Math 380, Spring 2018, Assignment 3

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:

1. Section 1.4.

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

1. Ideal (in $\mathsf{k}[x_1,\dots,x_n]$).
2. $\left\langle S\right\rangle$ (the ideal generated by a set $S$ of polynomials).
3. $\left\langle f_1,\dots,f_s\right\rangle$ (the ideal generated by the finite set $\{f_1,\dots,f_s\}$).
4. $\mathbb{I}(T)$ (the ideal of the set of points $T\subseteq\mathsf{k}^n$).

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

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

Solve the following problems:

1. Section 4, problems 1, 3, 5, 6(a), 6(b), 7, and 8. (In problem 6, the word "basis" means "generating set.")
2. Prove that, for any set $T\subseteq\mathsf{k}^n$, the set $\mathbb{I}(T)$ is always an ideal of $\mathsf{k}[x_1,\dots,x_n]$.
3. 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.")

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

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 $V$.

-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?

This is a deep question. The answer is provided by a famous theorem due to Hilbert, the "Nullstellensatz." It is a sort of "fundamental theorem of algebraic geometry." See Theorem 4.2.6, which we will hopefully study in April.

-Steven.Jackson (talk) 13:08, 13 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.

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.

-Steven.Jackson (talk) 10:08, 13 February 2018 (EST)

Endorsing also 1.4.7-1.4.8 and 2-3. (In 1.4.8(b), the "clearly impossible" part is not perfectly clear, but let's let this dog lie; at the moment clarity requires a very tedious argument, and we will have better tools for addressing this question after Chapter 2. The conclusion is correct.)

-Steven.Jackson (talk) 16:11, 13 February 2018 (EST)