Lemma 10.9.7. Let $R$ be a ring. Let $S \subset R$ a multiplicative subset. Let $M$, $N$ be $R$-modules. Assume all the elements of $S$ act as automorphisms on $N$. Then the canonical map

$\mathop{\mathrm{Hom}}\nolimits _ R(S^{-1}M, N) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(M, N)$

induced by the localization map, is an isomorphism.

Proof. It is clear that the map is well-defined and $R$-linear. Injectivity: Let $\alpha \in \mathop{\mathrm{Hom}}\nolimits _ R(S^{-1}M, N)$ and take an arbitrary element $m/s \in S^{-1}M$. Then, since $s \cdot \alpha (m/s) = \alpha (m/1)$, we have $\alpha (m/s) =s^{-1}(\alpha (m/1))$, so $\alpha$ is completely determined by what it does on the image of $M$ in $S^{-1}M$. Surjectivity: Let $\beta : M \rightarrow N$ be a given R-linear map. We need to show that it can be "extended" to $S^{-1}M$. Define a map of sets

$M \times S \rightarrow N,\quad (m,s) \mapsto s^{-1}\beta (m)$

Clearly, this map respects the equivalence relation from above, so it descends to a well-defined map $\alpha : S^{-1}M \rightarrow N$. It remains to show that this map is $R$-linear, so take $r, r' \in R$ as well as $s, s' \in S$ and $m, m' \in M$. Then

\begin{align*} \alpha (r \cdot m/s + r' \cdot m' /s') & = \alpha ((r \cdot s' \cdot m + r' \cdot s \cdot m') /(ss')) \\ & = (ss')^{-1}\beta (r \cdot s' \cdot m + r' \cdot s \cdot m') \\ & = (ss')^{-1} (r \cdot s' \beta (m) + r' \cdot s \beta (m')) \\ & = r \alpha (m/s) + r' \alpha (m' /s') \end{align*}

and we win. $\square$

Comment #6203 by Masa on

In the last equation block of the proof, some primes for $s$'s and $r$'s look strange. Also in the third line of this block, the last ")" should be added.

There are also:

• 4 comment(s) on Section 10.9: Localization

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).