The Stacks project

36.25 Künneth formula for Ext

Consider a cartesian diagram of schemes

\[ \xymatrix{ & X \times _ S Y \ar[ld]^ p \ar[rd]_ q \ar[dd]^ f \\ X \ar[rd]_ a & & Y \ar[ld]^ b \\ & S } \]

For $K \in D(\mathcal{O}_ X)$ and $M \in D(\mathcal{O}_ Y)$ in this section let us define

\[ K \boxtimes M = Lp^*K \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} Lq^*M \]

We claim there is a canonical map
\begin{equation} \label{perfect-equation-kunneth-ext} Ra_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') \otimes _{\mathcal{O}_ S}^\mathbf {L} Rb_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M') \longrightarrow Rf_*(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K \boxtimes M, K' \boxtimes M')) \end{equation}

for $K, K' \in D(\mathcal{O}_ X)$ and $M, M' \in D(\mathcal{O}_ Y)$. Namely, we can take the map adjoint to the map

\[ \begin{matrix} Lf^*\left(Ra_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') \otimes _{\mathcal{O}_ S}^\mathbf {L} Rb_* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M')\right) = \\ Lf^* Ra_* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} Lf^* Rb_* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M') = \\ Lp^* La^* Ra_* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} Lq^* Lb^* Rb_* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M') \to \\ Lp^* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} Lq^* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M') \to \\ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (Lp^*K, Lp^*K') \otimes _{\mathcal{O}_{X \times _ S Y}}^\mathbf {L} R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (Lq^*M, Lq^*M') \to \\ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K \boxtimes M, K' \boxtimes M') \end{matrix} \]

Here the first equality is compatibility of pullbacks with tensor products, Cohomology, Lemma 20.27.3. The second equality is $f = a \circ p = b \circ q$ and composition of pullbacks, Cohomology, Lemma 20.27.2. The first arrow is given by the adjunction maps $La^* Ra_* \to \text{id}$ and $Lb^* Rb_* \to \text{id}$ because pushforward and pullback are adjoint, Cohomology, Lemma 20.28.1. The second arrow is given by Cohomology, Remark 20.42.13. The third and final arrow is Cohomology, Remark 20.42.10. A simple special case of this is the following result.

Lemma 36.25.1. In the situation above, assume $a$ and $b$ are quasi-compact and quasi-separated and $X$ and $Y$ are tor independent over $S$. If $K$ is perfect, $K' \in D_\mathit{QCoh}(\mathcal{O}_ X)$, $M$ is perfect, and $M' \in D_\mathit{QCoh}(\mathcal{O}_ Y)$, then ( is an isomorphism.

Proof. In this case we have $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, K') = K' \otimes ^\mathbf {L} K^\vee $, $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, M') = M' \otimes ^\mathbf {L} M^\vee $, and

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K \boxtimes M, K' \boxtimes M') = (K' \otimes ^\mathbf {L} K^\vee ) \boxtimes (M' \otimes ^\mathbf {L} M^\vee ) \]

See Cohomology, Lemma 20.50.5 and we also use that being perfect is preserved by pullback and by tensor products. Hence this case follows from Lemma 36.23.1. (We omit the verification that with these identifications we obtain the same map.) $\square$

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 0FXX. Beware of the difference between the letter 'O' and the digit '0'.