Lemma 48.18.3 (0BZY): Makeshift base change

Table of contents
Part 3: Topics in Scheme Theory
Chapter 48: Duality for Schemes
Section 48.18: Base change for upper shriek
Lemma 48.18.3: Makeshift base change (cite)

Lemma 48.18.3 (Makeshift base change). In Situation 48.16.1 let

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]

be a cartesian diagram of $\textit{FTS}_ S$. Let $E \in D^+_\mathit{QCoh}(\mathcal{O}_ Y)$ be an object such that $Lg^*E$ is in $D^+(\mathcal{O}_ Y)$. If $f$ is flat, then $L(g')^*f^!E$ and $(f')^!Lg^*E$ restrict to isomorphic objects of $D(\mathcal{O}_{U'})$ for $U' \subset X'$ affine open mapping into affine opens of $Y$, $Y'$, and $X$.

Proof. By our assumptions we immediately reduce to the case where $X$, $Y$, $Y'$, and $X'$ are affine. Say $Y = \mathop{\mathrm{Spec}}(R)$, $Y' = \mathop{\mathrm{Spec}}(R')$, $X = \mathop{\mathrm{Spec}}(A)$, and $X' = \mathop{\mathrm{Spec}}(A')$. Then $A' = A \otimes _ R R'$. Let $E$ correspond to $K \in D^+(R)$. Denoting $\varphi : R \to A$ and $\varphi ' : R' \to A'$ the given maps we see from Remark 48.17.5 that $L(g')^*f^!E$ and $(f')^!Lg^*E$ correspond to $\varphi ^!(K) \otimes _ A^\mathbf {L} A'$ and $(\varphi ')^!(K \otimes _ R^\mathbf {L} R')$ where $\varphi ^!$ and $(\varphi ')^!$ are the functors from Dualizing Complexes, Section 47.24. The result follows from Dualizing Complexes, Lemma 47.24.6. $\square$

Comments (7)

Comment #4669 by Bogdan on November 14, 2019 at 22:19

Can't we use https://stacks.math.columbia.edu/tag/0E2W, https://stacks.math.columbia.edu/tag/0E9W and https://stacks.math.columbia.edu/tag/0B6U to pin down an isomorphism $L(g′)^\ast \circ f^!\to (f′)^!\circ Lg^\ast$ ?

Comment #4676 by Johan on November 15, 2019 at 06:59

Yes, we can. This takes a lot more work though and it is very hard to use the extra information (about canonicalness of the identifications) obtained in this manner. But maybe we should update the text with a comment saying what you said.

Comment #4800 by Johan on December 10, 2019 at 11:01

Thanks again. I have added a bit of discussion. See here.

Comment #5101 by Bogdan on May 15, 2020 at 03:15

I think one has to assume that $Lg^*E\in D^+(\mathcal O_{Y})$ since otherwise $(f')^!Lg^*E$ is not well-defined.

Comment #5102 by Bogdan on May 15, 2020 at 03:15

I think one has to assume that $Lg^*E\in D^+(\mathcal O_{Y'})$ since otherwise $(f')^!Lg^*E$ is not well-defined.

Comment #5103 by Johan on May 15, 2020 at 09:52

Argh! It seems you are correct. This just is a terrible lemma. Luckily it seems we only use it once!

Comment #5310 by Johan on June 17, 2020 at 10:05

Thanks and fixed here.

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