The Stacks project

17.15 Tensor product

Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules. We have briefly discussed the tensor product in the setting of change of rings in Sheaves, Sections 6.6 and 6.20. In exactly the same way we define first the tensor product presheaf

\[ \mathcal{F} \otimes _{p, \mathcal{O}_ X} \mathcal{G} \]

as the rule which assigns to $U \subset X$ open the $\mathcal{O}_ X(U)$-module $\mathcal{F}(U) \otimes _{\mathcal{O}_ X(U)} \mathcal{G}(U)$. Having defined this we define the tensor product sheaf as the sheafification of the above:

\[ \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G} = (\mathcal{F} \otimes _{p, \mathcal{O}_ X} \mathcal{G})^\# \]

This can be characterized as the sheaf of $\mathcal{O}_ X$-modules such that for any third sheaf of $\mathcal{O}_ X$-modules $\mathcal{H}$ we have

\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X} (\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}, \mathcal{H}) = \text{Bilin}_{\mathcal{O}_ X}(\mathcal{F} \times \mathcal{G}, \mathcal{H}). \]

Here the right hand side indicates the set of bilinear maps of sheaves of $\mathcal{O}_ X$-modules (definition omitted).

The tensor product of modules $M, N$ over a ring $R$ satisfies symmetry, namely $M \otimes _ R N = N \otimes _ R M$, hence the same holds for tensor products of sheaves of modules, i.e., we have

\[ \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G} = \mathcal{G} \otimes _{\mathcal{O}_ X} \mathcal{F} \]

functorial in $\mathcal{F}$, $\mathcal{G}$. And since tensor product of modules satisfies associativity we also get canonical functorial isomorphisms

\[ (\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}) \otimes _{\mathcal{O}_ X} \mathcal{H} = \mathcal{F} \otimes _{\mathcal{O}_ X} (\mathcal{G} \otimes _{\mathcal{O}_ X} \mathcal{H}) \]

functorial in $\mathcal{F}$, $\mathcal{G}$, and $\mathcal{H}$.

Lemma 17.15.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules. Let $x \in X$. There is a canonical isomorphism of $\mathcal{O}_{X, x}$-modules

\[ (\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G})_ x = \mathcal{F}_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{G}_ x \]

functorial in $\mathcal{F}$ and $\mathcal{G}$.

Proof. Omitted. $\square$

Lemma 17.15.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}'$, $\mathcal{G}'$ be presheaves of $\mathcal{O}_ X$-modules with sheafifications $\mathcal{F}$, $\mathcal{G}$. Then $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G} = (\mathcal{F}' \otimes _{p, \mathcal{O}_ X} \mathcal{G}')^\# $.

Proof. Omitted. $\square$

Lemma 17.15.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{G}$ be an $\mathcal{O}_ X$-module. If $\mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$ is an exact sequence of $\mathcal{O}_ X$-modules then the induced sequence

\[ \mathcal{F}_1 \otimes _{\mathcal{O}_ X} \mathcal{G} \to \mathcal{F}_2 \otimes _{\mathcal{O}_ X} \mathcal{G} \to \mathcal{F}_3 \otimes _{\mathcal{O}_ X} \mathcal{G} \to 0 \]

is exact.

Proof. This follows from the fact that exactness may be checked at stalks (Lemma 17.3.1), the description of stalks (Lemma 17.15.1) and the corresponding result for tensor products of modules (Algebra, Lemma 10.12.10). $\square$

Lemma 17.15.4. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ Y$-modules. Then $f^*(\mathcal{F} \otimes _{\mathcal{O}_ Y} \mathcal{G}) = f^*\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G}$ functorially in $\mathcal{F}$, $\mathcal{G}$.

Proof. Omitted. $\square$

Lemma 17.15.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules.

  1. If $\mathcal{F}$, $\mathcal{G}$ are locally generated by sections, so is $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}$.

  2. If $\mathcal{F}$, $\mathcal{G}$ are of finite type, so is $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}$.

  3. If $\mathcal{F}$, $\mathcal{G}$ are quasi-coherent, so is $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}$.

  4. If $\mathcal{F}$, $\mathcal{G}$ are of finite presentation, so is $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}$.

  5. If $\mathcal{F}$ is of finite presentation and $\mathcal{G}$ is coherent, then $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}$ is coherent.

  6. If $\mathcal{F}$, $\mathcal{G}$ are coherent, so is $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}$.

  7. If $\mathcal{F}$, $\mathcal{G}$ are locally free, so is $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}$.

Proof. We first prove that the tensor product of locally free $\mathcal{O}_ X$-modules is locally free. This follows if we show that $(\bigoplus _{i \in I} \mathcal{O}_ X) \otimes _{\mathcal{O}_ X} (\bigoplus _{j \in J} \mathcal{O}_ X) \cong \bigoplus _{(i, j) \in I \times J} \mathcal{O}_ X$. The sheaf $\bigoplus _{i \in I} \mathcal{O}_ X$ is the sheaf associated to the presheaf $U \mapsto \bigoplus _{i \in I} \mathcal{O}_ X(U)$. Hence the tensor product is the sheaf associated to the presheaf

\[ U \longmapsto (\bigoplus \nolimits _{i \in I} \mathcal{O}_ X(U)) \otimes _{\mathcal{O}_ X(U)} (\bigoplus \nolimits _{j \in J} \mathcal{O}_ X(U)). \]

We deduce what we want since for any ring $R$ we have $(\bigoplus _{i \in I} R) \otimes _ R (\bigoplus _{j \in J} R) = \bigoplus _{(i, j) \in I \times J} R$.

If $\mathcal{F}_2 \to \mathcal{F}_1 \to \mathcal{F} \to 0$ is exact, then by Lemma 17.15.3 the complex $\mathcal{F}_2 \otimes \mathcal{G} \to \mathcal{F}_1 \otimes \mathcal{G} \to \mathcal{F} \otimes \mathcal{G} \to 0$ is exact. Using this we can prove (5). Namely, in this case there exists locally such an exact sequence with $\mathcal{F}_ i$, $i = 1, 2$ finite free. Hence the two terms $\mathcal{F}_2 \otimes \mathcal{G}$ are isomorphic to finite direct sums of $\mathcal{G}$. Since finite direct sums are coherent sheaves, these are coherent and so is the cokernel of the map, see Lemma 17.12.4.

And if also $\mathcal{G}_2 \to \mathcal{G}_1 \to \mathcal{G} \to 0$ is exact, then we see that

\[ \mathcal{F}_2 \otimes _{\mathcal{O}_ X} \mathcal{G}_1 \oplus \mathcal{F}_1 \otimes _{\mathcal{O}_ X} \mathcal{G}_2 \to \mathcal{F}_1 \otimes _{\mathcal{O}_ X} \mathcal{G}_1 \to \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G} \to 0 \]

is exact. Using this we can for example prove (3). Namely, the assumption means that we can locally find presentations as above with $\mathcal{F}_ i$ and $\mathcal{G}_ i$ free $\mathcal{O}_ X$-modules. Hence the displayed presentation is a presentation of the tensor product by free sheaves as well.

The proof of the other statements is omitted. $\square$

Lemma 17.15.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. For any $\mathcal{O}_ X$-module $\mathcal{F}$ the functor

\[ \textit{Mod}(\mathcal{O}_ X) \longrightarrow \textit{Mod}(\mathcal{O}_ X) , \quad \mathcal{G} \longmapsto \mathcal{F} \otimes _\mathcal {O} \mathcal{G} \]

commutes with arbitrary colimits.

Proof. Let $I$ be a preordered set and let $\{ \mathcal{G}_ i\} $ be a system over $I$. Set $\mathcal{G} = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{G}_ i$. Recall that $\mathcal{G}$ is the sheaf associated to the presheaf $\mathcal{G}' : U \mapsto \mathop{\mathrm{colim}}\nolimits _ i \mathcal{G}_ i(U)$, see Sheaves, Section 6.29. By Lemma 17.15.2 the tensor product $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}$ is the sheafification of the presheaf

\[ U \longmapsto \mathcal{F}(U) \otimes _{\mathcal{O}_ X(U)} \mathop{\mathrm{colim}}\nolimits _ i \mathcal{G}_ i(U) = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}(U) \otimes _{\mathcal{O}_ X(U)} \mathcal{G}_ i(U) \]

where the equality sign is Algebra, Lemma 10.12.9. Hence the lemma follows from the description of colimits in $\textit{Mod}(\mathcal{O}_ X)$. $\square$

Comments (2)

Comment #5943 by Tim Campion on

Is it fair to omit the definition of a bilinear map of -modules? After all, a bilinear map of modules is a function such that for every point the map is linear, etc. The most naive tranlsation into sheaves would therefore say that is bilinear if for every global section of , the map is -linear, etc. This definition is clearly wrong. I think one wants to say that is -bilinear for every , but I'm not quite sure. I'm also not sure whether bilinearity can be checked on stalks, etc.

Comment #5944 by on

Yes, bilinear maps of sheaves of modules are those maps of sheaves of sets which give -bilinear maps on sets of sections over for all opens .

Equivalently you can ask certain diagrams of maps of sheaves of sets commute, immitating the usual axioms for bilinear maps of modules such as the axiom , etc.

If is a map of sheaves of sets and it induces a bilinar map of modules on stalks for all points of , then is a bilinear map of sheaves of modules as you can test whether local sections are equal by checking on stalks.

Let denote morphisms in the category of sheaves of sets on . Another way you could define the notion of a bilinear map is this: a map of sheaves of sets is bilinear if given any sheaf of sets the rule , is a bilinear map of modules over the ring . We don't usually take this point of view as it is easier to think about sets of local sections and it is clearly equivalent.

Another way to say the definition: is a ring object in the category of sheaves of sets and , , are module objects over this ring. Then a bilinear map can be defined for module objects over a ring object in any category. (To formulate what is a ring object and what is a module object over a ring object, and what is a bilinear map of such in a category it is pleasant --but not strictly necessary-- to assume the category has finite products; and this is true for the category of sheaves of sets.)

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