Remark 49.11.6. Let $\pi _ d : Y_ d \to X_ d$ be as in Example 49.11.2. Let $U_ d \subset X_ d$ be the maximal open over which $V_ d = \pi _ d^{-1}(U_ d)$ is finite syntomic as in Lemma 49.11.3. Then it is also true that $V_ d$ is smooth over $\mathbf{Z}$. (Of course the morphism $V_ d \to U_ d$ is not smooth when $d \geq 2$.) Arguing as in the proof of Lemma 49.11.4 this corresponds to the following deformation problem: given a small extension $C' \to C$ and a finite syntomic $C$-algebra $B$ with a section $B \to C$, find a finite syntomic $C'$-algebra $B'$ and a section $B' \to C'$ whose tensor product with $C$ recovers $B \to C$. By Lemma 49.11.1 we may write $B = C[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$ as a relative global complete intersection. After a change of coordinates we may assume $x_1, \ldots , x_ n$ are in the kernel of $B \to C$. Then the polynomials $f_ i$ have vanishing constant terms. Choose any lifts $f'_ i \in C'[x_1, \ldots , x_ n]$ of $f_ i$ with vanishing constant terms. Then $B' = C'[x_1, \ldots , x_ n]/(f'_1, \ldots , f'_ n)$ with section $B' \to C'$ sending $x_ i$ to zero works.

## 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Your email address will not be published.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.

## Comments (2)

Comment #7808 by Peng Du on

Comment #8038 by Stacks Project on