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I'm a little confused about the following exercise in A-M:

Let $f:A\to B$ be a homomorphism of rings and let $S$ be a multiplicatively closed subset of $A$. Let $T=f(S)$. Show that $S^{-1}B$ and $T^{-1}B$ are isomorphic as $S^{-1}A$-modules.

You would think the isomorphism is between $\frac{b}{s}\to \frac{b}{f(s)}$, but why is this an isomorphism. Doesn't that imply that f is injective? Why is this map invertible? Sorry if this is an obvious question, but why should these two rings be isomorphic intuitively?

Alvin Chen
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1 Answers1

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Recall the definitions.

$S^{-1}B$ is the equivalence classes of pairs $(b,s)\in B\times S$ respect to $(b,s)\sim (b',s')$ if and only if $(bs'-b's)u=0$ for some $u\in S$. Here we use the $A$-module structure on $B$ induced by $f$. Consequently, we may write $(bf(s')-b'f(s))f(u)=0$ for some $u\in S$.

$T^{-1}B$ is the equivalence classes of pairs $(b,t)\in B\times T$ respect to $(b,t)\sim(b',t')$ if and only if $(bt'-b't)v=0$ for some $v\in T$.

Consider the surjective function \begin{align} &g:B\times S\to B\times T& &(b,s)\mapsto(b,f(s)) \end{align} Then $g(b,s)\sim g(b',s')$ if and only if $(b,s)\sim(b',s')$. Consequently, $g$ induces a bijection $S^{-1}B\to T^{-1}B$.

Fabio Lucchini
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