Lemma 37.62.4. Let $f : X \to S$ be a local complete intersection morphism. Then

$f$ is locally of finite presentation,

$f$ is pseudo-coherent, and

$f$ is perfect.

Lemma 37.62.4. Let $f : X \to S$ be a local complete intersection morphism. Then

$f$ is locally of finite presentation,

$f$ is pseudo-coherent, and

$f$ 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 37.61.7. A smooth morphism is perfect as it is flat and locally of finite presentation, see Lemma 37.61.5. Finally a composition of perfect morphisms is perfect, see Lemma 37.61.4.
$\square$

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