2
$\begingroup$

I'm trying to understand the concept of the relative spectrum of a sheaf of quasicoherent algebras. Here is the situation: We are given a scheme $X$ and a quasicoherent sheaf of $\mathscr{O}_X$-algebras (i.e. a sheaf of algebras that is quasicoherent as an $\mathscr{O}_X$-module) $\mathscr{B}$.

Consider the contravariant functor $F = F_{\mathscr{B}, X}$ from schemes to sets that acts on objects by $$W \mapsto \{(\mu : W \to X, \varphi : \mathscr{B} \to \mu_* \mathscr{O}_W )\}$$ where $\mu$ is a morphism of schemes, $\varphi$ is a morphism of $\mathscr{O}_X$-algebras, and the right hand side is the set of all such pairs. The functor $F$ acts on morphisms $a : W' \to W$ by \begin{align} F(a) : F(W) &\to F(W')\\ (\mu, \varphi) &\mapsto (\mu' = \mu \circ a, \varphi' = a^* \varphi), \end{align} where $\varphi' = a^*\varphi : \mathscr{B} \to \mu_* \mathscr{O}_W \to {(\mu')}_* \mathscr{O}_{W'}$ in the composition of $\varphi$ with $\mu_* a^\#$, where $a^\# : \mathscr{O}_W \to a_* \mathscr{O}_{W'}$ (note $\mu_* a_* = (\mu \circ a)_* = (\mu')_*$)

The idea is that we get an affine morphism of schemes $\beta : \underline{\mathrm{Spec}}_X\mathscr{B} \to X$ and morphism of $\mathscr{O}_X$-algebras $\phi : \mathscr{B} \to \beta_* \mathscr{O}_{\underline{\mathrm{Spec}}_X\mathscr{B}}$ such that the pair $(\underline{\mathrm{Spec}}_X\mathscr{B}, (\beta, \phi))$ represents the functor $F$. That is, $F(-) = \mathrm{Hom}_{\mathrm{Sch}_{/X}}(-, \underline{\mathrm{Spec}}_X\mathscr{B})$ and $(\beta, \phi) \in F(\underline{\mathrm{Spec}}_X\mathscr{B})$ corresponds to the identity morphism under the bijection $F(\underline{\mathrm{Spec}}_X\mathscr{B}) \cong \mathrm{Hom}(\underline{\mathrm{Spec}}_X\mathscr{B}, \underline{\mathrm{Spec}}_X\mathscr{B})$. The scheme $\underline{\mathrm{Spec}}_X\mathscr{B}$ is called the relative spectrum of $\mathscr{B}$ over $X$.

How do we get this scheme and universal maps? As usual, we start with the affine case. Suppose $X = \mathrm{Spec}(A)$. Then since $\mathscr{B}$ is quasicoherent, it must be o.t.f. $\mathscr{B} = \tilde{B}$ for some $A$-algebra $B$. This gives a structure morphism $\beta : \mathrm{Spec}(B) \to \mathrm{Spec}(A)$. The multiplication map $B \otimes_A B \to B$ taking $b \otimes b' \mapsto b b'$ induces a morphism of qcoh $\mathscr{O}_{\mathrm{Spec}B}$-algebras $\beta^* \mathscr{B} = \widetilde{B \otimes_A B} \to \tilde{B} = \mathscr{O}_{\mathrm{Spec}B}$, and hence by adjunction a map $\mathscr{B} \to \beta_* \mathscr{O}_{\mathrm{Spec}B}$. We claim that this data represents the functor $F_{\mathscr{B}, \mathrm{Spec}(A)}$. This is proved in https://stacks.math.columbia.edu/tag/01LT, but there are some things I don't understand. The idea is fairly clear, We must show that for all schemes $W$, the map $$ \mathrm{Mor}_{\mathrm{Sch}}(W, \mathrm{Spec} B) \to \{(\mu, \varphi)\} \hspace{10pt} a \mapsto (a^* \beta = \beta \circ a, a^* \varphi) $$ is bijective. Let's call this map $f$. In order to do this, we can define an inverse map $f^{-1}$ and show that for all morphisms $a : W \to \mathrm{Spec}B$, $f^{-1}(f(a)) = a$ and for all pairs $(\mu, \varphi)$, $f(f^{-1}((\mu, \varphi))) = (\mu, \varphi)$. What I don't understand is how to define the inverse map. Specifically, in the lemma I cited above, what is the map $f^*$ ($\mu^*$ in my notation) supposed to be?

As a follow up question, suppose we have a pair $(\mu : W \to \mathrm{Spec} A, \varphi: \mu^* \mathscr{B} \to \mathscr{O}_W)$. Then by adjunction we get $\varphi : \mathscr{B} \to \mu_* \mathscr{O}_W$. Evaluating this map of $\mathscr{O}_{\mathrm{Spec}A}$-algebras on global sections, we have an $A$-algebra homomorphism $B \to \Gamma(W, \mathscr{O}_W)$. This corresponds to a morphism of schemes $W \to \mathrm{Spec}(B)$. If we define the inverse to be this map, does the proof go through? I'm having some trouble showing the maps are mutually inverse.

$\endgroup$

2 Answers 2

2
$\begingroup$

Your construction looks correct, and it is a natural bijection because it is constructed using adjunctions, which are natural bijective correspondences. Namely, since $\mathscr B=\widetilde B$, it follows from adjunction of $\widetilde{\phantom{B}}$ and $\Gamma$ that morphisms of $\mathscr O_{\mathrm{Spec}A}$-algebras $\mathscr B\to\mu_*\mathscr O_W$ correspond naturally to $A$-algebra morphisms $B\to\Gamma(\mathrm{Spec}A,\mu_*\mathscr O_W)\cong\Gamma(W,\mathscr O_W)$. By adjunction of Spec and $\Gamma$, these correspond naturally to morphisms $W\to\mathrm{Spec}B$ of schemes over $\mathrm{Spec}A$.

$\endgroup$
1
  • $\begingroup$ Ah, thank you! Of course we have the adjuction of $\widetilde{\cdot}$ and $\Gamma(-)$ since our base scheme $\mathrm{Spec}A$ is affine. $\endgroup$
    – ggg
    Jul 12, 2018 at 16:02
0
$\begingroup$

Here is an answer to my first question. Suppose $(\mu, \varphi)$ is a pair as above. Taking the construction of https://stacks.math.columbia.edu/tag/01I6, we get a ring homomorphism $$ B = \Gamma(\mathrm{Spec}B, \mathscr{B}) \overset{\mu^*}{\longrightarrow} \Gamma(W, \mu^* \mathscr{B}) \overset{\varphi}{\longrightarrow} \Gamma(W, \mathscr{O}_W) $$ inducing a morphism of schemes $W \to \mathrm{Spec}B$. As mentioned in my question, the map $\varphi$ is clear. I think the map $\mu^*$ is the natural map $\Gamma(Y, \mathscr{G}) \to \Gamma(X, \pi^*\mathscr{G})$ we get for any morphism $\pi : X \to Y$ and qcoh sheaf $\mathscr{G}$ on $Y$ applied to our situation ($\pi = \mu$ and $\mathscr{G} = \mathscr{B}$). The natural map is constructed using adjunction of the identity map on $\mathscr{G}$ to get $\mathscr{G} \to \pi_* \pi^* \mathscr{G}$, applying the global sections functor, and then composing with the identity map $\Gamma(Y, \pi_* \pi^* \mathscr{G}) \to \Gamma(X, \pi^* \mathscr{G})$.

$\endgroup$

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.