Lemma 92.25.3. In the situation above there is a canonical isomorphism

in $D(\mathcal{O}_ X)$.

Lemma 92.25.3. In the situation above there is a canonical isomorphism

\[ L_{X/\Lambda } = L\pi _!(Li^*\Omega _{\mathcal{O}/\underline{\Lambda }}) = L\pi _!(i^*\Omega _{\mathcal{O}/\underline{\Lambda }}) = L\pi _!(\Omega _{\mathcal{O}/\underline{\Lambda }} \otimes _\mathcal {O} \underline{\mathcal{O}}_ X) \]

in $D(\mathcal{O}_ X)$.

**Proof.**
We first observe that for any object $(U \to \mathbf{A})$ of $\mathcal{C}_{X/\Lambda }$ the value of the sheaf $\mathcal{O}$ is a polynomial algebra over $\Lambda $. Hence $\Omega _{\mathcal{O}/\underline{\Lambda }}$ is a flat $\mathcal{O}$-module and we conclude the second and third equalities of the statement of the lemma hold.

By Remark 92.25.2 the object $L\pi _!(\Omega _{\mathcal{O}/\underline{\Lambda }} \otimes _\mathcal {O} \underline{\mathcal{O}}_ X)$ is computed as the sheafification of the complex of presheaves

\[ U \mapsto \left(\Omega _{\mathcal{O}/\underline{\Lambda }} \otimes _\mathcal {O} \underline{\mathcal{O}}_ X\right)(\mathbf{A}_{\bullet , U}) = \Omega _{P_{\bullet , U}/\Lambda } \otimes _{P_{\bullet , U}} \mathcal{O}_ X(U) = L_{\mathcal{O}_ X(U)/\Lambda } \]

using notation as in Remark 92.25.2. Now Remark 92.18.5 shows that $L\pi _!(\Omega _{\mathcal{O}/\underline{\Lambda }} \otimes _\mathcal {O} \underline{\mathcal{O}}_ X)$ computes the cotangent complex of the map of rings $\underline{\Lambda } \to \mathcal{O}_ X$ on $X$. This is what we want by Lemma 92.24.3. $\square$

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