The Stacks project

Proof. Choose $r \geq 0$ and $y_1, \ldots , y_ r \in C$ which generate $C$ over $\mathop{\mathrm{Im}}(\varphi )$ and let $0 \leq s \leq r$ be such that $y_1, \ldots , y_ s$ are integral over $\mathop{\mathrm{Im}}(\varphi )$ such that $r - s = NF(C, \varphi , \tau ) > 0$. Since $B$ is finite over $A$, the image of $y_{s + 1}$ in $B$ satisfies a monic polynomial over $A$. Hence we can find $d \geq 1$ and $f_1, \ldots , f_ d \in k[x_1, \ldots , x_ n]$ such that

Since $\varphi : k[x_1, \ldots , x_ n] \to C$ is étale, we can find a nonzero and nonconstant polynomial $g \in k[T_1, \ldots , T_{n + 1}]$ such that

To see this you can use for example that $C \otimes _{\varphi , k[x_1, \ldots , x_ n]} k(x_1, \ldots , x_ n)$ is a finite product of finite separable field extensions of $k(x_1, \ldots , x_ n)$ (see Algebra, Lemmas 10.143.4) and hence $z$ satisfies a monic polynomial over $k(x_1, \ldots , x_ n)$. Clearing denominators we obtain $g$.

The existence of $g$ and Algebra, Lemma 10.115.2 produce integers $e_1, e_2, \ldots , e_ n \geq 1$ such that $z$ is integral over the subring $C'$ of $C$ generated by $t_1 = \varphi (x_1) + z^{pe_1}, \ldots , t_ n = \varphi (x_ n) + z^{pe_ n}$. Of course, the elements $\varphi (x_1), \ldots , \varphi (x_ n)$ are also integral over $C'$ as are the elements $y_1, \ldots , y_ s$. Finally, by our choice of $z$ the element $y_{s + 1}$ is integral over $C'$ too.

Consider the ring map

with image $C'$. Since $\text{d}(\varphi (x_ i)) = \text{d}(t_ i) = \text{d}(\varphi '(x_ i))$ in $\Omega _{C/k}$ (and this is where we use the characteristic of $k$ is $p > 0$) we conclude that $\varphi '$ is étale because $\varphi $ is étale, see Algebra, Lemma 10.151.9. Observe that $\varphi '(x_ i) - \varphi (x_ i) = t_ i - \varphi (x_ i) = z^{pe_ i}$ is in the kernel $J$ of the map $C \to C/\varphi (I)C \to B$ by our choice of $z$ as an element of $J$. Hence for $f \in I$ the element

is in $J$ as well. In other words, $\varphi '(I)C \subset J$ and we obtain a surjection

of algebras étale over $A$. Finally, the algebra $C$ is generated by the elements $\varphi (x_1), \ldots , \varphi (x_ n), y_1, \ldots , y_ r$ over $C' = \mathop{\mathrm{Im}}(\varphi ')$ with $\varphi (x_1), \ldots , \varphi (x_ n), y_1, \ldots , y_{s + 1}$ integral over $C' = \mathop{\mathrm{Im}}(\varphi ')$. Hence $NF(C, \varphi ', \tau ') < r - s = NF(C, \varphi , \tau )$. This finishes the proof. $\square$