The Stacks project

Proof. We have $H^0(X, \mathcal{O}_ X) = A$, see discussion following Situation 54.7.1. Then $\mathfrak m \subset H^0(X, \mathcal{I}) \subset H^0(X, \mathcal{O}_ X)$. The second inclusion is not an equality as $X_ s \not= \emptyset$. Thus $H^0(X, \mathcal{I}) = \mathfrak m$. As $\mathcal{I}^ n = \mathfrak m^ n\mathcal{O}_ X$ our Lemma 54.9.2 shows that $H^1(X, \mathcal{I}^ n) = 0$.

Choose generators $x_1, \ldots , x_{\mu + 1}$ of $\mathfrak m$. These define global sections of $\mathcal{I}$ which generate it. Hence a short exact sequence

Then $\mathcal{F}$ is a finite locally free $\mathcal{O}_ X$-module of rank $\mu$ and $\mathcal{F} \otimes \mathcal{I}$ is globally generated by Constructions, Lemma 27.13.9. Hence $\mathcal{F} \otimes \mathcal{I}^ n$ is globally generated for all $n \geq 1$. Thus for $n \geq 2$ we can consider the exact sequence

Applying the long exact sequence of cohomology using that $H^1(X, \mathcal{F} \otimes \mathcal{I}^{n - 1}) = 0$ by Lemma 54.9.2 we obtain that every element of $H^0(X, \mathcal{I}^ n)$ is of the form $\sum x_ i a_ i$ for some $a_ i \in H^0(X, \mathcal{I}^{n - 1})$. This shows that $H^0(X, \mathcal{I}^ n) = \mathfrak m^ n$ by induction. $\square$