0
$\begingroup$

Let $X$ be a noetherian scheme over base $S$ and $Y$ a closed subscheme of $X$ with arrow $j$ into $X$, $F,G$ two quasicoherent modules on $Y$. With $\boxtimes$ denote the exterior tensor product bifunctor (i.e.: pullback via the projections and tensor on the product scheme) on $Y$ resp. $X$.

Does one have a canonical iso

$(j\times j)_*(F \boxtimes_Y G) \rightarrow (j_*F) \boxtimes_X (j_*G$)?

Here the box on the left side means pullback via the projections of $Y\times_S Y$ and on the right via the projections of $X\times_SX$.

Or what further conditions does one need to have it?

$\endgroup$
  • $\begingroup$ Have you tried writing it down using affine charts? $\endgroup$ – user2035 Jan 27 '12 at 17:04
  • $\begingroup$ Yes, but it was very confusing... $\endgroup$ – Veen Jan 27 '12 at 21:10
  • 1
    $\begingroup$ Reduce to $S=\mathrm{Spec}(A)$, then both sides are just the tensor product over $A$. Grothendieck uses the notation $F\otimes_{O_S}G$ or $F\otimes_SG$ for the exterior tensor product (see EGA I (new edition), 3.3). $\endgroup$ – user2035 Jan 27 '12 at 21:51
2
$\begingroup$

This is indeed true, even more general: Let $i : Y \to X$, $j : Y' \to X'$ closed immersions of $S$-schemes and $F \in \mathrm{Qcoh}(Y)$, $G \in \mathrm{Qcoh}(Y')$. Then there is a canonical isomorphism $i_* F \boxtimes_{X,X'} j_* G \cong (i \times j)_* (F \boxtimes_{Y,Y'} G)$.

Proof: The commutative diagram

$$\matrix{Y \times_S Y' & \stackrel{i \times j}{\rightarrow} & X \times_S X' \\\\ \mathrm{pr}_1 \downarrow ~ & & ~ \downarrow \mathrm{pr}_1 \\\\ Y & \stackrel{i}{\rightarrow} & X}$$

gives a canonical homomorphism $\mathrm{pr}_1^* i_* F \to (i \times j)_* \mathrm{pr}_1^* F$ (which is not an isomorphism!), similarly for $G$. Thus we get a canonical homomorphism

$$i_* F \boxtimes_{X,X'} j_* G = \mathrm{pr}_1^* i_* F \otimes \mathrm{pr}_2^* j_* F \to (i \times j)_* \mathrm{pr}_1^* F \otimes (i \times j)_* \mathrm{pr}_2^* G$$

$$\to (i \times j)_* (\mathrm{pr}_1^* F \otimes \mathrm{pr}_2^* G) = (i \times j)_* (F \boxtimes_{Y,Y'} G).$$

This is an isomorphism: Since we have given it globally (and its formation commutes with (co)restriction of open subsets), we may assume that everything is affine, say $S=\mathrm{Spec}(k)$, $X=\mathrm{Spec}(A)$, $X'=\mathrm{Spec}(A')$, $Y=\mathrm{Spec}(A/I)$, $Y'=\mathrm{Spec}(A'/I')$ and $F = \tilde{M}$, $G = \tilde{M'}$. Then the homomorphism looks as follows:

$$(M \otimes_k A') \otimes_{A \otimes_k A'} (A \otimes_k M') \to (M \otimes_k A'/I') \otimes_{A/I \otimes_k A'/I'} (A/I \otimes_k M')$$

which maps $(m \otimes a') \otimes (a \otimes m') \mapsto (m \otimes [a']) \otimes ([a] \otimes m')$. In order to show that it is an isomorphism, it suffices to prove that $(m \otimes i') \otimes (a \otimes m')$ vanishes in the left tensor product for all $i' \in I'$ (similarly for $I$). But this is clear, since we can rewrite this tensor as $(1 \otimes i')(m \otimes 1) \otimes (a \otimes m') = (m \otimes 1) \otimes (a \otimes i'm')$ which vanishes because $M'$ is a $A'/I'$-module.

There should be a more global proof of this (valid in a topos context etc.), but I'm too lazy to write it down.

$\endgroup$
  • $\begingroup$ I'm completely happy with this excellent answer, thanks for the effort! $\endgroup$ – Veen Jan 28 '12 at 10:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.