4

Let $X$ be a scheme over a ring $S$, and $F$ some quasicoherent sheaf on $X$. The "functor of points" says that we can think of $X$ as the functor representing all of the $R$-points, for $R$ an $S$-algebra. This recovers the classical point of view, since it expresses the scheme as something which is locally the solution set to some equations over an $S$-algebra. Here locally means open subfunctors, and being a solution to some equations over $R$ would be like the $R$ points of $S[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$

I would like to understand how quasicoherent sheaves fit into this point of view. The natural guess is that they arise as compatible families of things which are locally modules over these sets of $R$ points. But precisely what functor to SET do they represent? It no longer makes sense (I think) to ask about morphisms from $Spec R$ to the quasicoherent sheaf $F$ on $X$.

One could consider the pullbacks, for each $R$ point $f: Spec R \to X$, we have the $R$ module $f^*(F)$. The equation $(f \circ g)^* \cong g^* \circ f^*$ suggests that this almost works, expect that this equation isn't literally an equality (I have heard that this causes problems). This gives just the fiber of $F$ over each $R$ point.

Is that the right thing to think? What is?

(I hope I am making sense.)

Elle Najt
  • 21,422
  • 2
    Yes, they are compatible families. Quasicoherent sheaves aren't themselves schemes, so I don't really understand the rest of your question. There is a stack (not sheaf) of quasicoherent sheaves. – Qiaochu Yuan Feb 20 '16 at 21:05
  • 3
    I discuss how to think of $\mathscr{O}_X$-modules as functors here. – Zhen Lin Feb 21 '16 at 08:17
  • The ambiguity of $(f\circ g)^\ast$ and $g^\ast\circ f^\ast$ (the fact that they are unequal—just naturally isomorphic) is directly related to the theory of stacks as alluded to in @QiaochuYuan's answer. Stacks are not valued in sets, but in the category of groupoids which roughly fixes the ambiguity you mentioned. That said, what I wanted to point out is that even though one might believe a lot of algebraic geometry extends to the theory of stacks, it really only extends to theory of algebraic (or Artin) stacks. And, unfortunately $\mathsf{Qcoh}$, while a stack – Alex Youcis Feb 21 '16 at 12:26
  • is not an algebraic stack. Not even something much tamer like the stack $\mathsf{Coh}$ of coherent sheaves (interpreted correctly) is an algebraic stack. One has to go all the way down to something like vector bundles to actually create an object well-behaved and 'geometric' enough to be algebraic, and thus amenable to algebro-geometric thinking. – Alex Youcis Feb 21 '16 at 12:27

0 Answers0