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Fix a prime number $p$. Let $A$ be a commutative ring, and consider an $A$-algbera $B$ of characteristic $p$. So we have a sequence of ring homomorphisms $$ A \to A/pA \to B. $$ Assume that we want to show that $B$ has flat dimension $\leq n$ (i.e., the $A$-module $B$ has Tor amplitude in $[-n,0]$). Then, is it enough to check the following condition (*)?

(*) $B$ has $p$-complete Tor amplitude in $[-n,0]$

Here (*) means, by definition (ref. Bhatt-Morrow-Scholze "Topological Hochschild homology and integral $p$-adic Hodge theory"), that the derived tensor product $B\otimes_A^L A/pA \in D(A/pA)$ has Tor amplitude in $[-n,0]$. That is, $B\otimes_A N \in D^{[-n,0]}(A/pA)$ for any $A/pA$-module $N$. On the other hand, our aim is to show that $B\otimes_A M \in D^{[-n,0]}(A)$ for any $A$-module $M$. Is this true?

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(You have to replace every usual tensor product $\otimes_A$ by derived tensor product $\otimes_A^L$ in the question.)

Yes. The key point is the $A/p$-module structure on $B$, which induces an $A/p$-structure on $B\otimes_A^LM\in D(A)$. In particular, the multiplication map by $p$ is zero on $B\otimes_A^L M$ in $D(A)$, and thus

$$B\otimes_A^L(M/^Lp)\simeq(B\otimes_A^LM)/^Lp\simeq B\otimes_A^LM\oplus B\otimes_A^LM[1]$$

in $D(A)$ (not in $D(A/p)$), where $N/^Lp:=\operatorname{cofib}(N\xrightarrow pN)$ for every $N\in D(A)$. By assumption, the left hand side is concentrated in (cohomological) degrees $[-(n+1),0]$ (since $M/^Lp\in D(A/p)$ is concentrated in degrees $[-1,0]$), and thus $B\otimes_A^LM$ is concentrated in degrees $[-(n+1),0]\cap[-n,1]=[-n,0]$.

The ring structure on $A$ is not used, and thus it can be extended to any $A/p$-module $B$, or even more generally, any $p$-torsion $A$-module of bounded $p^\infty$-torsion.

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  • $\begingroup$ Thank you. I'm not familiar with infinity categories, and so I'm not quite sure what the derived quotient is. But it seems to be the mapping cone in this situation. Even if so, I don't understand how to use the assumption: why "the left hand side is concentrated in (cohomological) degrees [βˆ’(𝑛+1),0]"? $\endgroup$
    – Zuka
    Commented Dec 8 at 2:51
  • $\begingroup$ @Zuka No infinity category is involved β€” here cofib is given by the triangulated structure. For the second, you first think of a simplified setup: If you take the derived tensor product of A with another module M, then this lies in the Tor-amplitude of M. In general, if M is any object concentrated in bounded degrees, then the tensor product is concentrated in the degree "amplified" by the Tor-amplitude of A. This is not hard to see, say, from induction of the "length" of the degree range of M. What we want is slightly different, but the same argument applies. $\endgroup$
    – Z. M
    Commented Dec 8 at 20:10
  • $\begingroup$ I understand. I’m sorry, I have another question. Why "𝑀/πΏπ‘βˆˆπ·(𝐴/𝑝) is concentrated in degrees [βˆ’1,0]"? I found that the cohomologies of 𝑀/𝐿𝑝 (I think of it as the mapping cone of p:𝑀→𝑀) is killed by p. But I doubt that this implies that 𝑀/πΏπ‘βˆˆπ·^[-1,0](𝐴/𝑝). $\endgroup$
    – Zuka
    Commented Dec 9 at 11:10
  • $\begingroup$ @Zuka I said that it is in D(A), not in D(A/p). This one is easy, and the vanishing is not used: you look at the long exact sequence associated to the distinguished triangle (M to M to M/p), and see that only nonzero cohomologies are -1 and 0. $\endgroup$
    – Z. M
    Commented Dec 9 at 20:42
  • $\begingroup$ Yes, I understand that 𝑀/𝐿𝑝, as an object of "𝐷(𝐴)", is concentrated in degrees [βˆ’1,0]. I thought it is in D(A/p), because otherwise we cannot use the assumption to conclude "the left hand side is concentrated in (cohomological) degrees [βˆ’(𝑛+1),0]". Do you claim that M/Lp is quasi-isomorphic to a complex of "A/p-modules" concentrated in [-1,0]? $\endgroup$
    – Zuka
    Commented Dec 10 at 3:52

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