The Stacks project

Lemma 76.54.2. Let $S$ be a scheme. Let $\{ g_ i : Y_ i \to Y\} $ be an fpqc covering of algebraic spaces over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces and set $X_ i = Y_ i \times _ Y X$ with projections $f_ i : X_ i \to Y_ i$ and $g'_ i : X_ i \to X$. Let $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$. Let $a, b \in \mathbf{Z}$. Then the following are equivalent

  1. $E$ has tor amplitude in $[a, b]$ as an object of $D(f^{-1}\mathcal{O}_ Y)$, and

  2. $L(g'_ i)^*E$ has tor amplitude in $[a, b]$ as a object of $D(f_ i^{-1}\mathcal{O}_{Y_ i})$ for all $i$.

Also true if “tor amplitude in $[a, b]$” is replaced by “locally finite tor dimension”.

Proof. Pullback preserves “tor amplitude in $[a, b]$” by Derived Categories of Spaces, Lemma 75.20.7 Observe that $Y_ i$ and $X$ are tor independent over $Y$ as $Y_ i \to Y$ is flat. Let us assume (2) and prove (1). We can compute tor dimension at stalks, see Cohomology on Sites, Lemma 21.46.10 and Properties of Spaces, Theorem 66.19.12. Let $\overline{x}$ be a geometric point of $X$. Choose an $i$ and a geometric point $\overline{x}_ i$ in $X_ i$ with image $\overline{x}$ in $X$. Then

\[ (L(g_ i')^*E)_{\overline{x}_ i} = E_{\overline{x}} \otimes _{\mathcal{O}_{X, \overline{x}}}^\mathbf {L} \mathcal{O}_{X_, \overline{x}_ i} \]

Let $\overline{y}_ i$ in $Y_ i$ and $\overline{y}$ in $Y$ be the image of $\overline{x}_ i$ and $\overline{x}$. Since $X$ and $Y_ i$ are tor independent over $Y$, we can apply More on Algebra, Lemma 15.61.2 to see that the right hand side of the displayed formula is equal to $E_{\overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}}^\mathbf {L} \mathcal{O}_{Y_ i, \overline{y}_ i}$ in $D(\mathcal{O}_{Y_ i, \overline{y}_ i})$. Since we have assume the tor amplitude of this is in $[a, b]$, we conclude that the tor amplitude of $E_{\overline{x}}$ in $D(\mathcal{O}_{Y, \overline{y}})$ is in $[a, b]$ by More on Algebra, Lemma 15.66.17. Thus (1) follows.

Using some elementary topology the case “locally finite tor dimension” follows too. $\square$


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