Lemma 67.48.10 (0825)—The Stacks project

Loading [MathJax]/extensions/tex2jax.js

Table of contents
Part 4: Algebraic Spaces
Chapter 67: Morphisms of Algebraic Spaces
Section 67.48: Relative normalization of algebraic spaces
Lemma 67.48.10 (cite)

previous lemma
next lemma

numberstags

history
statistics
3 tags refer to this tag

Lemma 67.48.10. Let $S$ be a scheme. Let $f : Y \to X$ be an integral morphism of algebraic spaces over $S$. Then the integral closure of $X$ in $Y$ is equal to $Y$.

Proof. By Lemma 67.45.7 this is a special case of Lemma 67.48.9. $\square$

Comments (0)

There are also:

1 comment(s) on Section 67.48: Relative normalization of algebraic spaces

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).

Comments (0)

Post a comment