I'm having trouble figuring out how to approach this problem and am wondering if anyone can give me a hint...

The problem is to show that for any ring $A$ and any localization $S^{-1}A$ there exists an ideal $I \subset A$ such that $S^{-1}A \cong$ a subring of $S_0^{-1}(A/I)$ where $S_0$ is the set of all non-zero divisors in $A$.

So far I think that it's at least clear enough to me that $S_0^{-1}(A/I)$ is a localization for some $S$. Since $S^{-1}(A/I)$ is the set of elements $\frac{a}{b}$ where $b$ is a non-zero divisor of $A/I$, which is to say that there does not exist $c \in A$ such that $bc \in I$. So if $I$ is a prime ideal then $b \in A-I$.

Am I headed down the wrong path? Any pointers?


Hint: $\mu:A \to S^{-1}A$ is injective only when $S$ does not contain zero divisors.

To see this, note that $\frac{a}{1}=\frac{0}{1} \implies au=0$ for some $u \in S$.

Edit: Sorry, either I am missing something, or was too unclear.

Alleged Hint: In the case that $S$ contains no zero divisors, $S \subseteq S_0$, so the canonical homomorphismm $\mu:A \to S_0^{-1}A$ also inverts $S$, and hence factors uniquely through $S^{-1}A$ by the universal property.

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  • $\begingroup$ So if I let $I$ be the set of all zero divisors, then clearly $S_0^{-1}(A/I) = \{ \frac{a}{b} | a \in (A-I) \cup \{ 0\}, b \in A-I\}$. It's still unclear to me though how $S^{-1}(A)$ could be isomorphic to a subring of this because we don't know what $S$ is. $\endgroup$ – TAS Sep 19 '17 at 19:40

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