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Let $R$ be a ring, and let $P$ be a prime ideal of $R$. Let $S = R \setminus P$, and set $R_p = S^{-1} R$.

What are the units in $R_p$? I'm trying to prove that $\frac{a}{b}$ is an unit if and only if $a \in S$, but I'm having trouble.

One direction is easy as if $a \in S$, $\frac{b}{a} \in R_p$, which is the desired inverse.

The other direction is giving me some trouble.

So we let $\frac{a}{b}$ be an unit, and we find its inverse $\frac{a'}{b'}$. Then $\frac{aa'}{bb'} = \frac{1}{1}$ This means there exists $t \in S$ such that $taa' = tbb'$, but I have no idea how to proceed from here.

Thanks for your help!

Eric Wofsey
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Phil
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1 Answers1

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Hint: Look at both sides of the equation $taa'=tbb'$. Each element of $R$ is either in $P$ or in $S$. With that in mind, what can you say about the different parts of this equation?

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Since $t,b,$ and $b'$ are all in $S$, so is $tbb'$. This means that $a\in S$, since otherwise $a$ would be in $P$ and so $taa'$ would be in $P$ as well.

Eric Wofsey
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