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I'm having trouble with this exercise from Elements of the Representation Theory of Associative Algebras I: Techniques of Representation Theory.

The exercise in question is from chapter IV.

So, let $k$ be an algebraically closed field and let $A$ be a finite-dimensional $k$-algebra, and $M$ a left $A$-module containing no projective direct summands.

Say $F = \underline{Hom}_{A^{op}}(M, -) $ and $G = Tor_1^{A^{op}}(M, -)$. The exercise asks to prove that $F \cong G$.

Now I have been thinking about this exercise for a while, and I have no idea how to prove it.

One thing to note, the functor $F$ is the projective stabilization of the $Hom$ functor, and it takes a module $N$ and sends it to the module $Hom(M, N) / \mathcal{P}(M, N)$, where $\mathcal{P}(M, N)$ is the submodule consisting of all morphisms $f: M \rightarrow N$ that factor through a projective module.

For the life of me, I don't know how to prove this. I have tried to construct a surjective morphism from $Hom(M, N)$ to $Tor(M, N)$ having kernel $\mathcal{P}(M, N)$. This seems like sort of a lot of work to build up $Tor(M, N)$ from scratch and then figuring out where to send a morphism $f$. And then there would still be the problem of showing that for some $g: N \rightarrow N'$ we have a commutative square which I can't quite format online using the following two maps as the top and bottom lines of the square.

$$Hom(M, N) / \mathcal{P}(M, N) \rightarrow Tor(M, N)$$ $$Hom(M, N') / \mathcal{P}(M, N') \rightarrow Tor(M, N')$$

Thanks!

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  • $\begingroup$ $M$ is not a right $A$-module, so something is wrong with the definition of the functor $F$. $\endgroup$
    – Dag Oskar Madsen
    Commented Mar 16, 2014 at 3:01
  • $\begingroup$ Hmm. Is there some way of fixing the setup so that F will be correctly defined? Is there a way of identifying $A^{op}$ and its modules with $A$ and its modules? Like, every left $A$-module is a right $A^{op}$-module? $\endgroup$
    – Samantha Y
    Commented Mar 16, 2014 at 14:44
  • $\begingroup$ It is probably a misprint. There is a functorial isomorphism $\underline{\operatorname{Hom}}_{A^{\rm{op}}}(\operatorname{Tr} M,-) \cong \operatorname{Tor}_1^{A^{\rm{op}}}(M,-)$. It could be what is meant. $\endgroup$
    – Dag Oskar Madsen
    Commented Mar 16, 2014 at 15:34
  • $\begingroup$ Ah that makes a bit more sense, since earlier in the chapter they spend some time trying to find a duality between mod $A$ and mod $A^{op}$ and instead settling for a duality between $\underline{mod}_{A}$ and $\underline{mod}_{A^{op}}$ induced by $M \mapsto$ Tr $M$. sorry for the sloppy latex-ing. I'll think about the question more, it should be a little less opaque if I actually have the correct functors. Thanks! $\endgroup$
    – Samantha Y
    Commented Mar 16, 2014 at 16:25

1 Answer 1

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Here is a proof of the functorial isomorphism $$\underline{\operatorname{Hom}}_{A^{\rm{op}}}(\operatorname{Tr} M,-) \cong \operatorname{Tor}_1^{A^{\rm{op}}}(M,-).$$

Since $M$ has no projective summands, we can choose a right $A$-module $X$ without projective summands such that $M \cong \operatorname{Tr} X$ and $X \cong \operatorname{Tr} M$, where $\operatorname{Tr}$ denotes the transpose. For ease of notation let $B=A^{\rm{op}}$.

Since $X$ has no projective summands, the Auslander-Reiten formula gives a functorial isomorphism $$\underline{\operatorname{Hom}}_B(X,-) \cong D \operatorname{Ext}^1_B(-,D \operatorname{Tr} X),$$ where $D$ denotes the duality $D=\operatorname{Hom}_k(-,k)$. For any right $B$-module $U$, there is an isomorphism $$D\operatorname{Tor}_1^B(U,-) \cong \operatorname{Ext}^1_B(-,DU).$$ In our case we get $$\underline{\operatorname{Hom}}_B(X,-) \cong D \operatorname{Ext}^1_B(-,D \operatorname{Tr} X) \cong \operatorname{Tor}_1^B(\operatorname{Tr} X,-),$$ which finishes the proof.

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  • $\begingroup$ We can let $X=\operatorname{Tr} M$. That part could have been formulated better. $\endgroup$
    – Dag Oskar Madsen
    Commented Mar 16, 2014 at 19:54
  • $\begingroup$ Do you have a reference for the proof of the isomorphism involving Ext and Tor? $\endgroup$
    – Samantha Y
    Commented Mar 16, 2014 at 21:59
  • $\begingroup$ Interestingly, I checked a paper by Auslander, Representation theory of artin algebras III Almost Split Sequences and he shows directly that $\underline{Hom}(M, - ) \cong Tor_1(Tr M, -)$ $\endgroup$
    – Samantha Y
    Commented Mar 17, 2014 at 1:15
  • $\begingroup$ In the edition I have of Rotman, An introduction to homological algebra the Ext-Tor-isomorphism is Theorem 9.51. Keep in mind that $k$ is an injective $k$-module. $\endgroup$
    – Dag Oskar Madsen
    Commented Mar 17, 2014 at 9:23
  • $\begingroup$ If you know 2 out of 3 of the Auslander-Reiten formula, the Ext-Tor-isomorphism, and the isomorphism in this question, you can always prove the third. $\endgroup$
    – Dag Oskar Madsen
    Commented Mar 17, 2014 at 9:29

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