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.

1. 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.

2. 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 20.46.5. 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 15.61.2 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 15.66.13. To see (2) observe that flatness of $\mathcal{G}$ is equivalent to the condition that $\mathcal{G}$ has tor amplitude in $[0, 0]$. Applying (1) we conclude. $\square$

Comment #2164 by on

After the equation in the proof, it should read independent, not indepdent.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).