# What are the sections of an ideal sheaf on a scheme?

Suppose $$X$$ is a scheme and $$f_1,...,f_n\in \Gamma(X,\mathcal O)$$ are global sections.
One often reads about the ideal sheaf $$\mathcal I=\mathcal (f_1,...,f_n)\subset \mathcal O$$, but I have never seen it defined.
The definition should be that $$\mathcal I$$ is the sheaf associated to the presheaf $$\mathcal I^-$$ whose value on an open subset $$U\subset X$$ is $$\mathcal I^-(U)=({f_1}{\vert_U},...,{f_n}{\vert_U})\cdot\mathcal O(U)$$. Right?
My question is simply: Is there a non-tautological sufficient condition on $$U$$ and the $$f_i$$'s guaranteeing that $$\mathcal I(U)=\mathcal I^-(U)$$?
The only sufficient condition I can think of is that $$n=1$$ and $$U$$ is an arbitrary open subset of an integral scheme $$X$$.

• A sufficient condition is that $U$ is affine. Indeed $\mathcal{I}$ fits in a s.e.s. $0\to \mathcal{K}\to \mathcal{O}^n\to \mathcal{I}\to 0$ and $H^1(U,\mathcal{K}) = 0$ for $U$ affine. Jul 13, 2020 at 16:08
• @Christ: Thank you very much for this fine answer: +1. Why not transform this comment in a genuine answer? Jul 14, 2020 at 11:26

A sufficient condition is that $$U$$ is affine. Indeed $$\mathcal I$$ fits in an s.e.s. $$0 \to \mathcal K \to \mathcal O^n \to \mathcal I \to 0$$ and $$H^1(U, \mathcal K) = 0$$ for $$U$$ affine.