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Recall that a commutative ring is Jacobson if every prime ideal is the intersection of the maximal ideals that contain it.

In the exercises of a commutative algebra course I gave I asked the students to show that a commutative ring is Jacobson if and only if every non-maximal prime ideal is the intersection of the prime ideals that strictly contain it. I now suspect that somewhere in the back of my mind I had imposed the condition that the ring should be Noetherian without actually saying this. Of course, Jacobson rings will always have this other property, and the converse is straightforward to prove if there is no strictly ascending chains of prime ideals. But is the result true in general?

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up vote 9 down vote accepted

The result is true in general.

We may assume a counterexample is given in the form of a domain $R$ satisfying the second property but with nontrivial Jacobson radical, i.e. the closed points of Spec $R$ are not dense. Let D be an affine open neighborhood of $(0)$ in Spec $R$ which contains no closed points. Since D is affine, there exists some $x\in$ D which is closed in D. That is,

$\overline{\lbrace x\rbrace}\setminus x\subset\text{Spec }R\setminus D$.

Since $D$ is open, this implies

$x\not\in \overline{\overline{\lbrace x\rbrace}\setminus x}$,

but this contradicts the requirements of the second property.

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Great! Permit me to rephrase this in algebra rather than geometry for the sake of algebraists. Pick $f$ in the Jacobson radical but non-zero and let $P$ be an ideal that is maximal amongst ideals disjoint from the set of powers of $f$. $P$ is prime but cannot be maximal so is (by condition 2) the intersection of the prime ideals that strictly contain it. But by definition of $P$ all these strictly bigger primes conatin $f$ thus $P$ contains $f$ a contradiction. Nice. – Simon Wadsley May 24 '10 at 11:30

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