1

Is $\mathbb{Z}\Bigl[\frac{1}{2},\frac{1}{3}\Bigr]$ a Dedekind Domain? Can anyone help me with a detailed reasoning?

Jonas Meyer
  • 53,602
rola
  • 63
  • There are many definitions and/or characterizations of Dedekind domains. Which do you have available? – Bill Dubuque Jul 24 '12 at 03:58
  • @Bill: Noetherian Integrally closed domain of Krull Dimension $1$. I am also familiar with the definition that every non zero ideal factors as prime ideals and every fractional ideal is invertible, though I don't consider myself handling these definitions very well:( – rola Jul 24 '12 at 04:00
  • 3
    Hint: it is a PID $\Rightarrow$ Dedekind. Generally localizations preserves PIDs. $\ \ $ – Bill Dubuque Jul 24 '12 at 04:11
  • @Bill: You mean to say this: We know, that $\mathbb{Z}$ is a P.I.D, so localizing gives $\mathbb{Z}\Bigl[\frac{1}{2},\frac{1}{3}\Bigr]$ as a P.I.D, and since it's a P.I.D its is a Dedikind domain. – rola Jul 24 '12 at 04:16
  • That's one route. – Bill Dubuque Jul 24 '12 at 04:17

1 Answers1

1

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.)

Community
  • 1
Lord_Farin
  • 17,743