Lemma 76.14.7 (04GA)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 76: More on Morphisms of Spaces
Section 76.14: Formally unramified morphisms
Lemma 76.14.7 (cite)

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Lemma 76.14.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The following are equivalent:

The morphism $f$ is unramified,
the morphism $f$ is locally of finite type and $\Omega _{X/Y} = 0$, and
the morphism $f$ is locally of finite type and formally unramified.

Proof. Choose a diagram

\[ \xymatrix{ U \ar[d] \ar[r]_\psi & V \ar[d] \\ X \ar[r]^ f & Y } \]

where $U$ and $V$ are schemes and the vertical arrows are étale and surjective. Then we see

\begin{align*} f\text{ unramified} & \Leftrightarrow \psi \text{ unramified} \\ & \Leftrightarrow \psi \text{ locally finite type and }\Omega _{U/V} = 0 \\ & \Leftrightarrow f\text{ locally finite type and }\Omega _{X/Y} = 0 \\ & \Leftrightarrow f\text{ locally finite type and formally unramified} \end{align*}

Here we have used Morphisms, Lemma 29.35.2 and Lemma 76.14.6. $\square$

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