The Stacks project

Lemma 36.22.6. Consider a cartesian square

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

of quasi-compact and quasi-separated schemes. Assume $g$ and $f$ Tor independent and $S = \mathop{\mathrm{Spec}}(R)$, $S' = \mathop{\mathrm{Spec}}(R')$ affine. For $M, K \in D(\mathcal{O}_ X)$ the canonical map

\[ R\mathop{\mathrm{Hom}}\nolimits _ X(M, K) \otimes ^\mathbf {L}_ R R' \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _{X'}(L(g')^*M, L(g')^*K) \]

in $D(R')$ is an isomorphism in the following two cases

  1. $M \in D(\mathcal{O}_ X)$ is perfect and $K \in D_\mathit{QCoh}(X)$, or

  2. $M \in D(\mathcal{O}_ X)$ is pseudo-coherent, $K \in D_\mathit{QCoh}^+(X)$, and $R'$ has finite tor dimension over $R$.

Proof. There is a canonical map $R\mathop{\mathrm{Hom}}\nolimits _ X(M, K) \to R\mathop{\mathrm{Hom}}\nolimits _{X'}(L(g')^*M, L(g')^*K)$ in $D(\Gamma (X, \mathcal{O}_ X))$ of global hom complexes, see Cohomology, Section 20.41. Restricting scalars we can view this as a map in $D(R)$. Then we can use the adjointness of restriction and $- \otimes _ R^\mathbf {L} R'$ to get the displayed map of the lemma. Having defined the map it suffices to prove it is an isomorphism in the derived category of abelian groups.

The right hand side is equal to

\[ R\mathop{\mathrm{Hom}}\nolimits _ X(M, R(g')_*L(g')^*K) = R\mathop{\mathrm{Hom}}\nolimits _ X(M, K \otimes _{\mathcal{O}_ X}^\mathbf {L} g'_*\mathcal{O}_{X'}) \]

by Lemma 36.5.3. In both cases the complex $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K)$ is an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Lemma 36.10.8. There is a natural map

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K) \otimes _{\mathcal{O}_ X}^\mathbf {L} g'_*\mathcal{O}_{X'} \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K \otimes _{\mathcal{O}_ X}^\mathbf {L} g'_*\mathcal{O}_{X'}) \]

which is an isomorphism in both cases by Lemma 36.10.9. To see that this lemma applies in case (2) we note that $g'_*\mathcal{O}_{X'} = Rg'_*\mathcal{O}_{X'} = Lf^*g_*\mathcal{O}_ X$ the second equality by Lemma 36.22.5. Using Lemma 36.10.4 and Cohomology, Lemma 20.45.4 we conclude that $g'_*\mathcal{O}_{X'}$ has finite Tor dimension. Hence, in both cases by replacing $K$ by $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K)$ we reduce to proving

\[ R\Gamma (X, K) \otimes ^\mathbf {L}_ A A' \longrightarrow R\Gamma (X, K \otimes ^\mathbf {L}_{\mathcal{O}_ X} g'_*\mathcal{O}_{X'}) \]

is an isomorphism. Note that the left hand side is equal to $R\Gamma (X', L(g')^*K)$ by Lemma 36.5.3. Hence the result follows from Lemma 36.22.5. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AA7. Beware of the difference between the letter 'O' and the digit '0'.