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. Observe that $A' \to A$ has nilpotent kernel and that by flatness of $R' \to A'$ we have $K = K' \otimes _{R'}^\mathbf {L} R$ (see Section 15.61). Hence the lemma follows by combining Lemmas 15.75.4 and 15.66.20. $\square$


Comments (4)

Comment #8792 by Shubhankar Sahai on

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

I guess that the slogan I suggested seems to be a bit stronger than what the lemma is doing (eg. as a -module 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

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 and therefore then one can show that tor-dimension can be checked after base change to without the seemingly implicit pseudocoherence assumption (apologies if this is already obvious in the proof above).

The reason is that if is an -module so that then is already an module. Therefore .

Now the equation seems to hold without passing through the explicit resolutions and because

Then one reasons as you do.

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

Comment #9289 by on

OK, this is such a technical lemma that a slogan does not work well, I think.

I like your and Yuchen Wu's suggestion of the additional lemma. I have added it here.


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