I}(T)$ is always an ''ideal'' of $\mathsf{k}[x_1,\dots,x_n]$.
# Working in $\mathbb{R}^1$, describe $\mathbb{V}(\mathbb{I}(\mathbb{Z}))$. ''(In topological language, this exercise shows that the "Zariski closure" of a set may be quite different from its ordinary closure. This is because the Zariski topology is extremely "coarse.")''

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

==Questions:==
This came up because of exercise 1.4.7 but is just a general question: Are we allowed to assume we're working in the smallest affine space that will fit around the polynomials concerned i.e: does the answer boil down to the same thing whether you do your computations in k[x,y] or k[x,y,z,..,w]?
:That is a very good question! For purposes of the class, yes, unless the problem explicitly states otherwise. So for problem 1.4.7 the varieties are subsets of $\mathsf{k}^2$. But: this particular problem makes sense, and its result is true, in $\mathsf{k}^n$ for any $n\geq2$.

:It is moderately enlightening to think about the problem for larger $n$. You will find the following, in general: suppose $f_1,\dots,f_s\in\mathsf{k}[x_1,\dots,x_n]$ and let $V=\mathbb{V}(f_1,\dots,f_s)$ (a subset of $\mathsf{k}^n$). But we could also think of $f_1,\dots,f_s$ as polynomials in $x_1,\dots,x_n,x_{n+1},\dots,x_{n+m}$ and then use them to define $W=\mathbb{V}(f_1,\dots,f_s)$ (a subset of $\mathsf{k}^{n+m}$). Then $W=V\times\mathsf{k}^m$. One says that $W$ is a ''cylinder over $V$''. (If you think about some cases where $n=2$ and $m=1$ you will understand why.)

:Cylinders are usually regarded as uninteresting; most geometric properties of $W$ are already carried by $V$, because most algebraic algorithms, carried out to answer geometric questions about $W$, will only involve $x_1,\dots,x_n$ and so they will automatically answer the same question about $V$.

::-[[User:Steven.Jackson|Steven.Jackson]] ([[User talk:Steven.Jackson|talk]]) 19:16, 12 February 2018 (EST)

==Solutions:==
'''1.4.1:'''

(a) From second equation, $y=1/x$. Plugging into the first, we have $x^2+\frac{1}{x^2}-1=0$ which we can write as $x^4-x^2+1=0$.

(b) Multiplying the second equation by $xy+1$, it becomes $x^2y^2-1=0$. If we multiply the first equation by $x^2$, we get $x^4+x^2y^2-x^2=0$ and we can now substitute $x^2y^2=1$ to obtain $x^4-x^2+1=0$, just as in part (a). Explicitly,
\[
x^4-x^2+1=x^2(x^2+y^2-1)-(xy+1)(xy-1)\;,
\]
where the coefficient polynomials are $x^2$ and $-(xy+1)$.

'''1.4.3:'''

In each case, to show equality we show containment in both directions.

(a) To show $\langle x+y, x-y\rangle\subseteq\langle x,y\rangle$, we need to show that each polynomial defining the ideal is an element of the other ideal. That is, we need $x+y\in\langle x,y\rangle$ and $x-y\in\langle x,y\rangle$. This is obvious: Since $x$ and $y$ are in $\langle x,y\rangle$, so is $x+y$ and $x-y$ because ideals are closed under consequences. Next, to show $\langle x+y, x-y\rangle\supseteq\langle x,y\rangle$, we need to show $x,y\in\langle x+y,x-y\rangle$. By inspection, we see
\[
x=\frac{1}{2}(x+y)+\frac{1}{2}(x-y)\quad \text{and}\quad y=\frac{1}{2}(x+y)-\frac{1}{2}(x-y)\;,
\]
so we are done.

(b) We trivially see $\langle x+xy,y+xy,x^2,y^2\rangle\subseteq\langle x,y\rangle$, the constituent polynomials of the left hand side are obvious consequences of the polynomials comprising the right. We will show $\langle x+xy,y+xy,x^2,y^2\rangle\supseteq\langle x,y\rangle$. After some algebraic annoyances we see

\[
x=(1-y)(x+xy)+y(y+xy)-y^2 \quad \text{and} \quad y=x(x+xy)+(1-x)(y+xy)-x^2\;.
\]

(c) $\langle 2x^2+3y^2-11,x^2-y^2-3\rangle\subseteq\langle x^2-4,y^2-1\rangle$:
\[
2x^2+3y^2-11=2(x^2-4)+3(y^2-1)\quad\text{and}\quad x^2-y^2-3=(x^2-4)-(y^2-1)\;.
\]

$\langle 2x^2+3y^2-11,x^2-y^2-3\rangle\supseteq\langle x^2-4,y^2-1\rangle$:
\[
x^2-4=\frac{1}{5}(2x^2+3y^2-11)+\frac{3}{5}(x^2-y^2-3)\quad \text{and}\quad y^2-1=\frac{1}{5}(2x^2+3y^2-11)-\frac{2}{5}(x^2-y^2-3)\;.
\]

'''1.4.5:''' As shown in class, the affine variety defined by a set of polynomials is equal to the affine variety defined by the ideal generated by its constituent polynomials. From problem 3b, we know the ideals are equal and so the varieties are also equal.

'''1.4.6a:''' If any vector space basis is infinite, ''every'' vector space basis is infinite. $I=\langle x\rangle\subseteq k[x]$ is the ideal generated by the polynomial $x$. The general form of any polynomial in this set is
\[
x(a_0+a_1x+a_2x^2+\cdots+a_nx^n)=a_0x+a_1x^2+\cdots+a_nx^{n+1}\;,
\]
where $n$ can be any finite number. We need a linearly independent spanning set for the ideal. The obvious choice is $\{x,x^2,x^3,\ldots\}$. This infinite set obviously spans the ideal. To show linear independence, we need to convince ourselves that $a_0x+a_1x^2+a_2x^3\cdots=0$ iff $a_i=0$ for all $i$. The forward direction is obvious. For the reverse, note that $x$ may be factored out of the left hand side. $x$ is a placeholder for an arbitrary field element and so may be divided out on both sides. This leaves $a_0+a_1x+a_2x^2+\cdots=0$. The constant term in this resulting polynomial must equal zero or the whole thing cannot equal zero. We are then left with the original situation with $a_i\to a_{i+1}$, so the argument repeats. Thus we have an infinite basis for the vectorspace $I$, proving that all bases for this set are infinite, even though the basis for the ideal is finite.

'''1.4.6b:''' $yx-xy=0$.

'''1.4.7:''' We are interested in the ideal $\mathbb{I}(\mathbb{V}(x^n,y^m))$ where $n$ and $m$ are arbitrary positive integers. Unpacking the definitions, $\mathbb{V}(x^n,y^m)$ is the set of points where the polynomials $x^n$ and $y^m$ vanish. This set is the singular point $\{0,0\}$. $\mathbb{I}(\{0,0\})$ is the set of all polynomials in k[x,y] which vanish at $\{0,0\}$. If the coefficient of the degree zero term vanishes, this is assured. A general term of the ideal may then be written
\[
a_{10}x+a_{01}y+a_{20}x^2+a_{11}xy+a_{02}y^2+a_{30}x^3+a_{21}x^2y+a_{12}xy^2+a_{03}y^3+\cdots
\]
We compare this with a general element of the ideal $\langle x,y\rangle=g_1x+g_2y$ where $g_1,g_2\in k[x,y]$. Let $g_1=b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\cdots+b_{ij}x^iy^j$ and $g_2=c_{00}+c_{10}x+c_{01}y+c_{20}x^2+c_{11}xy+c_{02}y^2+\cdots+c_{kl}x^ky^l$. Then
\[
\langle x,y\rangle = (b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\cdots+b_{ij}x^iy^j)x+(c_{00}+c_{10}x+c_{01}y+c_{20}x^2+c_{11}xy+c_{02}y^2+\cdots+c_{kl}x^ky^l)y\;.
\]
We can quite easily see that there is no constant term in the simplified result. It is clear that every other term is present, which could be demonstrated explicitly by associating to each $a_{ij}$ a sum of some $b_{kl}$ and $c_{mn}$. Thus we have the desired equality by explicit computation. (I didn't use the containment argument, but I hope this still demonstrates the fact.)

'''1.4.8:''' (a) We prove both directions. Suppose $f\in\mathbb{I}(\mathbb{V})$, then $f^m\in\mathbb{I}(\mathbb{V})$ because ideals are closed under consequences, that is, we can just multiply $f$ by the polynomial $f^{m-1}$ and it has to be in the ideal. Now suppose $f^m\in\mathbb{I}(\mathbb{V})$. $\mathbb{I}(\mathbb{V})$ is the set of polynomials which vanish on the variety $\mathbb{V}$. So $f^m=0$ at all points on $\mathbb{V}$. By simple algebra, it must be that $f=0$ also at these points, thus $f\in\mathbb{I}(\mathbb{V})$.

(b) By part (a) it suffices to show that $x\notin\langle x^2,y^2\rangle$. If it were, it must be that $x=g_1x^2+g_2y^2$ for $g_1,g_2\in k[x,y]$. This is clearly impossible. Thus $\langle x^2,y^2\rangle$ is not radical and is consequently not the ideal of any variety.

:Endorsing the solutions to 1.4.1-1.4.6 above. All correct, and beautifully written.

:$\quad$-[[User:Steven.Jackson|Steven.Jackson]] ([[User talk:Steven.Jackson|talk]]) 10:08, 13 February 2018 (EST)

Math 380, Spring 2018, Assignment 3

2018-02-13T17:15:44Z

John.DeBrota: /* 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''

==Read:==

# Section 1.4.

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

# Ideal (in $\mathsf{k}[x_1,\dots,x_n]$).
# $\left\langle S\right\rangle$ (the ideal generated by a set $S$ of polynomials).
# $\left\langle f_1,\dots,f_s\right\rangle$ (the ideal generated by the finite set $\{f_1,\dots,f_s\}$).
# $\mathbb{I}(T)$ (the ''ideal'' of the set of points $T\subseteq\mathsf{k}^n$).

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

# Theorem relating $\mathbb{V}(S)$ to $\mathbb{V}(\left\langle S\right\rangle)$.
# Theorem characterizing when $\left\langle f_1,\dots,f_s\right\rangle\subseteq\left\langle g_1,\dots,g_t\right\rangle$.
# Theorem characterizing when $\left\langle f_1,\dots,f_s\right\rangle=\left\langle g_1,\dots,g_t\right\rangle$.
# Theorem concerning the ''inclusion-reversing'' and ''inflationary'' character of the fundamental pairing $(\mathbb{V},\mathbb{I})$.

==Solve the following problems:==

# Section 4, problems 1, 3, 5, 6(a), 6(b), 7, and 8. ''(In problem 6, the word "basis" means "generating set.")''
# Prove that, for any set $T\subseteq\mathsf{k}^n$, the set $\mathbb{I}(T)$ is always an ''ideal'' of $\mathsf{k}[x_1,\dots,x_n]$.
# Working in $\mathbb{R}^1$, describe $\mathbb{V}(\mathbb{I}(\mathbb{Z}))$. ''(In topological language, this exercise shows that the "Zariski closure" of a set may be quite different from its ordinary closure. This is because the Zariski topology is extremely "coarse.")''

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

==Questions:==
This came up because of exercise 1.4.7 but is just a general question: Are we allowed to assume we're working in the smallest affine space that will fit around the polynomials concerned i.e: does the answer boil down to the same thing whether you do your computations in k[x,y] or k[x,y,z,..,w]?
:That is a very good question! For purposes of the class, yes, unless the problem explicitly states otherwise. So for problem 1.4.7 the varieties are subsets of $\mathsf{k}^2$. But: this particular problem makes sense, and its result is true, in $\mathsf{k}^n$ for any $n\geq2$.

:It is moderately enlightening to think about the problem for larger $n$. You will find the following, in general: suppose $f_1,\dots,f_s\in\mathsf{k}[x_1,\dots,x_n]$ and let $V=\mathbb{V}(f_1,\dots,f_s)$ (a subset of $\mathsf{k}^n$). But we could also think of $f_1,\dots,f_s$ as polynomials in $x_1,\dots,x_n,x_{n+1},\dots,x_{n+m}$ and then use them to define $W=\mathbb{V}(f_1,\dots,f_s)$ (a subset of $\mathsf{k}^{n+m}$). Then $W=V\times\mathsf{k}^m$. One says that $W$ is a ''cylinder over $V$''. (If you think about some cases where $n=2$ and $m=1$ you will understand why.)

:Cylinders are usually regarded as uninteresting; most geometric properties of $W$ are already carried by $V$, because most algebraic algorithms, carried out to answer geometric questions about $W$, will only involve $x_1,\dots,x_n$ and so they will automatically answer the same question about $V$.

::-[[User:Steven.Jackson|Steven.Jackson]] ([[User talk:Steven.Jackson|talk]]) 19:16, 12 February 2018 (EST)

==Solutions:==
'''1.4.1:'''

(a) From second equation, $y=1/x$. Plugging into the first, we have $x^2+\frac{1}{x^2}-1=0$ which we can write as $x^4-x^2+1=0$.

(b) Multiplying the second equation by $xy+1$, it becomes $x^2y^2-1=0$. If we multiply the first equation by $x^2$, we get $x^4+x^2y^2-x^2=0$ and we can now substitute $x^2y^2=1$ to obtain $x^4-x^2+1=0$, just as in part (a). Explicitly,
\[
x^4-x^2+1=x^2(x^2+y^2-1)-(xy+1)(xy-1)\;,
\]
where the coefficient polynomials are $x^2$ and $-(xy+1)$.

'''1.4.3:'''

In each case, to show equality we show containment in both directions.

(a) To show $\langle x+y, x-y\rangle\subseteq\langle x,y\rangle$, we need to show that each polynomial defining the ideal is an element of the other ideal. That is, we need $x+y\in\langle x,y\rangle$ and $x-y\in\langle x,y\rangle$. This is obvious: Since $x$ and $y$ are in $\langle x,y\rangle$, so is $x+y$ and $x-y$ because ideals are closed under consequences. Next, to show $\langle x+y, x-y\rangle\supseteq\langle x,y\rangle$, we need to show $x,y\in\langle x+y,x-y\rangle$. By inspection, we see
\[
x=\frac{1}{2}(x+y)+\frac{1}{2}(x-y)\quad \text{and}\quad y=\frac{1}{2}(x+y)-\frac{1}{2}(x-y)\;,
\]
so we are done.

(b) We trivially see $\langle x+xy,y+xy,x^2,y^2\rangle\subseteq\langle x,y\rangle$, the constituent polynomials of the left hand side are obvious consequences of the polynomials comprising the right. We will show $\langle x+xy,y+xy,x^2,y^2\rangle\supseteq\langle x,y\rangle$. After some algebraic annoyances we see

\[
x=(1-y)(x+xy)+y(y+xy)-y^2 \quad \text{and} \quad y=x(x+xy)+(1-x)(y+xy)-x^2\;.
\]

(c) $\langle 2x^2+3y^2-11,x^2-y^2-3\rangle\subseteq\langle x^2-4,y^2-1\rangle$:
\[
2x^2+3y^2-11=2(x^2-4)+3(y^2-1)\quad\text{and}\quad x^2-y^2-3=(x^2-4)-(y^2-1)\;.
\]

$\langle 2x^2+3y^2-11,x^2-y^2-3\rangle\supseteq\langle x^2-4,y^2-1\rangle$:
\[
x^2-4=\frac{1}{5}(2x^2+3y^2-11)+\frac{3}{5}(x^2-y^2-3)\quad \text{and}\quad y^2-1=\frac{1}{5}(2x^2+3y^2-11)-\frac{2}{5}(x^2-y^2-3)\;.
\]

'''1.4.5:''' As shown in class, the affine variety defined by a set of polynomials is equal to the affine variety defined by the ideal generated by its constituent polynomials. From problem 3b, we know the ideals are equal and so the varieties are also equal.

'''1.4.6a:''' If any vector space basis is infinite, ''every'' vector space basis is infinite. $I=\langle x\rangle\subseteq k[x]$ is the ideal generated by the polynomial $x$. The general form of any polynomial in this set is
\[
x(a_0+a_1x+a_2x^2+\cdots+a_nx^n)=a_0x+a_1x^2+\cdots+a_nx^{n+1}\;,
\]
where $n$ can be any finite number. We need a linearly independent spanning set for the ideal. The obvious choice is $\{x,x^2,x^3,\ldots\}$. This infinite set obviously spans the ideal. To show linear independence, we need to convince ourselves that $a_0x+a_1x^2+a_2x^3\cdots=0$ iff $a_i=0$ for all $i$. The forward direction is obvious. For the reverse, note that $x$ may be factored out of the left hand side. $x$ is a placeholder for an arbitrary field element and so may be divided out on both sides. This leaves $a_0+a_1x+a_2x^2+\cdots=0$. The constant term in this resulting polynomial must equal zero or the whole thing cannot equal zero. We are then left with the original situation with $a_i\to a_{i+1}$, so the argument repeats. Thus we have an infinite basis for the vectorspace $I$, proving that all bases for this set are infinite, even though the basis for the ideal is finite.

'''1.4.6b:''' $yx-xy=0$.

'''1.4.7:''' We are interested in the ideal $\mathbb{I}(\mathbb{V}(x^n,y^m))$ where $n$ and $m$ are arbitrary positive integers. Unpacking the definitions, $\mathbb{V}(x^n,y^m)$ is the set of points where the polynomials $x^n$ and $y^m$ vanish. This set is the singular point $\{0,0\}$. $\mathbb{I}(\{0,0\})$ is the set of all polynomials in k[x,y] which vanish at $\{0,0\}$. If the coefficient of the degree zero term vanishes, this is assured. A general term of the ideal may then be written
\[
a_{10}x+a_{01}y+a_{20}x^2+a_{11}xy+a_{02}y^2+a_{30}x^3+a_{21}x^2y+a_{12}xy^2+a_{03}y^3+\cdots
\]
We compare this with a general element of the ideal $\langle x,y\rangle=g_1x+g_2y$ where $g_1,g_2\in k[x,y]$. Let $g_1=b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\cdots+b_{ij}x^iy^j$ and $g_2=c_{00}+c_{10}x+c_{01}y+c_{20}x^2+c_{11}xy+c_{02}y^2+\cdots+c_{kl}x^ky^l$. Then
\[
\langle x,y\rangle = (b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\cdots+b_{ij}x^iy^j)x+(c_{00}+c_{10}x+c_{01}y+c_{20}x^2+c_{11}xy+c_{02}y^2+\cdots+c_{kl}x^ky^l)y\;.
\]
We can quite easily see that there is no constant term in the simplified result. It is clear that every other term is present, which could be demonstrated explicitly by associating to each $a_{ij}$ a sum of some $b_{kl}$ and $c_{mn}$. Thus we have the desired equality by explicit computation. (I didn't use the containment argument, but I hope this still demonstrates the fact.)

:Endorsing the solutions to 1.4.1-1.4.6 above. All correct, and beautifully written.

:$\quad$-[[User:Steven.Jackson|Steven.Jackson]] ([[User talk:Steven.Jackson|talk]]) 10:08, 13 February 2018 (EST)

Math 380, Spring 2018, Assignment 3

2018-02-13T04:50:18Z

John.DeBrota: /* Solutions: */

Math 380, Spring 2018, Assignment 3

2018-02-13T04:06:45Z

John.DeBrota: /* Solutions: */