The Stacks project

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. $\square$


Comments (5)

Comment #6203 by Masa on

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

Comment #8747 by Roy Shtoyer on

I beleive the statement is supposed to be and not .

Also, I think it's useful to add an explanation about what does "all the elements of S act as automorphisms on N" mean - with respect to what action? I find it a bit confusing.

Comment #9331 by on

@#8747. Note that by Lemma 10.9.5. So it's fine as stated and the proof gives the stated result. Going to leave as is.

There are also:

  • 6 comment(s) on Section 10.9: Localization

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