What is $I(\{(0,0)\})$?

Question

I have two (algebraic) sets:

1. $X_1 = Z(x) \subseteq \mathbb{A}^2$, ie, $X_1 = \{(0,y):y \in \mathbb{K}\} \subseteq \mathbb{A}^2$

2. $X_2 = Z(x+y^2) \subseteq \mathbb{A}^2$, ie, $X_2 = \{(-y^2,y):y \in \mathbb{K}\} \subseteq \mathbb{A}^2$.

We're assuming that $\mathbb{K}$ is algebraically closed (so we can use Hilbert's Nullstellensatz).

I want to prove that $I(X_1)+I(X_2) \neq I(X_1 \cap X_2)$.

My question is maybe stupid / too basic, but here I go:

Obviously, $X_1 \cap X_2 = \{(0,0)\}$.

Since $I(X_1)$ and $I(X_2)$ are radical ideals, $I(X_1) = <x>$ and $I(X_2) = <x+y^2>$.

Then, $I(X_1)+I(X_2) = <x>+<x+y^2>$.

So, $I(X_1)+I(X_2) = <x,y^2>$.

Therefore, we must show that $I(\{(0,0)\}) \neq <x,y^2>$.

But what is $I(\{(0,0)\})$?

It is $<x,x^2,\ldots,y,y^2,\ldots>$?

Answer 1

It is the vanishing ideal of the point $(0,0)$. This is precisely the ideal generated by $x$ and $y$, i.e. in your notation $\langle x, y \rangle$. Note that this is equal to the ideal $\langle x, x^2, \dots, y, y^2,\dots \rangle$ that you mention in your question. What this effectively says is that a polynomial vanishes on $(0,0)$ if and only if it does not have a constant term. It follows that $\langle x, y \rangle \neq \langle x,y^2 \rangle$.