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Questions: $\rm\color{#c00}{(1)}$ Is the $[\Longrightarrow]$ implication of $$ \frac{a}{b}\in\left(\mathbb{Z}_{(p)}\right)^{\times}\iff\frac{b}{a}\in\mathbb{Z}_{(p)} $$ obvious?

$\rm\color{#c00}{(2)}$ More generally, is the implication $$ \frac{a}{b}\in \left(S^{-1}R\right)^{\times}\implies\frac{b}{a}\in S^{-1}R $$ true?

It suffices to show $a\in S$. We have $$ \frac{a}{b}\in \left(S^{-1}R\right)^{\times}\iff\exists\frac{x}{y}\in S^{-1}R,\exists t\in S:t(ax-by)=0 $$

Now $tby\in S$, therefore $tax\in S$. We know $S$ is closed under multiplication, but how to infer $a\in S$?

In the case of $\mathbb{Z}_{(p)}$, $S=\mathbb{Z}\backslash p\mathbb{Z}$, hence we have $$ tax\not\in p\mathbb{Z}\implies a\not\in p\mathbb{Z}. $$

user26857
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Guest
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2 Answers2

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(2) If $R=\mathbb Z$ and $S=1+3\mathbb Z$, then $5/25\in S^{-1}R$ is invertible and $5\notin S$.

Actually $S$ is an example of a multiplicative set which is not saturated. (A multiplicative set $S$ is called saturated if $xy\in S\Leftrightarrow x\in S$ and $y\in S$.) The good new is that every multiplicative set is contained in a saturated multiplicative set $\widetilde S$ and $S^{-1}R\simeq\widetilde S^{-1}R$. (For instance, in the example above the saturation of $S$ is $\widetilde S=\mathbb Z-3\mathbb Z$.)

user26857
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  • Nice example! Since we know $R\backslash P$ is satured if (and only if) $P$ is a prime ideal, $R_{(P)}:=\left(R\backslash P\right)^{-1}R$ will be such that $\rm\color{#c00}{(2)}$ holds, right? – Guest Dec 03 '14 at 17:48
  • @Guest Yes, right. – user26857 Dec 03 '14 at 17:50
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If $a/b$ is a unit in the ring of $p$-local numbers then there is an element $c/d$ in $\mathbb{Z}_{(p)}$ such that $\frac{a}{b}\cdot \frac{c}{d}=1$, for integers $a,b,c,d$ with $p\nmid b$ and $p\nmid d$. This implies that $c/d=b/a$, so that $p\nmid a$, i.e., we have $b/a\in \mathbb{Z}_{(p)}$.

Dietrich Burde
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