0
$\begingroup$

Let $Y = \text{Spec}A$ be a noetherian affine scheme. Let $S$ be a graded $A$-algebra which is finitely generated by $S_{1}$ as an $S_{0}$ algebra. In other words, $S$ looks like $$ S = A[x_{0}, x_{1}, \ldots , x_{n}]/I $$ for some homogenous ideal $I$. Let $\mathcal{F}$ be the sheaf of algebras $\widetilde{S}$ on $\text{Spec}A$. I am trying to understand the projective space $\text{Proj}_{Y} \mathcal{F}$ via universal properties. My understanding so far is this: The scheme $\text{Proj}_{Y} \mathcal{F}$ is defined to be the $A$-scheme such for any other $A$-scheme $\mu: T \rightarrow \text{Spec}A$, to give an $A$-morphism from $T$ to $\text{Proj}_{Y} \mathcal{F}$ is to give a line bundle quotient $$ \mu^{*} \mathcal{F} \stackrel{q}{\longrightarrow} \mathcal{L} \longrightarrow 0. $$ Another universal property is to say that to give an $A$-morphism from $T$ to $\text{Proj}_{Y}\mathcal{F}$ is to give a line bundle $\mathcal{L}$ on $T$ along with a morphism of $A$-modules $\phi: S_{1} \rightarrow \Gamma(T, \mathcal{L})$ so that the image of $\phi$ generates $\mathcal{L}$.

My question is, why are these equivalent? More precisely,

Why is giving a surjection of sheaves $\mu^{*} \mathcal{F} {\rightarrow} \mathcal{L} \rightarrow 0$ equivalent to giving a morphism of $A$-modules $\phi: S_{1} \rightarrow \Gamma(T, \mathcal{L})$ whose image is a family of global sections of $\mathcal{L}$ which generates $\mathcal{L}$?

I am assuming the answer involves taking the given surjection $q: \mu^{*} \mathcal{F} \rightarrow \mathcal{L}$ and corresponding the adjunct morphism $q': \mathcal{F} \rightarrow \mu_{*} \mathcal{L}$. Taking global sections give us a morphism, $$ \Gamma(\text{Spec}A, \mathcal{F}) \longrightarrow \Gamma(T, \mathcal{L}). $$ Somehow I need to consider only the degree $1$ part of this to obtain a morphism of $A$-modules and show that it generates $\mathcal{L}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.