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

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  • $\begingroup$ In fact, I don't need to know what is $I(0)$, since we can argument that $y \in I(x_1 \cap X_2)$ and $y \not\in I(X_1)+I(X_2)$, but I find pedagogic to compute $I(0)$ $\endgroup$
    – Leafar
    Commented Mar 8, 2015 at 19:13

1 Answer 1

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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 $\langle x, y \rangle \neq \langle x,y^2 \rangle$.

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  • $\begingroup$ But isn't $<x,y> = \{ax + by : a,b \in \mathbb{K}\}$? So how do we have $x^2 \in <x>$? $\endgroup$
    – Leafar
    Commented Mar 8, 2015 at 20:31
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    $\begingroup$ @Leafar No, $\langle x,y\rangle$ is equal to all polynomial combinations of $x$ and $y$. $\endgroup$ Commented Mar 8, 2015 at 23:28
  • $\begingroup$ @FredrikMeyer Thanks $\endgroup$
    – Leafar
    Commented Mar 8, 2015 at 23:31

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