Lemma 37.60.4. Let $f : X \to S$ be a local complete intersection morphism. Then

1. $f$ is locally of finite presentation,

2. $f$ is pseudo-coherent, and

3. $f$ is perfect.

Proof. Since a perfect morphism is pseudo-coherent (because a perfect ring map is pseudo-coherent) and a pseudo-coherent morphism is locally of finite presentation (because a pseudo-coherent ring map is of finite presentation) it suffices to prove the last statement. Being perfect is a local property, hence we may assume that $f$ factors as $\pi \circ i$ where $\pi$ is smooth and $i$ is a Koszul-regular immersion. A Koszul-regular immersion is perfect, see Lemma 37.59.7. A smooth morphism is perfect as it is flat and locally of finite presentation, see Lemma 37.59.5. Finally a composition of perfect morphisms is perfect, see Lemma 37.59.4. $\square$

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