$\xymatrix{ Z \ar[r]_ i \ar[rd]_ j & X \ar[d] \\ & Y }$

be a commutative diagram of schemes where $i$ and $j$ are formally unramified. Then there is a canonical exact sequence

$\mathcal{C}_{Z/Y} \to \mathcal{C}_{Z/X} \to i^*\Omega _{X/Y} \to 0$

where the first arrow comes from Lemma 37.7.5 and the second from Lemma 37.7.10.

Proof. Denote $Z \to Z'$ the universal first order thickening of $Z$ over $X$. Denote $Z \to Z''$ the universal first order thickening of $Z$ over $Y$. By Lemma 37.7.10 here is a canonical morphism $Z' \to Z''$ so that we have a commutative diagram

$\xymatrix{ Z \ar[r]_{i'} \ar[rd]_{j'} & Z' \ar[r] \ar[d] & X \ar[d] \\ & Z'' \ar[r] & Y }$

Apply Morphisms, Lemma 29.32.18 to the left triangle to get an exact sequence

$\mathcal{C}_{Z/Z''} \to \mathcal{C}_{Z/Z'} \to (i')^*\Omega _{Z'/Z''} \to 0$

As $Z''$ is formally unramified over $Y$ (see Lemma 37.7.4) we have $\Omega _{Z'/Z''} = \Omega _{Z/Y}$ (by combining Lemma 37.6.7 and Morphisms, Lemma 29.32.9). Then we have $(i')^*\Omega _{Z'/Y} = i^*\Omega _{X/Y}$ by Lemma 37.7.9. $\square$

## 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Your email address will not be published.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 067V. Beware of the difference between the letter 'O' and the digit '0'.