The Stacks project

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 $E \in D^+_\mathit{QCoh}(\mathcal{O}_ Y)$. 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 48.17.4 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 (3)

Comment #4669 by Bogdan on

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 ?

Comment #4676 by on

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 on

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


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