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Let $R$ be an algebra of finite type over a field $k$. I want to show that $R$ is Jacobson, i.e. that any prime ideal $\mathfrak{p}$ in $R$ is an intersection of maximal ideals. I am not getting very far however.

I know that the Jacobson radical of an algebra of finite type over a field $k$ is the nilradical, or in other words, the intersection of all maximal ideals is the same as the intersection of all prime ideals. So my first thought was to try to show that any prime ideal $\mathfrak{p}$ is the intersection of all maximal ideals which contain $\mathfrak{p}$. However if this is the case, I am not sure how to show the intersection of all such maximal ideals is actually $\subseteq \mathfrak{p}$.

Any help would be appreciated with this, thanks.

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This is a general form of the Nullstellensatz so is not trivial to prove (that is it doesn't follow from a simple syntactic manipulation of the definitions). Do you know any proofs of the Nullstellensatz you can adapt? –  Qiaochu Yuan May 3 '12 at 22:34
    
I have some, so I can try! –  Paul Slevin May 3 '12 at 22:50
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@PaulSlevin: It is enough to prove that an integral domain of finite type over a field has trivial Jacobson radical, and this follows from lemma 5.3.6 of these notes. But actually something stronger is true: any algebra of finite type over a Jacobson ring is again a Jacobson ring: see [Eisenbud, Commutative algebra, Thm 4.19]. –  Zhen Lin May 3 '12 at 23:14
    
@ZhenLin Why does the integral domain $R$ having trivial Jacobson radical imply that every prime $\mathfrak{p} \subseteq R$ is an intersection of maximal ideals? –  Paul Slevin May 4 '12 at 1:48
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@PaulSlevin: It's a standard reduction argument. The primes (and maximal ideals!) of $R / \mathfrak{p}$ correspond with the primes (and maximal ideals) of $R$ containing $\mathfrak{p}$. –  Zhen Lin May 4 '12 at 9:56

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