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I'd like a little help on how to begin this problem.

Show that a PID $R$ is Jacobson-semisimple $\Leftrightarrow$ $R$ is a field or $R$ contains infinitely many nonassociate irreducible elements.


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up vote 2 down vote accepted

If $R$ is a PID and has infinitely many nonassociated irreducible elements, then given any nonunit $x\in R$ you can find an irreducible element that does not divide $x$; can you find a maximal ideal that does not contain $x$? If so, you will have proven that $x$ is not in the Jacobson radical of $R$. The case where $R$ is a field is pretty easy as well.

Conversely, suppose $R$ is a PID that is not a field, but contains only finitely many nonassociated primes; can you exhibit an element that will necessarily lie in every maximal ideal of $R$?

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Yes I can...Thanks. – Nana May 5 '11 at 13:06

HINT $\ $ A PID has a nonzero element divisible by every prime iff it has finitely many primes.

NOTE $\ $ This holds much more generally. We have the following generalization to Krull domains (e.g. UFDs and Noetherian integrally-closed domains, e.g. Dedekind domains, number rings)

THEOREM $\ $ A Krull domain has a nonzero element in every nonzero prime ideal iff it is a PID with finitely many primes.

For a proof see e.g. Theorem 1 in Gilmer: The pseudo-radical of a commutative ring.

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Thanks for the alternative. – Nana May 5 '11 at 13:09

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