Is the hom in derived category of a dg-algebra compatible with base field extension?

Question

Let $k$ be a field and $A$ be an ordinary $k$-algebra. Let $M$ and $N$ be two left $A$-modules then we could consider the Ext group $\mathrm{Ext}^i_A(M,N)$ which is actually a $k$-vector space. Let $l/k$ be a field extension and we consider $A_l$, $M_l$, and $N_l$ be the base-change algebras and modules, i.e. $A_l=A\otimes_kl$, etc. We can also consider $\mathrm{Ext}^i_{A_l}(M_l,N_l)$ and we have the natural map $$ \mathrm{Ext}^i_A(M,N)\otimes_kl\to \mathrm{Ext}^i_{A_l}(M_l,N_l).$$ We know it is an isomorphism if $M$ is a finite present $A$-module.

Now let $A$ be a differential graded (dg) $k$-algebra and consider the derived category of dg-$A$-modules $D(A)$. Let $M$ and $N$ be two objects in $D(A)$. For a field extension $l/k$ we can also form $A_l$, $M_l$, and $N_l$ in the similar way.

My question is

Is the natural map $$ Hom_{D(A)}(M,N)\otimes_kl\to Hom_{D(A_l)}(M_l,N_l) $$ an isomorphism for $M$ a perfect complex of dg-$A$-modules and arbitrary $N$?

It seems to be similar to the above result but I cannot find any reference.

Answer 1

Yes.

The full subcategory consisting of those $M$ for which this natural map is an isomorphism (for all $N$) is a thick subcategory by a five lemma argument, and the thick subcategory generated by $A$ is precisely the category of perfect complexes, so it is sufficient to prove it for $M=A$.

But $\text{Hom}_{D(A)}(A,N)=H^0(N)$, so for $M=A$ the map is the natural map $$H^0(N)\otimes_kl\to H^0(N_l),$$ which is an isomorphism.