Lemma 92.26.4 (08VG)—The Stacks project

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Table of contents
Part 6: Deformation Theory
Chapter 92: The Cotangent Complex
Section 92.26: The cotangent complex of a morphism of algebraic spaces
Lemma 92.26.4 (cite)

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Lemma 92.26.4. Let $\Lambda$ be a ring. Let $X$ be an algebraic space over $\Lambda$ . Then

$L_{X/\mathop{\mathrm{Spec}}(\Lambda )} = L_{\mathcal{O}_ X/\underline{\Lambda }}$

where $\underline{\Lambda }$ is the constant sheaf with value $\Lambda$ on $X_{\acute{e}tale}$ .

Proof. Let $p : X \to \mathop{\mathrm{Spec}}(\Lambda )$ be the structure morphism. Let $q : \mathop{\mathrm{Spec}}(\Lambda )_{\acute{e}tale}\to (*, \Lambda )$ be the obvious morphism. By the distinguished triangle of Lemma 92.22.3 it suffices to show that $L_ q = 0$ . To see this it suffices to show (Properties of Spaces, Theorem 66.19.12) for a geometric point $\overline{t} : \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(\Lambda )$ that

$(L_ q)_{\overline{t}} = L_{\mathcal{O}_{\mathop{\mathrm{Spec}}(\Lambda )_{\acute{e}tale}, \overline{t}}/\Lambda }$

(Lemma 92.18.9) is zero. Since $\mathcal{O}_{\mathop{\mathrm{Spec}}(\Lambda )_{\acute{e}tale}, \overline{t}}$ is a strict henselization of a local ring of $\Lambda$ (Properties of Spaces, Lemma 66.22.1) this follows from Lemma 92.8.4. $\square$

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