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$F$ is a field and $F[X^2, X^3]$ is a subring of $F[X]$, the polynomial ring. I need to show that nonzero prime ideals of $F[X^2, X^3]$ are maximal.

A classmate suggested taking a nonzero prime ideal $\mathfrak{p}$ of $F[X^2, X^3]$ and embedding $F[X^2,X^3]/\mathfrak{p} \hookrightarrow F[X]/(\mathfrak{p})$ and ultimately showing that $R/\mathfrak{p}$ was a field, but the proof we discussed was quite convoluted and it's not even apparent to me that that is an injective map. I'm inclined to think there's a shorter, more direct approach.

Does anyone see one? It's not imperative that I find a shorter or nicer proof, but seeing another approach can't hurt and I'm sure the grader would appreciate an easy-to-follow proof.

Thank you kindly.

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Any integral domain of Krull dimension $1$ has the property you are looking for. It follows from both answers below that $\dim F[X^2,X^3]=1$. –  user26857 Jan 23 '13 at 15:43
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2 Answers

The ring $F[X]$ is an integral extension of $F[X^2,X^3]$ because $X$ is the root of the polynomial $t^2 - X^2 \in (F[X^2,X^3])[t]$. Now let $\mathfrak{p}$ be a prime ideal of $F[X^2,X^3]$. By the Lying - over theorem (Proposition 5.10 of Atiyah - Macdonald) we get that there is $\mathfrak{q}$ a prime ideal of $F[X]$ such that $\mathfrak{q} \cap F[X^2,X^3] = \mathfrak{p}$. However $F[X]$ is a PID hence $\mathfrak{q}$ is maximal and so $\mathfrak{p}$ is also maximal.

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How does this always happen to me. –  JSchlather Jan 23 '13 at 6:07
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@JacobSchlather Both of us had exactly the same idea! –  fpqc Jan 23 '13 at 6:08
    
@uncookedfalcon No need to use going-up; look at Corollary 5.8 of Atiyah - Macdonald. –  fpqc Jan 23 '13 at 6:16
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a good point - thanks man :p (+1) –  uncookedfalcon Jan 23 '13 at 6:17
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This is probably overkill. Notice that the extension $F[X^2,X^3] \subset F[X]$ is an integral extension. This follows because $X$ satisfies the equation $t^2-X^2 \in F[X^2,X^3][t]$ and clearly $F[X]=F[X^2,X^3][X]$. Now suppose that some non-zero prime ideal $\mathfrak p$ of $F[x^2,x^3]$ is not maximal then we have a proper chain of ideals $(0) \subsetneq \mathfrak p \subsetneq \mathfrak q$. Which would imply by the Going Up theorem that we can find a chain of prime ideals $(0) \subsetneq \mathcal P \subsetneq \mathcal Q$ in $F[X]$. This is clearly impossible.

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+1 For your nice answer. –  fpqc Jan 23 '13 at 6:09
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@BenjaLim, Thanks I like yours as well. It reminds me of something, but I'm not quite sure what. –  JSchlather Jan 23 '13 at 6:10
    
Our answers upto isomorphism are the same, because IIRC the Going-up Theorem is just an extension of the Lying-over theorem. –  fpqc Jan 23 '13 at 6:13
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I've never heard of the Going Up theorem/property before, but I like it! Thank you to both of you! –  JoeDub Jan 23 '13 at 6:19
    
@JoeDub Do you have Atiyah - Macdonald? All the results that Jacob and I have used can be found in Chapter 5. –  fpqc Jan 23 '13 at 6:20
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