Ext and Tor duality

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 $\mathbb{Z}$?

Solution 1:

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.