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?


1 Answer 1


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 $(D\otimes_A N)/p = D/p\otimes_A N$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.