Lemma 58.20.1. Let $(A, \mathfrak m)$ be a Noetherian local ring. Set $X = \mathop{\mathrm{Spec}}(A)$ and let $U = X \setminus \{ \mathfrak m\} $. Let $\pi : Y \to X$ be a finite morphism such that $\text{depth}(\mathcal{O}_{Y, y}) \geq 2$ for all closed points $y \in Y$. Then $Y$ is the spectrum of $B = \mathcal{O}_ Y(\pi ^{-1}(U))$.
Proof. Set $V = \pi ^{-1}(U)$ and denote $\pi ' : V \to U$ the restriction of $\pi $. Consider the $\mathcal{O}_ X$-module map
where $j : U \to X$ is the inclusion morphism. We claim Divisors, Lemma 31.5.11 applies to this map. If so, then $B = \Gamma (Y, \mathcal{O}_ Y)$ and we see that the lemma holds. Let $x \in X$ be the closed point. It suffices to show that $\text{depth}((\pi _*\mathcal{O}_ Y)_ x) \geq 2$. Let $y_1, \ldots , y_ n \in Y$ be the points mapping to $x$. By Algebra, Lemma 10.72.11 it suffices to show that $\text{depth}(\mathcal{O}_{Y, y_ i}) \geq 2$ for $i = 1, \ldots , n$. Since this is the assumption of the lemma the proof is complete. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)