Lemma 26.7.2. Let $(X, \mathcal{O}_ X) = (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ be an affine scheme. There are canonical isomorphisms

1. $\widetilde{M \otimes _ R N} \cong \widetilde M \otimes _{\mathcal{O}_ X} \widetilde N$, see Modules, Section 17.16.

2. $\widetilde{\text{T}^ n(M)} \cong \text{T}^ n(\widetilde M)$, $\widetilde{\text{Sym}^ n(M)} \cong \text{Sym}^ n(\widetilde M)$, and $\widetilde{\wedge ^ n(M)} \cong \wedge ^ n(\widetilde M)$, see Modules, Section 17.21.

3. if $M$ is a finitely presented $R$-module, then $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\widetilde M, \widetilde N) \cong \widetilde{\mathop{\mathrm{Hom}}\nolimits _ R(M, N)}$, see Modules, Section 17.22.

First proof. Using Lemma 26.7.1 and Modules, Lemma 17.10.5 we see that the functor $M \mapsto \widetilde M$ can be viewed as $\pi ^*$ for a morphism $\pi$ of ringed spaces. And pulling back modules commutes with tensor constructions by Modules, Lemmas 17.16.4 and 17.21.3. The morphism $\pi : (X, \mathcal{O}_ X) \to (\{ *\} , R)$ is flat for example because the stalks of $\mathcal{O}_ X$ are localizations of $R$ (Lemma 26.5.4) and hence flat over $R$. Thus pullback by $\pi$ commutes with internal hom if the first module is finitely presented by Modules, Lemma 17.22.5. $\square$

Second proof. Proof of (1). By Lemma 26.7.1 to give a map $\widetilde{M \otimes _ R N}$ into $\widetilde M \otimes _{\mathcal{O}_ X} \widetilde N$ we have to give a map on global sections $M \otimes _ R N \to \Gamma (X, \widetilde M \otimes _{\mathcal{O}_ X} \widetilde N)$ which exists by definition of the tensor product of sheaves of modules. To see that this map is an isomorphism it suffices to check that it is an isomorphism on stalks. And this follows from the description of the stalks of $\widetilde{M}$ (either in Lemma 26.5.4 or in Modules, Lemma 17.10.5), the fact that tensor product commutes with localization (Algebra, Lemma 10.12.16) and Modules, Lemma 17.16.1.

Proof of (2). This is similar to the proof of (1), using Algebra, Lemma 10.13.5 and Modules, Lemma 17.21.2.

Proof of (3). Since the construction $M \mapsto \widetilde{M}$ is functorial there is an $R$-linear map $\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\widetilde{M}, \widetilde{N})$. The target of this map is the global sections of $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\widetilde M, \widetilde N)$. Hence by Lemma 26.7.1 we obtain a map of $\mathcal{O}_ X$-modules $\widetilde{\mathop{\mathrm{Hom}}\nolimits _ R(M, N)} \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\widetilde M, \widetilde N)$. We check that this is an isomorphism by comparing stalks. If $M$ is finitely presented as an $R$-module then $\widetilde M$ has a global finite presentation as an $\mathcal{O}_ X$-module. Hence we conclude using Algebra, Lemma 10.10.2 and Modules, Lemma 17.22.4. $\square$

Third proof of part (1). For any $\mathcal{O}_ X$-module $\mathcal{F}$ we have the following isomorphisms functorial in $M$, $N$, and $\mathcal{F}$

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\widetilde{M} \otimes _{\mathcal{O} _ X} \widetilde{N}, \mathcal{F}) & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\widetilde{M}, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O} _ X} (\widetilde{N}, \mathcal{F})) \\ & = \mathop{\mathrm{Hom}}\nolimits _ R(M, \Gamma (X, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\widetilde{N}, \mathcal{F})) \\ & = \mathop{\mathrm{Hom}}\nolimits _ R(M, \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\widetilde{N}, \mathcal{F})) \\ & = \mathop{\mathrm{Hom}}\nolimits _ R(M, \mathop{\mathrm{Hom}}\nolimits _ R(N, \Gamma (X,\mathcal{F}))) \\ & = \mathop{\mathrm{Hom}}\nolimits _ R(M \otimes _ R N, \Gamma (X, \mathcal{F})) \\ & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(\widetilde{M \otimes _ R N}, \mathcal{F}) \end{align*}

The first equality is Modules, Lemma 17.22.1. The second equality is the universal property of $\widetilde{M}$, see Lemma 26.7.1. The third equality holds by definition of $\mathop{\mathcal{H}\! \mathit{om}}\nolimits$. The fourth equality is the universal property of $\widetilde{N}$. Then fifth equality is Algebra, Lemma 10.12.8. The final equality is the universal property of $\widetilde{M \otimes _ R N}$. By the Yoneda lemma (Categories, Lemma 4.3.5) we obtain (1). $\square$

Comment #1833 by Keenan Kidwell on

In the last sentence of the first text block, is the parenthetical remark really supposed to say "as a functor?" We haven't defined the stalks of a functor, right? I feel like it should be "as an $\mathscr{O}_X$-module," but then it seems unnecessary altogether.

Comment #1870 by on

Comment #1907 by Keenan Kidwell on

I think it's better now!

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