Lemma 35.23.28. The property $\mathcal{P}(f) =$“$f$ is unramified” is fpqc local on the base. The property $\mathcal{P}(f) =$“$f$ is G-unramified” is fpqc local on the base.
Proof. A morphism is unramified (resp. G-unramified) if and only if it is locally of finite type (resp. finite presentation) and its diagonal morphism is an open immersion (see Morphisms, Lemma 29.35.13). We have seen already that being locally of finite type (resp. locally of finite presentation) and an open immersion is fpqc local on the base (Lemmas 35.23.11, 35.23.10, and 35.23.16). Hence the result follows formally. $\square$
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