Lemma 15.83.8 (0DJG)—The Stacks project

Lemma 15.83.8. Let $R' \to A'$ be a flat ring map of finite presentation. Let $R' \to R$ be a surjective ring map whose kernel is a nilpotent ideal. Set $A = A' \otimes _{R'} R$. Let $K' \in D(A')$ and set $K = K' \otimes _{A'}^\mathbf {L} A$ in $D(A)$. If $K$ is $R$-perfect, then $K'$ is $R'$-perfect.

Proof. We can represent $K$ by a bounded above complex of finite free $A$-modules $E^\bullet $, see Lemma 15.64.5. By Lemma 15.75.3 we conclude that $K'$ is pseudo-coherent because it can be represented by a bounded above complex $P^\bullet $ of finite free $A'$-modules with $P^\bullet \otimes _{A'} A = E^\bullet $. Observe that this also means $P^\bullet \otimes _{R'} R = E^\bullet $ (since $A = A' \otimes _{R'} R$).

Let $I = \mathop{\mathrm{Ker}}(R' \to R)$. Then $I^ n = 0$ for some $n$. Choose $[a, b]$ such that $K$ has tor amplitude in $[a, b]$ as a complex of $R$-modules. We will show $K'$ has tor amplitude in $[a, b]$. To do this, let $M'$ be an $R'$-module. If $IM' = 0$, then

\[ K' \otimes _{R'}^\mathbf {L} M' = P^\bullet \otimes _{R'} M' = E^\bullet \otimes _ R M' = K \otimes _ R^\mathbf {L} M' \]

(because $A'$ is flat over $R'$ and $A$ is flat over $R$) which has nonzero cohomology only for degrees in $[a, b]$ by choice of $a, b$. If $I^{t + 1}M' = 0$, then we consider the short exact sequence

\[ 0 \to IM' \to M' \to M'/IM' \to 0 \]

with $M = M'/IM'$. By induction on $t$ we have that both $K' \otimes _{R'}^\mathbf {L} IM'$ and $K' \otimes _{R'}^\mathbf {L} M'/IM'$ have nonzero cohomology only for degrees in $[a, b]$. Then the distinguished triangle

\[ K' \otimes _{R'}^\mathbf {L} IM' \to K' \otimes _{R'}^\mathbf {L} M' \to K' \otimes _{R'}^\mathbf {L} M'/IM' \to (K' \otimes _{R'}^\mathbf {L} IM')[1] \]

proves the same is true for $K' \otimes _{R'}^\mathbf {L} M'$. This proves the desired bound for all $M'$ and hence the desired bound on the tor amplitude of $K'$. $\square$

Comments (3)

Comment #8792 by Shubhankar Sahai on February 04, 2024 at 00:18

I apologise if this is my misunderstanding, but this seems to be a useful technical lemma. Maybe one can add a slogan 'Perfect complexes lift along nilpotent immersions with same tor-amplitude'

Comment #8793 by Shubhankar Sahai on February 04, 2024 at 00:34

I guess that the slogan I suggested seems to be a bit stronger than what the lemma is doing (eg. as a $k[x]/x^e$ -module $k$ has unbounded tor-dimension). Maybe a better slogan is that 'perfectness and tor-amplitude can be checked after base change along a nilpotent surjection'

Comment #8795 by shubhankar on February 04, 2024 at 15:52

One final comment (which I learnt from Yuchen Wu, all mistakes are solely mine).

It seems that at least in the absolute case i.e. when $A'=R'$ and therefore $A=R$ then one can show that tor-dimension can be checked after base change to $R$ without the seemingly implicit pseudocoherence assumption (apologies if this is already obvious in the proof above).

The reason is that if $M'$ is an $R'$ -module so that $IM'=0$ then $M'$ is already an $R$ module. Therefore $M'=R\otimes^L_R M'$ .

Now the equation $K'\otimes^L_{R'}M'=K\otimes^L_{R}M'$ seems to hold without passing through the explicit resolutions $P$ and $E$ because $K'\otimes^L_{R'}M'=K'\otimes^L_{R'}(R\otimes^L_R M')=(K'\otimes^L_{R'}R)\otimes^L_RM'=K\otimes_R^L M'.$

Then one reasons as you do.

If this is fine, I would suggest adding this in tag 0651.

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