Lemma 29.35.3. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

1. The morphism $f$ is unramified (resp. G-unramified).

2. For every affine open $U \subset X$, $V \subset S$ with $f(U) \subset V$ the ring map $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is unramified (resp. G-unramified).

3. There exists an open covering $S = \bigcup _{j \in J} V_ j$ and open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that each of the morphisms $U_ i \to V_ j$, $j\in J, i\in I_ j$ is unramified (resp. G-unramified).

4. There exists an affine open covering $S = \bigcup _{j \in J} V_ j$ and affine open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that the ring map $\mathcal{O}_ S(V_ j) \to \mathcal{O}_ X(U_ i)$ is unramified (resp. G-unramified), for all $j\in J, i\in I_ j$.

Moreover, if $f$ is unramified (resp. G-unramified) then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $f|_ U : U \to V$ is unramified (resp. G-unramified).

Proof. This follows from Lemma 29.14.3 if we show that the property “$R \to A$ is unramified” is local. We check conditions (a), (b) and (c) of Definition 29.14.1. These properties are proved in Algebra, Lemma 10.151.3. $\square$

Comment #3027 by Anon on

In item (2), please change "For every affine opens" to "For every affine open."

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