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!

  • 1
    $\begingroup$ What have you tried so far? Can you please write down a few principal ideals and check? $\endgroup$
    – Arkady
    Jun 28 at 0:38
  • 2
    $\begingroup$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.