defining an ideal sheaf of a scheme

Question

Let $X$ be a scheme. Let $A$ be an ideal of $\Gamma(X, O_X)$. I am wondering what does it mean by $Q$ is the $O_X$ ideal generated by $A$? Also how does one show that ideal sheaf defined in such a way is quasi-coherent? Thank you.

Answer 1

Suppose $A$ is generated by $\{f_i\}_{i\in I}$ as an ideal of $\Gamma(X,O_X)=R$. Then we can define a map $O_X^{\oplus I}\to O_X$ by multiplying the $i^{th}$ coordinate by $f_i$ and summing. The (sheaf) image of this map is $Q$, the requested sheaf of ideals.

In order to show that this is quasicoherent, consider the corresponding map of rings $R^{\oplus I}\to A$. This has a kernel $K$, and we can choose generators for $K$ as an $R$-module which gives us a map $R^{\oplus J}\to K$. Taking the composite, we get a map $R^{\oplus J}\to R^{\oplus I}\to A$ which demonstrates that $A$ is the cokernel of a map of free modules. Using the same map on $O_X^{\oplus J}\to O_X^{\oplus I}\to Q$, we see that $Q$ is again (globally) a cokernel of free modules and thus quasicoherent.

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In response to the request for additional clarification:

For any open set $U\subset X$, the restriction homomorphism $res_U:R=O_X(X)\to O_X(U)$ makes $O_X(U)$ in to an $R$-algebra, and thus $O_X^{\oplus I}(U)$ and $O_X^{\oplus J}(U)$ are both $R$-algebras in a natural way. In particular, this means that the maps $O_X^{\oplus J}(U)\to O_X^{\oplus I}(U)$ and $O_X^{\oplus I}\to O_X(U)$ given by $res_U$ applied to the entries of the matrix of the map $R^{\oplus J}\to R^{\oplus I}$ and $R^{\oplus I}\to R$ are also ring homomorphisms, and the fact that the various restriction maps are compatible ring homomorphisms imply that they combine to determine morphisms of sheaves $O_X^{\oplus J}\to O_X^{\oplus I}\to O_X$.

As $Q$ is defined to be the sheaf image of this last map, we get a sequence of maps $O_X^{\oplus J}\to O_X^{\oplus I}\to Q\to 0$, which is exact at $Q$. If we can show that this sequence is exact at $O_X^{\oplus I}$, then we will have verified the definition of $Q$ being quasicoherent (for any point $x\in X$, the open neighborhood is $X$ and the relevant exact sequence is just the same exact sequence we've been talking about above). Showing exactness at $O_X^{\oplus I}$ directly can be a bit of a chore, though - there's a way we can shortcut this rather nicely.

For any scheme $X$, there's a scheme called the affinization of $X$ which is exactly $\operatorname{Spec} O_X(X)$. There's a natural map $X\to \operatorname{Spec} O_X(X)$ coming from the adjunction between global sections and $\operatorname{Spec}$ as functors between rings and schemes (the adjunction means that $\operatorname{Hom}(X,\operatorname{Spec} R) = \operatorname{Hom}(R,\Gamma(X))$ for any scheme $X$ and any ring $R$, so we pick $R=\Gamma(X)$ and take the map on the left side corresponding to $id_R$ on the right side). Let $Y=\operatorname{Spec} R$ be the affinization of $X$. Then as $R^{\oplus J}\to R^{\oplus I} \to A \to 0$ is exact, we get that the corresponding short exact sequence of sheaves on $Y$ given by $O_Y^{\oplus J}\to O_Y^{\oplus I} \to \widetilde{A} \to 0$ is exact, and as pullback is right-exact, the pullback to $X$ of this sequence is precisely $O_X^{\oplus J}\to O_X^{\oplus I}\to Q\to 0$ and this is still exact.