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?

  • $\begingroup$ 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

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.

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

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