Let $X$ be a scheme and $\mathscr{I}$ a sheaf of ideals. Let $U$ be an open set of $X$ and $s_1,\dots,s_n \in \Gamma(U,\mathscr{I})$. I am seeking a clarification of what we mean by "the ideal sheaf $\mathscr{I}$' generated by the $s_i$'s" (e.g., Hartshorne refers to this in the second part of the proof of Theorem II.8.17). It seems to me that this is the extension to $X$ by zero (in the sense of Exercise II.1.19 of Hartshorne) of the subsheaf $\mathscr{J}$ of $\mathscr{I}|_U$ generated by the $s_i$'s. This is the sheaf of $X$ associated to the presheaf $V \mapsto \mathscr{J}(V)$ for $V \subseteq U$ and $\mathscr{J}(V)=0$ otherwise. Using the fact that a section of that sheaf is a collection of compatible local sections of the presheaf and the following algebraic fact we finally get that $\mathscr{I}'$ is an ideal sheaf. Algebraic fact: let $A$ be ring and $f_j \in A$ such that $(f_1,…,f_m)=A$. Let $s_j \in A_{f_j}$ such that $s_j = s_k$ in $A_{f_j f_k}$. Then there is an $a \in A$ such that $s_j = a$ in $A_{f_j}$ for every $j$. Do i have it right?
Ideal sheaf generated by local sections of an ideal sheaf
algebraic-geometryquasicoherent-sheaves
Best Answer
No, this isn't the intended meaning in that proof. Here's the relevant section of the text:
What Hartshorne is really doing here is that he's restricting to an open neighborhood $U$ of $y$ where $\ker\varphi$ is free and generated by the sections $x_i$. The ideal sheaf $\mathscr{I}'$ is an ideal sheaf constructed on the open subscheme $U$ generated by the sections $x_i$, and all of the conclusions of the proof hold locally - one can then patch everything together to get the global version of what we want.