2
$\begingroup$

I've seen the definition of a quasicoherent sheaf $\mathcal{F}$ on an arbitrary functor $$ X : CRing \to Sets $$ as a specification of an $R$-module $\mathcal{F}(x)$ for each $R$-point $x \in X(R)$ with a collection of isomorphisms $\alpha_{x,x'}: R' \otimes_R \mathcal{F}(x) \to \mathcal{F}(x') $ where $f : R\to R'$ is a morphism of commutative rings with $x' = X(f)x$.

There is a third condition, which is that if we have $f : R \to R' $ and $g: R' \to R$ and $x \in X(R)$ with $x' = X(f)x$ and $x'' = X(gf)x$ that then $\alpha_{x,x''}$ is given by composing the maps $$ R''\otimes_R \mathcal{F}(x) \to R'' \otimes_{R'} (R' \otimes_R \mathcal{F}(x)) \to R'' \otimes_{R'} \mathcal{F}(x') \to \mathcal{F}(x''), $$ where the second and third maps are induced by $\alpha_{x,x'}$ and $\alpha_{x',x''}$ respectively.

My question is how does one recover the notion of a quasicoherent sheaf ona scheme from the above notion applied to a schemes functor of points, that is if we have a quasicoherent sheaf on the functor of points $h_X$ of a scheme $X$ how does one naturally define a quasicoherent sheaf on $X$.

I know how to go in the opposite direction, you just pull back th sheaf along $R$-points to get a quasicoherent sheaf on $h_X$ from one on $X$, and this suggests to me that to get a quasicoherent sheaf on $X$ from one on $h_X$ one should just push forward all the $R$-modules $\mathcal{F}(x)$ along the $R$-points $x$, and glue these together somehow?

Any help on this would be really appreciated!

$\endgroup$
  • $\begingroup$ Yes, you can glue sheaves together: see Exercise 1.22 in [Hartshorne, Ch. II]. Quasicoherence is more or less automatic, since it is a local property. $\endgroup$ – Zhen Lin Feb 5 '16 at 13:57
2
$\begingroup$

In the general definition, you forgot to mention the condition that $\alpha_{x,x}$ is the canonical isomorphism.

Given a scheme $X$ and $R$-modules $M|_x$ for every $R$-point $x : \mathrm{Spec}(R) \to X$ equipped with coherence isomorphisms, choose an open affine covering $(u_i : \mathrm{Spec}(R_i) \to X)$ and consider the $R_i$-modules $ M|_{u_i}$. These induce quasi-coherent modules $M_i:=\widetilde{M|_{u_i}}$ on $\mathrm{Spec}(R_i)$. We have isomorphisms $M_i|_{\mathrm{Spec}(R_i) \cap \mathrm{Spec}(R_j)} \cong M_j|_{\mathrm{Spec}(R_i) \cap \mathrm{Spec}(R_j)}$ satisfying the cocycle condition. To see this, cover $\mathrm{Spec}(R_i) \cap \mathrm{Spec}(R_j)$ by open affines $\mathrm{Spec}(R_{ijk})$ and use the compatibility of the given modules for $R_i \to R_{ijk}$ and $R_j \to R_{ijk}$. Then we may glue the $M_i$ to some quasi-coherent module $M$ on $X$ with $u_i^* M \cong M_i = \widetilde{M|_{u_i}}$. We still need to check $x^* M \cong \widetilde{M|_x}$ holds for any $x : \mathrm{Spec}(R) \to X$. To see this, cover $\mathrm{Spec}(R)$ with open affines $\mathrm{Spec}(S_i)$ such that $x|_{\mathrm{Spec}(S_i)}$ factors through $\mathrm{Spec}(R_i)$, and use compatibility with respect to $R_i \to S_i$. Of course, there are many things to check here, but every step is rather straight forward.

$\endgroup$

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.