Lemma 10.9.7 (07K0)—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 May 02, 2021 at 22:21

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.

Comment #6343 by Johan on July 13, 2021 at 12:14

Thanks and fixed here.

Comment #6344 by Johan on July 13, 2021 at 12:21

Thanks and fixed here.

Comment #8747 by Roy Shtoyer on December 10, 2023 at 14:10

I beleive the statement is supposed to be $\mathop{\mathrm{Hom}}\nolimits _ {S^{-1}R} (S^{-1}M, N) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ and not $\mathop{\mathrm{Hom}}\nolimits _ {R} (S^{-1}M, N)$ .

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 Stacks project on June 02, 2024 at 09:20

@#8747. Note that $\mathop{\mathrm{Hom}}\nolimits _ R (S^{-1}M, N) = \mathop{\mathrm{Hom}}\nolimits _ {S^{-1}R} (S^{-1}M, N)$ by Lemma 10.9.5. So it's fine as stated and the proof gives the stated result. Going to leave as is.

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