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Let $A$ be a commutative ring and $S$ a multiplicative closed subset of $A$. If $A$ is a PID, show that $S^{-1}A$ is a PID.

I've taken an ideal $I$ of $S^{-1}A$ and I've tried to see that is generated by one element; the ideal $I$ has the form $S^{-1}J$ with $J$ an ideal of $A$. $J$ is generated by one element but I can't see why $I$ has to be generated by one element, maybe I'm wrong.

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    $\begingroup$ PIDs are required to be non-zero. So the question is No. ;) You have to assume that $0 \notin S$. $\endgroup$ – Martin Brandenburg Oct 23 '13 at 7:57
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If $I$ is an ideal of $S^{-1}A$, then there exists an ideal $J$ of $A$ such that $I = JS^{-1}A$. Now there exists $r \in A$ such that $J = (r)$. We show that $$ I = (r) $$ where $r$ is now thought of as an element of $S^{-1}A$.

Firstly, if $x \in (r)$, then $x = \frac{a}{b}r$ where $\frac{a}{b} \in S^{-1}A$ and so $x \in JS^{-1}A = I$.

Secondly, if $y \in I = JS^{-1}A$, then $y = nr\frac{a}{b}$, where $nr \in J$ and $\frac{a}{b} \in S^{-1}A$, so $y = n\frac{a}{b}r \in (r)$.

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  • $\begingroup$ Dear Sir , $x\in (r)=J$ J is ideal of A So how $x=a/br$ I do not understand .Please Can you tell me Thanks a lot $\endgroup$ – idon'tknow Apr 21 '19 at 15:38
  • $\begingroup$ @idon'tknow observe that when we say that $I = (r)$ then it is meant that $I$ is generated by $\frac r 1.$ So if $x \in \left (\frac r 1 \right )$ then there exists $\frac a b \in S^{-1} A$ such that $x = \frac a b \cdot \frac r 1 = \frac {ar} {b},$ as required. $\endgroup$ – Anacardium Oct 29 '20 at 5:43
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An arbitrary element $x$ of $S^{-1}J$ is of the form $$ x=\sum_is_i^{-1}j_i $$ with $s_i\in S, j_i\in J$ for all $i$. If $j$ is a generator of $J$ as an ideal, then $j_i=a_ij$ for some $a_i\in A$. Can you now write $x$ in the form $x=x'j$ for some $x'\in S^{-1}A$?

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