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In the appendix of this paper of Felix, Halperin and Thomas, Proposition $A.6$ is the following:

Let $R$ be a differential graded algebra, and let $M$ be an $R$-module. Then
$$\text{Tor}^R(\mathbb{k},M)^{\vee} \cong \text{ Ext}_R(\mathbb{k},M^{\vee}),$$ where $M^{\vee} = \text{Hom}(M,\mathbb{k})$ denotes the dual.

Where can I find a proof of this? Moreover, is this only true when $\mathbb{k}$ is a field, or is it also true over the integers $\mathbb{Z}$?

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    $\begingroup$ I am not too familiar with differential graded algebras, but I am getting strong Tensor- Hom adjunction vibes here. $\endgroup$ Commented Jun 3, 2020 at 20:06
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    $\begingroup$ I guess something like: start with a free resolution $P_{\bullet}\xrightarrow{\epsilon} k$, then $Tor(k,M) = H_{n}(P_{\bullet} \otimes M)$. Now dualising, we find $Hom(H_{n}(P_{\bullet}\otimes M),k) = H_{n}(Hom(P_{\bullet}\otimes M,k) \cong H_{n}(Hom(P_{\bullet},Hom(M,k))$, where the last isomorphism is the adjunction, and the last term is isomorphic to $Ext(k, M^{*})$. $\endgroup$ Commented Jun 3, 2020 at 20:21
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    $\begingroup$ One thing which requires explanation is why homology commutes with Hom, but I cannot think why that works off the top of my head (in case it does work). $\endgroup$ Commented Jun 3, 2020 at 20:25
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    $\begingroup$ @Locallyunskillful : well this $Hom$ is over $k$, which is a field (if you assume that it is), and it is therefore exact; in particular it commutes with homology $\endgroup$ Commented Jun 7, 2020 at 20:02

2 Answers 2

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It says in the attached document that all Homs and tensors are with respect to the field $\mathbb{k}$. As a module over itself $\mathbb{k}$ is injective so $\text{Hom}(-,\mathbb{k})$ commutes with homology - this is the crucial point.

For a proof, let $P\to M$ be a semiprojective resolution. Then there are isomorphisms $$ \begin{align*} \text{Ext}_{R}(\mathbb{k},N^{\vee})&= H(\text{Hom}_{R}(\mathbb{k},\text{Hom}_{\mathbb{k}}(P,\mathbb{k})))\,\, \text{by the definitions,}\\ &\simeq H(\text{Hom}_{\mathbb{k}}(\mathbb{k}\otimes_{R}P,\mathbb{k})) \,\, \text{by Hom-Tensor adjunction,} \\ &\simeq \text{Hom}_{\mathbb{k}}(H(\mathbb{k}\otimes P),\mathbb{k})\,\, \text{since $\mathbb{k}$ is self-injective}, \\ &=\text{Tor}_{R}(\mathbb{k},M)^{\vee}. \end{align*} $$ You could even replace the first $\mathbb{k}$ with any $R$-module and it would still hold.

In terms of references, you could use Proposition 12.10.12 (derived Hom-Tensor adjunction) in A. Yekuteili's book on derived categories, since this is related to DG modules over DGAs. Here is a proof along these lines, where $\mathbb{k}$ being self-injective gives an isomorphism $\text{Hom}_{\mathbb{k}}(-,\mathbb{k})\simeq \text{RHom}_{\mathbb{k}}(-,\mathbb{k})$ in $\textbf{D}(\mathbb{k})$:

$$ \begin{align*} \text{Ext}_{R}(\mathbb{k},N^{\vee})&\simeq H(\text{RHom}_{R}(\mathbb{k},\text{RHom}_{\mathbb{k}}(N,\mathbb{k}))) \\ &\simeq H(\text{RHom}_{\mathbb{k}}(\mathbb{k}\otimes_{R}^{\text{L}}N,\mathbb{k})) \\ &\simeq \text{Hom}_{\mathbb{k}}(H(\mathbb{k}\otimes_{R}^{\text{L}}N),\mathbb{k})\\ &\simeq\text{Tor}_{R}(\mathbb{k},N)^{\vee}. \end{align*} $$

Alternatively you could look at Section 10.8.2 of Weibel. This also looks at derived Hom-tensor adjunction, although you would need to be more careful since he only uses (partially) bounded complexes.

The result is also true over any ring if you don't care about grading. See, for example, Theorem 3.2.1 in Enochs and Jenda's Relative Homological Algebra, which says that if $A$ is an $R$-module, $B$ is an $(R,S)$-bimodule and $C$ is an injective $S$-module then there are isomorphisms $$\text{Ext}_{R}^{n}(A,\text{Hom}_{S}(B,C))\simeq \text{Hom}_{S}(\text{Tor}_{n}^{R}(A,B),C)$$ for all $n\geq 0$.

As you can see, the key is always the injectivity of $C$.

Edit: for the second question, which I somehow missed

In terms of whether this is true over the integers, the answer will, in general, be no because $\mathbb{Z}$ is not self-injective. I think I have a counterexample for the ring $\mathbb{Z}[x]$. It is clear that $\text{Hom}_{\mathbb{Z}}(\text{Tor}_{1}^{\mathbb{Z}[x]}(\mathbb{Z},\mathbb{Z}[x]),\mathbb{Z})=0$. On the other hand $\text{Hom}_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\simeq \mathbb{Z}[[x]]$, and then $$\text{Ext}_{\mathbb{Z}[x]}^{1}(\mathbb{Z},\mathbb{Z}[[x]])\simeq \mathbb{Z}[[x]]/(x)\simeq\mathbb{Z},$$ so the two sides are not equal. If there is an error in this, please let me know.

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  • $\begingroup$ Just a note; the last isomorphism mentioned can also be found in Rotman's homological algebra book (Corollary 10.63); the spectral sequence $\operatorname{Ext}^p_S(\operatorname{Tor}^R_q(A,B),C) \Rightarrow \operatorname{Ext}^{n}_R(A,\operatorname{Hom}_S(B,C))$ collapses due to the injectivity of $C$. $\endgroup$ Commented Aug 19, 2020 at 17:42
  • $\begingroup$ Wicked. Bounty awarded. $\endgroup$
    – Matt
    Commented Aug 20, 2020 at 10:20
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Not easy to find references in August. Anyway, the result, in the local case, is general enough and you can find a description of the argument, for example in http://homepages.math.uic.edu/~bshipley/huneke.pdf, Example 3.6, page 12. As you can see, the duality states that $$ Tor_R^i(M, N)^\vee \simeq Ext_i^R(M, N^\vee) $$ with $R$ local and Noetherian and $M$ and $N$ $R$-modules. I don't know if this gives you a hint for the graded case...

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  • $\begingroup$ Sorry - I have not had time to digest this yet - but to clarify, are you saying that the statement does hold and that the proof in your reference is a hint to that, or are you saying that you're genuinely not sure? :) $\endgroup$
    – Matt
    Commented Aug 18, 2020 at 23:23
  • $\begingroup$ Not sure yet. I keep looking. $\endgroup$
    – Daniel N
    Commented Aug 19, 2020 at 3:29
  • $\begingroup$ Okay, thank you for your answer. I won't mark this as an answer since it does not answer the right question, but this is very informative nonetheless :) $\endgroup$
    – Matt
    Commented Aug 19, 2020 at 7:26

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