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.



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


