TITLE: $S^{-1}B$ and $T^{-1}B$ isomorphic for $T=f(S)$
QUESTION [8 upvotes]: Let $f:A\to B$ be a homomorphism of rings, $S$ be a multiplicatively closed subset of $A$ and $T=f(S)$. Then $S^{-1}B$ and $T^{-1}B$ are isomorphic as $S^{-1}A$-modules.

First we define the obvious homomorphism $\phi:S^{-1}B\to T^{-1}B$ by $\frac{b}{s}\mapsto \frac{b}{f(s)}$. I can derive the facts that $\phi$ is a homomorphism as well as surjectivity, but I am unsure about injectivity.
Suppose $\frac{b}{f(s)}=\frac{0}{1}$, then $\exists\ t=f(s')\in T$ such that $tb=f(s')b=0$. And now I am stuck, because what I need is $s''b=0$ for some $s''\in S$, but $f(s')\in T$, so it doesn't really help me.

REPLY [6 votes]: Injectivity. Let $\frac{r_1}{s_1},\frac{r_2}{s_2}\in S^{-1}B$, and suppose 
that
\begin{align*}
\phi\left(\frac{r_1}{s_1}\right)&=\phi\left(\frac{r_2}{s_2}\right) \\
\frac{r_1}{f(s_1)}&=\frac{r_2}{f(s_2)}
\end{align*}
So $\exists$ $t\in T$ such that
\begin{equation*}
t\left(f(s_2)r_1-f(s_1)r_2\right)=0
\end{equation*}
Since $t\in T=f(S)$, $\exists$ $s\in S$ such that $f(s)=t$, and
\begin{align*}
f(s)\left(f(s_2)r_1-f(s_1)r_2\right)&=0
\end{align*}
Since $B$ is an $A$ module via $f$
\begin{equation*}
s\cdot (s_2\cdot r_1-s_1\cdot r_2)=0
\end{equation*}
and thus $\frac{r_1}{s_1}=\frac{r_2}{s_2}$. So 
$\operatorname{ker}(\phi)=0$.