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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.

user264745
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3 Answers3

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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 $y = n\frac{a}{b}r \in (r)$.

Prahlad Vaidyanathan
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  • 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 – Curious student Apr 21 '19 at 15:38
  • @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. – Anacardium Oct 29 '20 at 05:43
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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 $x'\in S^{-1}A$?

Jyrki Lahtonen
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PID means principal ideal ring and integral domain. I will use notation $R$ instead of $A$.

Let $J$ be an ideal in $S^{-1}R$. Then $\exists I$ ideal in $R$ such that $J=S^{-1}I$. Since $R$ is principal ideal and commutative with identity, we have $I=\langle a\rangle=Ra$ for some $a\in R$. We claim $S^{-1}I=S^{-1} \langle a\rangle = \langle a/s \rangle $ for any $s\in S$. Let $r/t\in S^{-1}I$. Then $r/t=r_1a/t_1$ for some $r_1\in R$ and $t_1\in S$. Note $$r/t=r_1a/t_1= r_1as/t_1s =(r_1s/t_1) (a/s).$$ Thus $r/t\in (S^{-1}R)(a/s)= \langle a/s \rangle $ and $S^{-1}I\subseteq \langle a/s \rangle$. Let $r/t \in \langle a/s \rangle$. Then $r/t=(r_1/t_1)(a/s)$ for some $r_1/t_1\in S^{-1}R$. Clearly $$r/t=(r_1/t_1)(a/s)=r_1a/t_1s\in S^{-1}I$$ since $r_1a\in \langle a\rangle =I$. Thus $\langle a/s \rangle \subseteq S^{-1}I$. Hence $S^{-1}I = \langle a/s \rangle$.

Now we show $S^{-1}R$ is integral domain. $S^{-1}R$ is commutative ring with identity. We claim $S^{-1}A$ has no zero divisor. Assume towards contradiction, $\exists a/s,b/t \in S^{-1}R -\{0/s\}$ such that $(a/s)(b/t)=ab/st=0/s$. So $\exists s_1\in S$ such that $s_1(abs-st0)=abss_1=0$. Since $R$ has no zero divisor and $0\notin S$, we have $a=0$ or $b=0$. WLOG, assume $a=0$. Then $a/s=0/s$. Thus we reach contradiction. Hence $S^{-1}R$ has no zero divisor. Therefore $S^{-1}R$ is integral domain.

user264745
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