Let $A$ be a finite-dimensional $k$-algebra, $\def\H{\operatorname{Hom}} D=\H_k(-,k)$. How to show that $ R\H_A (DA,A) \cong D(DA \otimes^{\mathbb{L}}_A DA)$?
$D(DA \otimes^{\mathbb{L}}_A DA) = \H_k(DA \otimes^{\mathbb{L}}_A DA, k) =\!\!\!\!? \ \H_k(DA,R\H_A(DA,k)) = ?$
Thank you.
By Lemma 15.72.2 of https://stacks.math.columbia.edu/tag/0A5W, if we choose any free finitely generated resolution $A_i$ of $DA$ (with differentials $A_i\rightarrow A_{i-1}$, $i\geq 0$, $A_{-1}=DA$), $RHom_A(DA,A)$ is represented by the complex $Hom(A_i,A)$ (where the differentials go from $i \geq 0$ to $i+1$ and the complex is zero in negative degree).
That’s always possible because by working recursively, we can ensure that all the involved kernels have finite dimension over $k$ and are thus finitely generated over $A$.
By Definition 15.58.15 in https://stacks.math.columbia.edu/tag/06XY, $DA \otimes_A^{\mathbb{L}} DA$ is represented by the complex $DA \otimes A_i$ (differentials from $i$ to $i-1$, zero in negative degree), and thus $D(DA \otimes_A^{\mathbb{L}} DA)$ is represented by the complex $D(DA \otimes_A A_i)$ (differentials from $i$ to $i+1$, zero in negative degree).
So all that remains to be proved is that if $M$ is a free $A$-module of finite rank, there is a natural isomorphism $D(DA \otimes_A M) \rightarrow Hom_A(M,A)$. It is enough to show that there is a perfect natural $k$-duality $\langle \cdot,\, \cdot\rangle: (DA \otimes_A M) \times Hom_A(M,A) \rightarrow k$.
We define it as $(\alpha \otimes m,f) \longmapsto \alpha(f(m))$. Note that the two spaces involved have the same finite dimension over $k$, so it’s enough to show that if $\langle \cdot,\, f\rangle=0$, then $f=0$.
Now, let $f: M \rightarrow A$ be such that $\langle \cdot,\, f \rangle=0$. This means that for every $m \in M$, $f(m)$ is in the kernel of every $\alpha \in DA$. But $A$ is a $k$-vector space, so the intersection of all the kernels of the elements of $DA$ is zero, so $f=0$, and we are done.