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If $p$ is prime number, prove that $$\mathbb{Z}_{(p)}=\left\{\frac{a}{b}\in\mathbb{Q}: \text{$p$ doesn't divide $b$}\right\}$$ is a PID.

So, first step is to show that $\mathbb{Z}_{(p)}$ is an Ideal Domain and then show that every ideal in it is principal.

I know that $\mathbb{Z}_{(p)}$ is the subring of $\mathbb{Q}$ then it follows that $\mathbb{Z}_{(p)}$ is commutative. Then I'm lost what to do. Any help is appreciated :)

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marked as duplicate by Dietrich Burde, Daniel W. Farlow, user91500, Alex S, user26857 abstract-algebra Apr 7 '16 at 21:01

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  • $\begingroup$ Try to think a little bit backwards: You know that if there is only one maximal ideal, such ideal contains exactly those element that are not invertible. Do you know which kind of fractions in $\mathbb{Z}_{(p)}$ are not invertible? $\endgroup$ – Darío G Apr 7 '16 at 11:19
  • $\begingroup$ How is "$p$ doesn't divide $n$" a property of $\frac ab$? There are not even any variables in common. So what you're defining is either all of $\mathbb Q$ or the empty set, depending on whether $p$ divides $n$ or not. $\endgroup$ – Henning Makholm Apr 7 '16 at 11:19
  • $\begingroup$ @HenningMakholm You are right, I was suggesting help without even noticing this! $\endgroup$ – Darío G Apr 7 '16 at 11:21
  • $\begingroup$ @Henning, Sorry I have edited the question. It's $p$ doesn't divide $b$ $\endgroup$ – El Qanas Apr 7 '16 at 11:24
  • $\begingroup$ @Wore, elements in $\mathbb{Z}_{(p)}$ which is not invertible is the fractions where the numerator is the multiple of $p$ $\endgroup$ – El Qanas Apr 7 '16 at 11:26