The Stacks project

Lemma 90.8.1. Let $A \to A' \to B$ be ring maps such that $B = B \otimes _ A^\mathbf {L} A'$. Then $L_{B/A} = L_{B/A'}$ in $D(B)$.

Proof. According to the discussion above (i.e., using Lemma 90.6.1) and Lemma 90.4.3 we have to show that the sheaf given by the rule $(P \to B) \mapsto \Omega _{P/A} \otimes _ P B$ on $\mathcal{C}_{B/A}$ is the pullback of the sheaf given by the rule $(P \to B) \mapsto \Omega _{P/A'} \otimes _ P B$. The pullback functor $g^{-1}$ is given by precomposing with the functor $u : \mathcal{C}_{B/A} \to \mathcal{C}_{B/A'}$, $(P \to B) \mapsto (P \otimes _ A A' \to B)$. Thus we have to show that

\[ \Omega _{P/A} \otimes _ P B = \Omega _{P \otimes _ A A'/A'} \otimes _{(P \otimes _ A A')} B \]

By Algebra, Lemma 10.131.12 the right hand side is equal to

\[ (\Omega _{P/A} \otimes _ A A') \otimes _{(P \otimes _ A A')} B \]

Since $P$ is a polynomial algebra over $A$ the module $\Omega _{P/A}$ is free and the equality is obvious. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08QZ. Beware of the difference between the letter 'O' and the digit '0'.