Lemma 90.4.3. Let $A \to B$ be a ring map. Let $\pi$ and $i$ be as in (90.4.0.1). There is a canonical isomorphism

$L_{B/A} = L\pi _!(Li^*\Omega _{\mathcal{O}/A}) = L\pi _!(i^*\Omega _{\mathcal{O}/A}) = L\pi _!(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B})$

in $D(B)$.

Proof. For an object $\alpha : P \to B$ of the category $\mathcal{C}$ the module $\Omega _{P/A}$ is a free $P$-module. Thus $\Omega _{\mathcal{O}/A}$ is a flat $\mathcal{O}$-module. Hence $Li^*\Omega _{\mathcal{O}/A} = i^*\Omega _{\mathcal{O}/A}$ is the sheaf of $\underline{B}$-modules which associates to $\alpha : P \to A$ the $B$-module $\Omega _{P/A} \otimes _{P, \alpha } B$. By Lemma 90.4.2 we see that the right hand side is computed by the value of this sheaf on the standard resolution which is our definition of the left hand side (Definition 90.3.2). $\square$

There are also:

• 2 comment(s) on Section 90.4: Simplicial resolutions and derived lower shriek

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).