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Let our ring be $\mathbb C[x,y]$

and say I want to determine vanishing ideals of below subsets of $\mathbb A^2(\mathbb C)$- affine complex plane, with finitely many polynomials(since every ideal is finitely generated).

    1. $V=\{(0,0),(1,1)\}\subset A^2(\mathbb C)$
    1. $U=\{(n,n)\mid n\in\mathbb Z\}\subset A^2(\mathbb C)$

Definition of vanishing ideal : Let $V\subset A^2(\mathbb C)$ then $$I(V)=\{f\in \mathbb C[x,y] \mid f(x,y)=0, \forall (x,y)\in V\}$$

For $1.$ I know $I(\{(a,b)\})=\langle x-a,y-b\rangle $ $$I(\{(0,0),(1,1)\})=I(0,0)\cap I(1,1)\\=\langle x,y\rangle \cap \langle x-1,y-1\rangle =\langle xy(x-1)(y-1)\rangle $$

For $2.$ Similarly: $$I(\{(n,n)\mid n\in\mathbb Z\})=\bigcap_{n\in \mathbb Z} I(n,n)\\=\left\langle\prod_{n\in\mathbb Z}(x-n)(y-n)\right\rangle$$

Are my approaches correct or in 2. the product not defined (because is not polynomial?)?

And what is the good/proper approach/algorithm to calculate vanishing ideals for any polynomial ring $k[x_1,x_2,\cdots, x_n]$ where $k$ need not to be algebraicly closed.

KReiser
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2 Answers2

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Both of your attempts at solutions need some help here.

First, $(x,y)\cap (x-1,y-1) = (x(x-1),x(y-1),y(x-1),y(y-1))$, not the principal ideal you've written in your post.

Second, you are right to be suspicious of an infinite product. That's not an element of $\Bbb C[x,y]$, so it cannot be the generator of your ideal. Instead, one should observe that $x-y$ is in all of those ideals and conjecture that this is a generator: since $V(x-y)\cong \Bbb A^1$ and the closed sets of $\Bbb A^1$ are finite sets of points, the empty set, and the whole space, the closure of $\{(n,n)\}_{n\in\Bbb Z}$ must be all of $V(x-y)$.

In general, an algorithm for computing the vanishing ideal of a set depends on how you're given the set. If one is given a map of affine varieties and one wants to find equations for the image, the best tools come from Grobner bases & elimination theory. If all you have is a collection of points, then you're looking for the kernel of the map $k[V]\to\prod_p k(p)$, where the map $k[V]\to k(p)$ is evaluation of a regular function at $p$. That can be hard to compute sometimes, but that's life.

KReiser
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For things like the second one, it's often helpful to reduce the number of variables.

If $p(x,y)$ is a polynomial in two variables such that $p(n,n) = 0$ for all $n \in \mathbb{Z}$, then $q(x) = p(x,x)$ is a polynomial in one variable which vanishes for all $n \in \mathbb{Z}$.

But of course a polynomial in one variable can only equal zero infinitely many times if it is actually the zero polynomial. So $q(x) = p(x,x) = 0$ as a polynomial, and we see that actually $p(x,y)$ vanishes on the entire line $y=x$.

Nate
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