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Let $p$ be a nonzero prime ideal of $A=k[x^2,x^3]$. I want to show $p$ is maximal.

My trial is that $A/p$ contains $k$ and since $k$ is a field, if I can show that $A/p$ is integral over $k$ then it should be a field, too. But is it true that $A/p$ is integral over $k$?

Actually I know other ways of this. (For someone interested: link1 link2)

Gobi
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2 Answers2

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An integral extension of rings preserves the Krull dimension. Your ring $A$ has the ring $k[x]$ as an integral extension. So the Krull dimension of $A$ is one and every nonzero prime ideal is maximal.

Because of this, it follows from Zariski's Lemma that $A/p$ is a finite degree field extension of $k$, so is integral over $k$.

(I worry about people getting tired of me constantly referencing my own lecture notes, but if it helps: all of the facts I mentioned above can be found in my commutative algebra notes.)

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Pete L. Clark
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Alternative solution: $k[x]$ is an integral extension of $k[x^2,x^3]$. Then by the lying over theorem, there is a prime ideal $P$ of $k[x]$ such that

$$P \cap k[x^2,x^3] = p.$$

But now $k[x]$ is a PID and so $P$ is maximal. Thus $p$ too is a maximal ideal.