most recent 30 from http://mathoverflow.net 2013-06-20T10:32:52Z http://mathoverflow.net/feeds/question/25474 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25474/primes-in-a-commutative-jacobson-ring Simon Wadsley 2010-05-21T12:12:43Z 2010-05-23T07:23:47Z

<p>Recall that a commutative ring is Jacobson if every prime ideal is the intersection of the maximal ideals that contain it. </p> <p>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?</p>

http://mathoverflow.net/questions/25474/primes-in-a-commutative-jacobson-ring/25633#25633 Kevin Ventullo 2010-05-23T07:23:47Z 2010-05-23T07:23:47Z

<p>The result is true in general. </p> <p>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,</p> <p>$\overline{\lbrace x\rbrace}\setminus x\subset\text{Spec }R\setminus D$.</p> <p>Since $D$ is open, this implies</p> <p>$x\not\in \overline{\overline{\lbrace x\rbrace}\setminus x}$,</p> <p>but this contradicts the requirements of the second property.</p>