Lemma 16.3.4. Let $R \to A$ be a smooth ring map. Then there exists a smooth $R$-algebra map $A \to B$ with a retraction such that $B$ is standard smooth over $R$, i.e.,

$B \cong R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$

and $\det (\partial f_ j/\partial x_ i)_{i, j = 1, \ldots , c}$ is invertible in $B$.

Proof. Apply Lemma 16.3.3 to get a smooth $R$-algebra map $A \to C$ with a retraction such that $C = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ is a relative global complete intersection over $R$. As $C$ is smooth over $R$ we have a short exact sequence

$0 \to \bigoplus \nolimits _{j = 1, \ldots , c} C f_ j \to \bigoplus \nolimits _{i = 1, \ldots , n} C\text{d}x_ i \to \Omega _{C/R} \to 0$

Since $\Omega _{C/R}$ is a projective $C$-module this sequence is split. Choose a left inverse $t$ to the first map. Say $t(\text{d}x_ i) = \sum c_{ij} f_ j$ so that $\sum _ i \frac{\partial f_ j}{\partial x_ i} c_{i\ell } = \delta _{j\ell }$ (Kronecker delta). Let

$B' = C[y_1, \ldots , y_ c] = R[x_1, \ldots , x_ n, y_1, \ldots , y_ c]/(f_1, \ldots , f_ c)$

The $R$-algebra map $C \to B'$ has a retraction given by mapping $y_ j$ to zero. We claim that the map

$R[z_1, \ldots , z_ n] \longrightarrow B',\quad z_ i \longmapsto x_ i - \sum \nolimits _ j c_{ij} y_ j$

is étale at every point in the image of $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(B')$. In $\Omega _{B'/R[z_1, \ldots , z_ n]}$ we have

$0 = \text{d}f_ j - \sum \nolimits _ i \frac{\partial f_ j}{\partial x_ i} \text{d}z_ i \equiv \sum \nolimits _{i, \ell } \frac{\partial f_ j}{\partial x_ i} c_{i\ell } \text{d}y_\ell \equiv \text{d}y_ j \bmod (y_1, \ldots , y_ c)\Omega _{B'/R[z_1, \ldots , z_ n]}$

Since $0 = \text{d}z_ i = \text{d}x_ i$ modulo $\sum B'\text{d}y_ j + (y_1, \ldots , y_ c)\Omega _{B'/R[z_1, \ldots , z_ n]}$ we conclude that

$\Omega _{B'/R[z_1, \ldots , z_ n]}/ (y_1, \ldots , y_ c)\Omega _{B'/R[z_1, \ldots , z_ n]} = 0.$

As $\Omega _{B'/R[z_1, \ldots , z_ n]}$ is a finite $B'$-module by Nakayama's lemma there exists a $g \in 1 + (y_1, \ldots , y_ c)$ that $(\Omega _{B'/R[z_1, \ldots , z_ n]})_ g = 0$. This proves that $R[z_1, \ldots , z_ n] \to B'_ g$ is unramified, see Algebra, Definition 10.151.1. For any ring map $R \to k$ where $k$ is a field we obtain an unramified ring map $k[z_1, \ldots , z_ n] \to (B'_ g) \otimes _ R k$ between smooth $k$-algebras of dimension $n$. It follows that $k[z_1, \ldots , z_ n] \to (B'_ g) \otimes _ R k$ is flat by Algebra, Lemmas 10.128.1 and 10.140.2. By the critère de platitude par fibre (Algebra, Lemma 10.128.8) we conclude that $R[z_1, \ldots , z_ n] \to B'_ g$ is flat. Finally, Algebra, Lemma 10.143.7 implies that $R[z_1, \ldots , z_ n] \to B'_ g$ is étale. Set $B = B'_ g$. Note that $C \to B$ is smooth and has a retraction, so also $A \to B$ is smooth and has a retraction. Moreover, $R[z_1, \ldots , z_ n] \to B$ is étale. By Algebra, Lemma 10.143.2 we can write

$B = R[z_1, \ldots , z_ n, w_1, \ldots , w_ c]/(g_1, \ldots , g_ c)$

with $\det (\partial g_ j/\partial w_ i)$ invertible in $B$. This proves the lemma. $\square$

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).