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This is a proposition in Hartshorne's Algebraic Geometry.

Let $X$ be a scheme. Then an $\mathcal{O}_X$ module $\mathcal{F}$ is quasi coherent then for every open affine subset $U=\text{Spec}(A)$ of $X$, there is an $A$-module $M$ such that $\mathcal{F}|_U\cong \widetilde{M}$.

Given open affine subset $U=\text{Spec}(A)$ of $X$, I am searching for an $A$ module $M$ such that $\mathcal{F}|_U\cong\widetilde{M}$. If that is the case, in particular we should have $\mathcal{F}|_U(U)=\widetilde{M}(U)=M$. So, if there is some module $M$ it has to be isomorphic to $\mathcal{F}(U)$.

Let $\{D(g_i):1\leq i\leq r\}$ be an open cover for $U=\text{Spec}(A)$.

It remains to show that $\mathcal{F}|_U(D(g_i))=\widetilde{M}(D(g_i))$. We have $\widetilde{M}(D(g_i))=M_{g_i}$ for every $1\leq i\leq r$.

As $\mathcal{F}|_U=\mathcal{G}$ is quasi coherent on affine scheme, we have $\mathcal{G}(D(g_i))\cong\mathcal{G}(U)_{g_i}=M_{g_i}$. Thus, $\mathcal{F}|_U\cong \widetilde{M}$.

Let me know if this justification is sufficient to conclude what I have concluded. Hartshorne's book has given an argument which looks a bit complicated. Is it really straightforward or am I missing something?

  • I don't really see any difference between what you write and what Hartshorne writes, except that Hartshorne explicitly mentions the canonical map $\tilde{M}\to \mathscr{F}$ (resp. $\tilde{M}\to \mathscr{F}\vert_U$). You may be aware of this and implicitly used this, but just to be sure: In order to prove that $\mathscr{F}\vert_U = \tilde{M}$ you first need a map between those sheaves and then prove that it is an isomorphism, which you do on the open covering $D(g_i)$. – user363120 Jul 17 '17 at 22:11
  • @user363120 actually I tried this by myself and then saw what hartshorne said.. he reduced to the case when X itself is affine whose necessity is what I was confused about.. I am not giving any map from whole of $\mathcal{F}|_U$ to $\widetilde{M}$. It happened that there is an isomorphism on each element of the open cover so I said it glues to get an isomorphism on global sections... may be I should check they agree on intersections.. I think it would then be same thing as understanding that alpha map he has given.. yes now I think it is more or less same.. –  Jul 17 '17 at 22:23
  • @user363120 btw thanks for double help.. –  Jul 17 '17 at 22:38

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