Lemma 36.22.8. Consider a cartesian square of schemes

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

Assume $g$ and $f$ Tor independent.

If $E \in D(\mathcal{O}_ X)$ has tor amplitude in $[a, b]$ as a complex of $f^{-1}\mathcal{O}_ S$-modules, then $L(g')^*E$ has tor amplitude in $[a, b]$ as a complex of $f^{-1}\mathcal{O}_{S'}$-modules.

If $\mathcal{G}$ is an $\mathcal{O}_ X$-module flat over $S$, then $L(g')^*\mathcal{G} = (g')^*\mathcal{G}$.

**Proof.**
We can compute tor dimension at stalks, see Cohomology, Lemma 20.45.5. If $x' \in X'$ with image $x \in X$, then

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

Let $s' \in S'$ and $s \in S$ be the image of $x'$ and $x$. Since $X$ and $S'$ are tor independent over $S$, we can apply More on Algebra, Lemma 15.60.2 to see that the right hand side of the displayed formula is equal to $E_ x \otimes _{\mathcal{O}_{S, s}}^\mathbf {L} \mathcal{O}_{S', s'}$ in $D(\mathcal{O}_{S', s'})$. Thus (1) follows from More on Algebra, Lemma 15.65.13. To see (2) observe that flatness of $\mathcal{G}$ is equivalent to the condition that $\mathcal{G}[0]$ has tor amplitude in $[0, 0]$. Applying (1) we conclude.
$\square$

## Comments (2)

Comment #2164 by Pieter Belmans on

Comment #2193 by Johan on