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 with 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.

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## Comments (1)

Comment #7808 by Peng Du on