4
$\begingroup$

Let $R$ be a ring and $\text{Mod}\,R$ the category of $R$ modules. For two $R$-modules $X,Y$ one can define $\text{Ext}_R^n(X,Y)$ as follows. We take an injective resolution $0\rightarrow Y\rightarrow I_0 \rightarrow I_1 \rightarrow \dots,$ throw away $Y$ and apply $\text{Hom}_R(X,-)$ to obtain the cochain complex $0 \rightarrow \text{Hom}_R(X,I_0) \rightarrow \text{Hom}_R(X,I_1) \rightarrow \dots,$ denoted by $C_Y$. Then, we define $\text{Ext}_R^n(X,Y)$ as the $n$'th cohomology of $C_Y$. The horseshoe lemma shows, that a short exact sequence $0\rightarrow L \rightarrow M \rightarrow N \rightarrow 0$ of $R$-modules induces a short exact sequence of cochain complexes $0\rightarrow C_L \rightarrow C_M \rightarrow C_N \rightarrow 0,$ which induces a long exact sequence $0\rightarrow \text{Ext}^0_R(X,L) \rightarrow \text{Ext}^0_R(X,M)\rightarrow \text{Ext}^0_R(X,N)\rightarrow \text{Ext}^1_R(X,L)\rightarrow \text{Ext}^1_R(X,M)\rightarrow \text{Ext}^1_R(X,N) \rightarrow ...,$ where $\text{Ext}^0_R(X,Y) = \text{Hom}_R(X,Y)$.

Now my question is: What happens, if we do the same with a projective resolution?

That is, we take a projective resolution $\dots \rightarrow P_1\rightarrow P_0 \rightarrow Y \rightarrow 0$ throw away $Y$ and apply $\text{Hom}_R(X,-)$ to obtain the chain complex $\dots \rightarrow \text{Hom}_R(X,P_1) \rightarrow \text{Hom}_R(X,P_0) \rightarrow 0,$ denoted by $D_Y$. Then we define $F^n (Y)$ as the $n$'th homology of $D_Y$. Again the horseshoe lemma shows, that a short exact sequence $0\rightarrow L \rightarrow M \rightarrow N \rightarrow 0$ of $R$-modules induces a short exact sequence of chain complexes $0\rightarrow D_L \rightarrow D_M \rightarrow D_N \rightarrow 0$. In fact, we only need that $\text{Hom}_R(X,-)$ is an additive functor for that. The short exact sequence of chain complexes induces a long exact sequence $\dots \rightarrow F^1(L) \rightarrow F^1(M)\rightarrow F^1(N)\rightarrow F^0(L) \rightarrow F^0(M)\rightarrow F^0(N)\rightarrow 0$. However, this time we have $F^0(Y) \neq \text{Hom}_R(X,Y)$ in general. We had equality before because $\text{Hom}_R(X,-)$ is left-exact but this time we would need right-exactness. Nevertheless, one can show that this construction induces a natural transformation $F^0 \rightarrow \text{Hom}_R(X,-)$.

Now my question is: Can we describe $F^n$ better? In particular for $n=0$? Is this contrsuction useful in any way?

$\endgroup$
5
  • $\begingroup$ Why is this well-defined? $\endgroup$ – John Palmieri Apr 11 at 15:13
  • $\begingroup$ It should be well-defined, because two different projective resolutions are homotopy equivalent, so applying $\text{Hom}(X,-)$ should yield again two homotopy equivalent chain complexes since the functor is additive. $\endgroup$ – kevkev1695 Apr 11 at 15:21
  • 1
    $\begingroup$ The widely used dual functor of $\text{Ext}$ is $\text{Tor}$, which differs from your definition by changing $\text{Hom}_R(X,-)$ into $-\otimes_{R}X$ and swaps $X$ and $Y$. $\endgroup$ – Zerox Apr 11 at 15:40
  • $\begingroup$ Yes, in this case $F^0 $ equals the tensor product again, because it is right-exact. $\endgroup$ – kevkev1695 Apr 11 at 15:41
  • $\begingroup$ Similar question (without answer): mathoverflow.net/questions/387352/… $\endgroup$ – Jochen Wengenroth Apr 12 at 5:24
8
$\begingroup$

Suppose for now that $X$ is finitely presented. Set $X^\vee = \operatorname{Hom}_R(X,R)$, so that there is a natural transformation $X^\vee\otimes_R M\to \operatorname{Hom}_R(X,M)$ which is an isomorphism for $M$ a projective $R$-module: For finite rank free modules, this is immediate, for general free modules this follows since both sides commute with filtered colimits (this uses the finite presentation hypothesis), and for general projective modules this follows since they are retracts of free modules. Thus the complex $D_Y(X)$ is isomorphic to the complex $...\to X^\vee\otimes_R P_1\to X^\vee\otimes_R P_0\to 0$, so that its homology is given by $F^n(X) = \operatorname{Tor}^n_R(X^\vee,Y)$.

A general $R$-module $X$ is the filtered colimit of the diagram of maps from a finitely presented module $f:X'\to X$. Pulling out the colimit gives an isomorphism of chain complexes $D_Y(X)\cong\varprojlim_{f:X'\to X,X'\text{ f.p.}} D_Y(X')$. There is a map $H_n(D_Y(X))\to \varprojlim_{f:X'\to X,X'\text{ f.p.}} F^n(X')$; however, as Denis-Charles Cisinski mentions filtered limits are not exact, so that this map does not have to be an isomorphism. The map to $\operatorname{Hom}_R(X,Y)$ factors through it via the limit of the maps $(X')^\vee\otimes_R Y\to\operatorname{Hom}(X',Y)$.

$\endgroup$
3
  • 4
    $\begingroup$ The case where $X$ is of finite presentation looks good, but aren't you optimistic about the general case? The filtered colimit is pulled out as a cofiltered limit, and limits are not exact, so that I do not see why your general description of $F^n$ is valid. $\endgroup$ – Denis-Charles Cisinski Apr 11 at 20:06
  • $\begingroup$ Perhaps there are a few intended extra hypotheses, like that $R$ is self-injective and X has only countably many f.g. submodules. Then you show that $R^n lim_{X'\subseteq X f.g.} \hom_R(X,R)$ vanishes, for $n>0$, by reducing the countable cofiltered limit to a limit over a countable infinite ordinal, i.e., a sequential lim, where the Mittag-Leffler condition (which you get if R is self-injective, since the inclusions of submodules being monic) ensures that the limit is acyclic. This is just an off-the-cuff idea which I did not try to write out, so perhaps there is a foolish mistake in it. $\endgroup$ – A.S. Apr 11 at 20:53
  • 1
    $\begingroup$ In the case where $R$ is noetherian and $Y$ is of finite type, one may choose each $P_i$ of finite type, so that the functors $(-)\otimes_R P_i$ commute with small limits. This imply that the Formula $F^n(X)\cong \mathrm{Tor}^n_R(X^\vee,Y)$ will still be true for arbitrary $X$ under these extra finiteness hypothesises on $R$ and $Y$. $\endgroup$ – Denis-Charles Cisinski Apr 13 at 23:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.