If R is a semilocal Dedekind domain, then R is a PID.

Question

On page 447 of Groups, Rings, Modules by Maurice Auslander and David Buchsbaum:

Proposition 1.3. For a semilocal integral domain $R$, if $M$ is a finitely generated projective $R$-module, then $M$ is a free $R$-module.

Corollary 1.4. If $R$ is a semilocal Dedekind domain, then $R$ is a PID.

I understand Proposition 1.3, but couldn't understand why Proposition 1.3 implies Corollary 1.4.

Thank you very much for reading.

score 3 · Accepted Answer · answered Jul 22 '19 at 05:05

3

In a Dedekind domain, ideals are projective modules. So a nonzero ideal is free in your set-up. It can't have rank $\ge2$, as then tensoring up with the quotient field $K$ would give a $K^2$ contained in $K$. So it has rank zero or one, and has a single generator.

answered Jul 22 '19 at 05:05

Angina Seng

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Thank you very much! I really appreciate the answer. – Jul 22 '19 at 05:33

If R is a semilocal Dedekind domain, then R is a PID.

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