Is $\mathbb{Z}\Bigl[\frac{1}{2},\frac{1}{3}\Bigr]$ a Dedekind Domain? Can anyone help me with a detailed reasoning?
Is $\mathbb{Z}\Bigl[\frac{1}{2},\frac{1}{3}\Bigr]$ a Dedekind domain?
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abstract-algebra
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0That's one route. – 2012-07-24
1 Answers
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This answer is distilled from Gone's and rola's comments:
By this question, and since $\Bbb Z \subset \Bbb Z[\frac12,\frac13] \subset \Bbb Q = Q(\Bbb Z)$, we conclude $\Bbb Z[\frac12,\frac13]$ is a PID. PIDs are Dedekind, so we are done.
(More generally, any localization of a PID is again a PID.)