The Stacks project

Lemma 15.32.6. Let $A \to B \to C$ be ring maps. Assume $B \to C$ is a local complete intersection homomorphism. Choose a presentation $\alpha : A[x_ s, s \in S] \to B$ with kernel $I$. Choose a presentation $\beta : B[y_1, \ldots , y_ m] \to C$ with kernel $J$. Let $\gamma : A[x_ s, y_ t] \to C$ be the induced presentation of $C$ with kernel $K$. Then we get a canonical commutative diagram

\[ \xymatrix{ 0 \ar[r] & \Omega _{A[x_ s]/A} \otimes C \ar[r] & \Omega _{A[x_ s, y_ t]/A} \otimes C \ar[r] & \Omega _{B[y_ t]/B} \otimes C \ar[r] & 0 \\ 0 \ar[r] & I/I^2 \otimes C \ar[r] \ar[u] & K/K^2 \ar[r] \ar[u] & J/J^2 \ar[r] \ar[u] & 0 } \]

with exact rows. In particular, the six term exact sequence of Algebra, Lemma 10.133.4 can be completed with a zero on the left, i.e., the sequence

\[ 0 \to H_1(\mathop{N\! L}\nolimits _{B/A} \otimes _ B C) \to H_1(L_{C/A}) \to H_1(L_{C/B}) \to \Omega _{B/A} \otimes _ B C \to \Omega _{C/A} \to \Omega _{C/B} \to 0 \]

is exact.

Proof. The only thing to prove is the injectivity of the map $I/I^2 \otimes C \to K/K^2$. By assumption the ideal $J$ is Koszul-regular. Hence we have $IA[x_ s, y_ j] \cap K^2 = IK$ by Lemma 15.31.5. This means that the kernel of $K/K^2 \to J/J^2$ is isomorphic to $IA[x_ s, y_ j]/IK$. Since $I/I^2 \otimes _ A C = IA[x_ s, y_ j]/IK$ this provides us with the desired injectivity of $I/I^2 \otimes _ A C \to K/K^2$ so that the result follows from the snake lemma, see Homology, Lemma 12.5.17. $\square$


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Your email address will not be published.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 07D4. Beware of the difference between the letter 'O' and the digit '0'.