Lemma 93.17.2. Let $k$ be a field. Let $1 \leq c \leq n$ be integers. Let $f_1, \ldots , f_ c \in k[x_1, \ldots x_ n]$ be elements. Let $a_{ij}$, $0 \leq i \leq n$, $1 \leq j \leq c$ be variables. Consider
\[ g_ j = f_ j + a_{0j} + a_{1j}x_1 + \ldots + a_{nj}x_ n \in k[a_{ij}][x_1, \ldots , x_ n] \]
Denote $Y \subset \mathbf{A}^{n + c(n + 1)}_ k$ the closed subscheme cut out by $g_1, \ldots , g_ c$. Denote $\pi : Y \to \mathbf{A}^{c(n + 1)}_ k$ the projection onto the affine space with variables $a_{ij}$. Then there is a nonempty Zariski open of $\mathbf{A}^{c(n + 1)}_ k$ over which $\pi $ is smooth.
Proof.
Recall that the set of points where $\pi $ is smooth is open. Thus the complement, i.e., the singular locus, is closed. By Chevalley's theorem (in the form of Morphisms, Lemma 29.22.2) the image of the singular locus is constructible. Hence if the generic point of $\mathbf{A}^{c(n + 1)}_ k$ is not in the image of the singular locus, then the lemma follows (by Topology, Lemma 5.15.15 for example). Thus we have to show there is no point $y \in Y$ where $\pi $ is not smooth mapping to the generic point of $\mathbf{A}^{c(n + 1)}_ k$. Consider the matrix of partial derivatives
\[ (\frac{\partial g_ j}{\partial x_ i}) = (\frac{\partial f_ j}{\partial x_ i} + a_{ij}) \]
The image of this matrix in $\kappa (y)$ must have rank $< c$ since otherwise $\pi $ would be smooth at $y$, see discussion in Smoothing Ring Maps, Section 16.2. Thus we can find $\lambda _1, \ldots , \lambda _ c \in \kappa (y)$ not all zero such that the vector $(\lambda _1, \ldots , \lambda _ c)$ is in the kernel of this matrix. After renumbering we may assume $\lambda _1 \not= 0$. Dividing by $\lambda _1$ we may assume our vector has the form $(1, \lambda _2, \ldots , \lambda _ c)$. Then we obtain
\[ a_{i1} = - \frac{\partial f_ j}{\partial x_1} - \sum \nolimits _{j = 2, \ldots , c} \lambda _ j(\frac{\partial f_ j}{\partial x_ i} + a_{ij}) \]
in $\kappa (y)$ for $i = 1, \ldots , n$. Moreover, since $y \in Y$ we also have
\[ a_{0j} = -f_ j - a_{1j}x_1 - \ldots - a_{nj}x_ n \]
in $\kappa (y)$. This means that the subfield of $\kappa (y)$ generated by $a_{ij}$ is contained in the subfield of $\kappa (y)$ generated by the images of $x_1, \ldots , x_ n, \lambda _2, \ldots , \lambda _ c$, and $a_{ij}$ except for $a_{i1}$ and $a_{0j}$. We count and we see that the transcendence degree of this is at most $c(n + 1) - 1$. Hence $y$ cannot map to the generic point as desired.
$\square$
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