Let $S$ be a multiplicatively closed subset of a commutative noetherian ring $A$. Let $M$ and $N$ be finitely generated $A$-modules. If $M_S$ is isomorphic to $N_S$, show that $M_t$ is isomorphic to $N_t$ for some $t \in S.$

  • $\begingroup$ Welcome to Mathematics! But I'm afraid there is no real question in that. What do you want? What have you tried? $\endgroup$
    – vonbrand
    Apr 14, 2013 at 23:48
  • $\begingroup$ In another words, if S^{-1}M is isomorphic to S^{-1}N, then there exists an element t in the multiplicative set S such that M_t is isomorphic to N_t $\endgroup$
    – user72578
    Apr 15, 2013 at 0:03
  • 8
    $\begingroup$ I understand that we would prefer questions to be phrased as such, instead of as imperatives like this one is, but it strikes me as highly dishonest to play dumb and say there's no question here. Even worse so to vote closure because you don't like the way a new user phrased their very first question. $\endgroup$
    – Jim
    Apr 15, 2013 at 0:16

2 Answers 2


Since $M$ is finitely presented, we have that $$\text{Hom}_A(M,N)_S \cong \text{Hom}_{A_S}(M_S, N_S) \text{ (*)}$$

via the map which takes $f/s$ to the product of the constant map $1/s$ and the map from $M_S$ to $N_S$ induced by $f$, by THM 7.11 in Matsumura's "Commutative Ring Theory". Take $f \in \text{Hom}_{A_S}(M_S, N_S)$ to be an isomorphism. Suppose $g/s \in \text{Hom}_R(M,N)_S$ maps to $f$ under isomorphism (*). Then $g$ must induce an isomorphism from $M_S$ to $N_S$.

We have an exact sequence

$$0 \rightarrow \ker g \rightarrow M \rightarrow N \rightarrow \text{coker }g \rightarrow 0.$$

Notice that $(\ker g)_S=0=(\text{coker } g)_S$. Since $\ker g$ and $\text{coker } g$ are finitely generated, we may choose $a \in \text{Ann}_A(\ker g) \cap S$ and $b \in \text{Ann}_A(\text{coker } g) \cap S$. Now, take $t:=ab$.

  • $\begingroup$ Thanks a lot for your help. $\endgroup$
    – user72578
    Apr 15, 2013 at 17:59

There is a more general geometric statement whose proof I find a little bit clearer.

Let $F,G$ sheaves of modules over some ringed space $X$. If $F$ is of finite presentation, then $\underline{\hom}_{\mathcal{O}_X}(F,G)_x \to \hom_{\mathcal{O}_{X,x}}(F_x,G_x)$ is an isomorphism (one easily reduces to the case $F=\mathcal{O}_X$). Thus, if $F,G$ are of finite presentation, and $F_x \cong G_x$, then there are local sections of $\underline{\hom}(F,G)$ and $\underline{\hom}(G,F)$ at $x$ which compose to the identity in $\underline{\hom}(F,F)_x$ and $\underline{\hom}(G,G)_x$. Both equalities hold actually as local sections of $\underline{\hom}(F,F)$ and $\underline{\hom}(G,G)$ for some small open neighborhood $U$ of $x$. This shows $F|_U \cong G|_U$.

If we apply this to quasi-coherent sheaves on an affine scheme, we get that if $M_{\mathfrak{p}} \cong N_{\mathfrak{p}}$ for finitely presented modules over some commutative ring $A$ (not assumed to be noetherian) for some prime ideal $\mathfrak{p}$, then there is some $f \notin \mathfrak{p}$ with $M_f \cong N_f$. If more generally $M_S \cong N_S$ for some multiplicative subset $S$, then we may assume $0 \notin S$ and localize at a prime ideal $\mathfrak{p}$ disjoint from $S$. Then we are in the situation before.


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