## 48.18 Base change for upper shriek

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 in $\textit{FTS}_ S$ such that $X$ and $Y'$ are Tor independent over $Y$. Our setup is currently not sufficient to construct a base change map $L(g')^* \circ f^! \to (f')^! \circ Lg^*$ in this generality. The reason is that in general it will not be possible to choose a compactification $j : X \to \overline{X}$ over $Y$ such that $\overline{X}$ and $Y'$ are tor independent over $Y$ and hence our construction of the base change map in Section 48.5 does not apply1.

A partial remedy will be found in Section 48.28. Namely, if the morphism $f$ is flat, then there is a good notion of a relative dualizing complex and using Lemmas 48.28.9 48.28.6, and 48.17.8 we may a canonical base change isomorphism. If we ever need to use this, we will add precise statements and proofs later in this chapter.

Lemma 48.18.1. 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 $\textit{FTS}_ S$ with $g$ flat. Then there is an isomorphism $L(g')^* \circ f^! \to (f')^! \circ Lg^*$ on $D_\mathit{QCoh}^+(\mathcal{O}_ Y)$.

Proof. Namely, because $g$ is flat, for every choice of compactification $j : X \to \overline{X}$ of $X$ over $Y$ the scheme $\overline{X}$ is Tor independent of $Y'$. Denote $j' : X' \to \overline{X}'$ the base change of $j$ and $\overline{g}' : \overline{X}' \to \overline{X}$ the projection. We define the base change map as the composition

$L(g')^* \circ f^! = L(g')^* \circ j^* \circ a = (j')^* \circ L(\overline{g}')^* \circ a \longrightarrow (j')^* \circ a' \circ Lg^* = (f')^! \circ Lg^*$

where the middle arrow is the base change map (48.5.0.1) and $a$ and $a'$ are the right adjoints to pushforward of Lemma 48.3.1 for $\overline{X} \to Y$ and $\overline{X}' \to Y'$. This construction is independent of the choice of compactification (we will formulate a precise lemma and prove it, if we ever need this result).

To finish the proof it suffices to show that the base change map $L(g')^* \circ a \to a' \circ Lg^*$ is an isomorphism on $D_\mathit{QCoh}^+(\mathcal{O}_ Y)$. By Lemma 48.4.4 formation of $a$ and $a'$ commutes with restriction to affine opens of $Y$ and $Y'$. Thus by Remark 48.6.1 we may assume that $Y$ and $Y'$ are affine. Thus the result by Lemma 48.6.2. $\square$

Lemma 48.18.2. In Situation 48.16.1 let $f : X \to Y$ be an étale morphism of $\textit{FTS}_ S$. Then $f^! \cong f^*$ as functors on $D^+_\mathit{QCoh}(\mathcal{O}_ Y)$.

Proof. We are going to use that an étale morphism is flat, syntomic, and a local complete intersection morphism (Morphisms, Lemma 29.35.10 and 29.35.12 and More on Morphisms, Lemma 37.54.8). By Lemma 48.17.8 it suffices to show $f^!\mathcal{O}_ Y = \mathcal{O}_ X$. By Lemma 48.17.10 we know that $f^!\mathcal{O}_ Y$ is an invertible module. Consider the commutative diagram

$\xymatrix{ X \times _ Y X \ar[r]_{p_2} \ar[d]_{p_1} & X \ar[d]^ f \\ X \ar[r]^ f & Y }$

and the diagonal $\Delta : X \to X \times _ Y X$. Since $\Delta$ is an open immersion (by Morphisms, Lemmas 29.34.13 and 29.35.5), by Lemma 48.17.1 we have $\Delta ^! = \Delta ^*$. By Lemma 48.16.3 we have $\Delta ^! \circ p_1^! \circ f^! = f^!$. By Lemma 48.18.1 applied to the diagram we have $p_1^!\mathcal{O}_ X = p_2^*f^!\mathcal{O}_ Y$. Hence we conclude

$f^!\mathcal{O}_ X = \Delta ^!p_1^!f^!\mathcal{O}_ Y = \Delta ^*(p_1^*f^!\mathcal{O}_ Y \otimes p_1^!\mathcal{O}_ X) = \Delta ^*(p_2^*f^!\mathcal{O}_ Y \otimes p_1^*f^!\mathcal{O}_ Y) = (f^!\mathcal{O}_ Y)^{\otimes 2}$

where in the second step we have used Lemma 48.17.8 once more. Thus $f^!\mathcal{O}_ Y = \mathcal{O}_ X$ as desired. $\square$

In the rest of this section, we formulate some easy to prove results which would be consequences of a good theory of the base change map.

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 48.17.4 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. $\square$

Lemma 48.18.4. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Assume $f$ is flat. Set $\omega _{X/Y}^\bullet = f^!\mathcal{O}_ Y$ in $D^ b_{\textit{Coh}}(X)$. Let $y \in Y$ and $h : X_ y \to X$ the projection. Then $Lh^*\omega _{X/Y}^\bullet$ is a dualizing complex on $X_ y$.

Proof. The complex $\omega _{X/Y}^\bullet$ is in $D^ b_{\textit{Coh}}$ by Lemma 48.17.8. Being a dualizing complex is a local property. Hence by Lemma 48.18.3 it suffices to show that $(X_ y \to y)^!\mathcal{O}_ y$ is a dualizing complex on $X_ y$. This follows from Lemma 48.17.6. $\square$

 The reader who is well versed with derived algebraic geometry will realize this is not a “real” problem. Namely, taking $\overline{X}'$ to be the derived fibre product of $\overline{X}$ and $Y'$ over $Y$, one can argue exactly as in the proof of Lemma 48.18.1 to define this map. After all, the Tor independence of $X$ and $Y'$ guarantees that $X'$ will be an open subscheme of the derived scheme $\overline{X}'$.

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