The Stacks project

Lemma 27.4.2. In Situation 27.3.1. Let $F$ be the functor associated to $(S, \mathcal{A})$ above. If $S$ is affine, then $F$ is representable by the affine scheme $\mathop{\mathrm{Spec}}(\Gamma (S, \mathcal{A}))$.

Proof. Write $S = \mathop{\mathrm{Spec}}(R)$ and $A = \Gamma (S, \mathcal{A})$. Then $A$ is an $R$-algebra and $\mathcal{A} = \widetilde A$. The ring map $R \to A$ gives rise to a canonical map

\[ f_{univ} : \mathop{\mathrm{Spec}}(A) \longrightarrow S = \mathop{\mathrm{Spec}}(R). \]

We have $f_{univ}^*\mathcal{A} = \widetilde{A \otimes _ R A}$ by Schemes, Lemma 26.7.3. Hence there is a canonical map

\[ \varphi _{univ} : f_{univ}^*\mathcal{A} = \widetilde{A \otimes _ R A} \longrightarrow \widetilde A = \mathcal{O}_{\mathop{\mathrm{Spec}}(A)} \]

coming from the $A$-module map $A \otimes _ R A \to A$, $a \otimes a' \mapsto aa'$. We claim that the pair $(f_{univ}, \varphi _{univ})$ represents $F$ in this case. In other words we claim that for any scheme $T$ the map

\[ \mathop{\mathrm{Mor}}\nolimits (T, \mathop{\mathrm{Spec}}(A)) \longrightarrow \{ \text{pairs } (f, \varphi )\} ,\quad a \longmapsto (f_{univ} \circ a, a^*\varphi _{univ}) \]

is bijective.

Let us construct the inverse map. For any pair $(f : T \to S, \varphi )$ we get the induced ring map

\[ \xymatrix{ A = \Gamma (S, \mathcal{A}) \ar[r]^{f^*} & \Gamma (T, f^*\mathcal{A}) \ar[r]^{\varphi } & \Gamma (T, \mathcal{O}_ T) } \]

This induces a morphism of schemes $T \to \mathop{\mathrm{Spec}}(A)$ by Schemes, Lemma 26.6.4.

The verification that this map is inverse to the map displayed above is omitted. $\square$


Comments (6)

Comment #7016 by on

Just before the construction of the inverse map : is that instead of ?

Comment #7121 by Elías Guisado on

Who is exactly the map in the description of the inverse?

Comment #7122 by Elías Guisado on

Is it , the unit of the pushforward-pullback adjunction of sheaves of modules?

Comment #7236 by on

@#7016. Thanks and fixed here.

@#7121 and #7122. Yes. Discussion. Say we have a continuous map of topological spaces and we have a sheaf on , then we have a map often called a pullback map. Now it is indeed true that you can write and then you can get the map by appying the adjunction map . Another method, is to say that , use the fact that is a functor, and use that . Here is the singleton sheaf AKA the final object of the category of sheaves of sets. These constructions give the same thing.

Of course, if we have a map of ringed spaces and is a sheaf of modules, then we also have a pullback map . It can be constructed by either method discussed above or its existence can be deduced from the construction of the map for sheaves of sets.

Comment #8436 by on

I would propose writing the whole proof in this way: I'm going to adopt the POV of representability of instead of , see https://stacks.math.columbia.edu/tag/01LQ#comment-8435 . We have (where just means regarded as an -module, notation introduced in the paragraph before 10.14.3). To show that represents we check that the identity map is a universal element for . Fix . We have a map , induced by . To construct the inverse, for an -linear map (an element of ), we have a map on global sections , which is -linear. Therefore, it induces a morphism over , by Schemes, 26.6.4. On the one hand, it is easy to see that the composite is the identity (use uniqueness in Schemes, 26.6.4). On the other hand, to see that is the identity, we note that all the involved maps are compatible with restrictions to distinguished open subsets of (i.e., given distinguished open, one restricts to and to ). Hence, it suffices to see that commutes. But this is easy.

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  • 14 comment(s) on Section 27.4: Relative spectrum as a functor

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