3
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.