$S^{-1}Ext^n_R(A,B) \cong Ext^n_{S^{-1}R}(S^{-1}A,S^{-1}B)$

Question

I am trying to understand a proof from the book An Introduction to Homological Algebra, Weibel.

Let $A$ be a finitely generated module over a noetherian ring $R$. Then for every multiplicative set $S$, all modules B, and all n, $S^{-1}Ext^n_R(A,B) \cong Ext^n_{S^{-1}R}(S^{-1}A,S^{-1}B)$.

PROOF:

Choose a resolution $F\longrightarrow A$ by finitely generated free $R$-modules. Then $S^{-1}F \longrightarrow S^{-1}A$ is a resolution by finitely generated free $S^{-1}R$-modules.

Because $S^{-1}$ is an exact functor from $R$-modules to $S^{-1}R$-modules, we have:

$S^{-1}Ext^n_R(A,B) = S^{-1}(H^* Hom_R(F,B)) \cong H^*(S^{-1}Hom_R(F,B)) \cong H^*Hom_{S^{-1}R}(S^{-1}F, S^{-1}B) = Ext^n_{S^{-1}R}(S^{-1}A,S^{-1}B)$

QUESTIONS:

1. I have seen in my lectures that $Ext_R^n(M,N)$ are the right derived functor cohomologies of some module $N$ with respect to $h^M$. And you need a complex $0 \longrightarrow M \longrightarrow I_0 \longrightarrow I_1 \longrightarrow ...$, where $I_i$ are injective objects, to compute it. But here he has a resolution $.... \longrightarrow F_0\longrightarrow A \longrightarrow 0$ (I mean, it goes the other way). Help?

2. I do not understand why $S^{-1}(H^* Hom_R(F,B)) \cong H^*(S^{-1}Hom_R(F,B))$. Help?

3. In case that $A$ is not finitely generated, is this true? I guess it is not. But how could I prove it?

4. Why do we need to consider the resolution $S^{-1}F \longrightarrow S^{-1}A$ if we only computate $Ext^n_R(A,B)$?

Thank you very much.

Answer 1

One of the things one learns about the $\mathrm{Ext}$ functor is that it is 'balanced', in the sense that to compute $\mathrm{Ext}^*(M,N)$ for two left $R$-modules $M$ and $N$, one can either resolve $M$ by projectives $P_*\to M$ or $N$ by injectives $N\to I^*$. What one then shows is that the two morphisms of chain complexes $\hom(P_*,N)\longleftarrow \hom(P_*,I^*) \longrightarrow \hom(M,I^*)$ obtained by pre- or post-composition with the augmentation maps for $M$ and $N$ are quasi-isomorphisms, so this shows the definitions agree.

Since localization is exact, it commutes with homology. A general fact of life is exact functors commute with homology. In this case, if $C$ is a complex of $R$-modules, then we have a map of chain complexes $S^{-1}H(C) \to H(S^{-1}C)$ that sends $s^{-1}[z]$ to $[s^{-1}z]$, and this is an isomorphism. This follos from the fact that if $F$ is flat then $F \otimes H(C) \to H(F\otimes C)$ is an isomorphism, for example by the Kunneth formula.

The finite generation is required to commute localization and homs, that is, there is in general a map $S^{-1}\hom_R(M,N) \to \hom_{S^{-1}R}(S^{-1}M,S^{-1}N)$ but this fails to be an isomorphism for $M$ and $N$ arbitrary. If $M$ is finitely presented, it is an isomorphism. In case $R$ is noetherian, this is the same as $M$ being finitely generated. Since Ext is Hom at 0, the general claim is false.

Finally, you want that resolution because it computes the right hand side of your formula, i.e. the Ext over the localized ring with arguments the localized modules.

Answer 2

1. The Ext defined using injectives is equivalent to the one defined using projectives, it is called balanced Ext.

2. There is a natural map $S^{-1}\operatorname{Hom}_R(M,N)\to \operatorname{Hom}_{S^{-1}R}(S^{-1}M,S^{-1}N)$ by $\frac{f}{s}\mapsto \left[\frac{x}{t}\mapsto \frac{f(x)}{st}\right]$. Note that when $M$ is free of finite rank, the map is isomorphism (see 3.). When $M$ is finitely presented (if $R$ is noetherian, then any finitely generated modules are finitely presented), Let $$R^{m}\to R^n\to M\to 0$$ Be right short exact sequence. Consider the following diagram $$\begin{array}{ccccccc} 0 & \to &S^{-1}\operatorname{Hom}_R(M,N)& \to & S^{-1}\operatorname{Hom}_R(R^n,N)&\to& S^{-1}\operatorname{Hom}_R(R^m,N)\\ &&\downarrow && \downarrow&& \downarrow\\ 0& \to &\operatorname{Hom}_{S^{-1}R}(S^{-1}M,S^{-1}N) & \to & \operatorname{Hom}_{S^{-1}R}(S^{-1}R^n,S^{-1}N)& \to & \operatorname{Hom}_{S^{-1}R}(S^{-1}R^{m},S^{-1}N) \end{array}$$ The right two columns of maps are isomorphic, thus so is the first one.

3. The problem occurs when it involves infinite products. For example, $$\operatorname{Hom}_{S^{-1}R}(S^{-1}R^{\oplus \Lambda},S^{-1}N)=(S^{-1}N)^{\Lambda}\qquad S^{-1}\operatorname{Hom}_R(R^{\oplus \Lambda},N)=S^{-1}\big(N^{ \Lambda}\big)$$ They are quite different for infinite $\Lambda$. For example, $R=\mathbb{Z}=N$, and $S=\{1,2,4,8\ldots \}$ and $\Lambda= \mathbb{Z}_{\geq 0}$, then there is no element of $S^{-1}(N^{\Lambda})$ corresponding to $$\left(1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots \right)\in (S^{-1}N)^{\Lambda}$$

4. We didn't, you need to check the proof again.