# derived completion and flat base change

Let $$f:A \to B$$ be a flat morphism of commutative $$p$$-adic completely rings. We denote by $$D_{\text{comp}}(A)$$ the derived category of complexes over $$A$$, which is derived $$p$$-adic complete.

For a complex $$L$$ over $$A$$, there is a natural morphism $$(L \otimes^{\mathbb{L}}_{A} B)^{\wedge}_p \to (L^{\wedge}_p \otimes^{\mathbb{L}}_{A} B)^{\wedge}_p,$$ where $$L^{\wedge}_p$$ is the derived $$p$$-adic completion of $$L$$.

My question: When is the morphism isomorphism?

It is always a (quasi-)isomorphism.

In fact, because you are deriving the tensor product you do not need to assume $$B$$ is flat, and you do not need any assumptions on $$A,B,f$$, or even to assume that $$B$$ is a ring.

The general statement is that if $$M,N$$ are complexes over $$A$$, then $$(M\otimes^L_A N)^\wedge_p\to (M^\wedge_p\otimes^L_A N)^\wedge_p$$ is a quasi-isomorphism.

I'll write a proof below, and remove the $$L$$'s from the tensor product, but I still mean derived.

Lemma: A map $$f:M\to N$$ induces a quasi-isomorphim $$M^\wedge_p\to N^\wedge_p$$ is and only if it induces a quasi-isomorphism $$M/p\to N/p$$, where $$-/p$$ means derived cokernel.
Proof: One direction follows from the fact that $$M\to M^\wedge_p$$ induces a quasi-isomorphism after $$-/p$$.

For the other direction, by induction and using the natural cofiber sequence $$M/p\to M/p^n\to M/p^{n-1}$$, one sees that $$M/p^n\to N/p^n$$ is a quasi-isomorphism, and therefore so is $$M^\wedge_p=\lim_n M/p^n\to \lim_n N/p^n=N^\wedge_p$$ , where $$\lim_n$$ is implicitly derived.

Now the general statement follows immediately: you need to prove that $$(M\otimes_A N)/p\to (M^\wedge_p\otimes_A N)/p$$ is an quasi-iso, but the exactness of $$\otimes_A$$ implies that $$(D\otimes_A N)/p = D/p\otimes_A N$$.