Lemma 37.6.6 (02HD)—The Stacks project

Loading [MathJax]/extensions/tex2jax.js

Table of contents
Part 2: Schemes
Chapter 37: More on Morphisms
Section 37.6: Formally unramified morphisms
Lemma 37.6.6 (cite)

previous lemma
next lemma

numberstags

history
statistics

Lemma 37.6.6. Let $f : X \to S$ be a morphism of schemes. Assume $X$ and $S$ are affine. Then $f$ is formally unramified if and only if $\mathcal{O}_ S(S) \to \mathcal{O}_ X(X)$ is a formally unramified ring map.

Proof. This is immediate from the definitions (Definition 37.6.1 and Algebra, Definition 10.148.1) by the equivalence of categories of rings and affine schemes, see Schemes, Lemma 26.6.5. $\square$

Comments (0)

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