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Let a non-zero ideal $P \in \operatorname{Spec}(K[x,y])$ be a non-maximal ideal, where $K$ is an algebraically closed field. Can I say that $P \cap K[x] = (0)$ or $P \cap K[y] = (0)$?

If not, can you think of any counterexamples? Any help is welcome. Thanks!

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    $\begingroup$ What have you tried so far? Can you please write down a few principal ideals and check? $\endgroup$
    – Arkady
    Jun 28 at 0:38
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    $\begingroup$ You mean a, not the, right? $\endgroup$ Jun 28 at 0:43
  • $\begingroup$ It's "a" instead of "the". Sorry! $\endgroup$
    – Santos
    Jun 28 at 0:50

1 Answer 1

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Yes, you can say this.

Let us suppose $K[x] \cap P$ is nontrivial. Since $K$ is algebraically closed and $P$ is prime, there is some $a \in K$ such that $x - a \in P$.

And if $K[y] \cap P$ is also nontrivial, then there is some $b \in K$ such that $y - b \in P$.

Let $R = K[x, y] / (x - a, y - b) \cong K$, and let $\pi : K[x, y] \to R$ be the canonical map. Then $R / \pi(P) \cong K[x, y]/P$.

Now since $P$ is prime, we see that $K[x, y] / P \cong R/\pi(P)$ is an integral domain. Therefore, $\pi(P)$ is prime, hence maximal. Then $P$ is maximal.

Obviously, this generalises to polynomials over any finite number of variables.

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