The Stacks project

Lemma 92.5.8. Let $A \to B \to C$ be ring maps. If $B$ is a polynomial algebra over $A$, then there is a distinguished triangle $L_{B/A} \otimes _ B^\mathbf {L} C \to L_{C/A} \to L_{C/B} \to L_{B/A} \otimes _ B^\mathbf {L} C[1]$ in $D(C)$.

Proof. We will use the observations of Remark 92.5.5 without further mention. Choose a resolution $\epsilon : P_\bullet \to C$ of $C$ over $B$ (for example the standard resolution). Since $B$ is a polynomial algebra over $A$ we see that $P_\bullet $ is also a resolution of $C$ over $A$. Hence $L_{C/A}$ is computed by $\Omega _{P_\bullet /A} \otimes _{P_\bullet , \epsilon } C$ and $L_{C/B}$ is computed by $\Omega _{P_\bullet /B} \otimes _{P_\bullet , \epsilon } C$. Since for each $n$ we have the short exact sequence $0 \to \Omega _{B/A} \otimes _ B P_ n \to \Omega _{P_ n/A} \to \Omega _{P_ n/B}$ (Algebra, Lemma 10.138.9) and since $L_{B/A} = \Omega _{B/A}[0]$ (Lemma 92.4.7) we obtain the result. $\square$


Comments (1)

Comment #8989 by Nico on

Tiny typo. The short exact sequence "0\to\Omega_{B/A}\otimes_BP_n\to\Omega_{P_n/A}\to\Omega_{P_n/B}" needs a at the end.


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Your email address will not be published.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 08SA. Beware of the difference between the letter 'O' and the digit '0'.