The Stacks project

Proof. Say $N = \{ \nu _ j\} $ with $\nu _ j = (\nu _{j1}, \ldots , \nu _{jn})$. Let $A_ i = \max _ j \nu _{ji} - \min _ j \nu _{ji}$. If for each $i$ we have $e_{i - 1} > A_ ie_ i + A_{i + 1}e_{i + 1} + \ldots + A_ ne_ n$ then the lemma holds. For suppose that $e \cdot (\nu - \nu ') = 0$. Then for $n \ge 2$,

We may assume that $(\nu _1 - \nu '_1) \ge 0$. If $(\nu _1 - \nu '_1) > 0$, then

This contradiction implies that $\nu '_1 = \nu _1$. By induction, $\nu '_ i = \nu _ i$ for $2 \le i \le n$. $\square$

Proof. Write $g = \sum _{\nu \in N} a_\nu x^\nu $ with $a_\nu \in R$ not zero. Here $N$ is a finite set of multi-indices as in Lemma 10.115.1 and $x^\nu = x_1^{\nu _1} \ldots x_ n^{\nu _ n}$. Note that the leading term in

Hence the lemma follows from Lemma 10.115.1 which guarantees that there is exactly one nonzero term $a_\nu x^\nu $ of $g$ which gives rise to the leading term of $g(x_1 + x_ n^{e_1}, x_2 + x_ n^{e_2}, \ldots , x_{n - 1} + x_ n^{e_{n - 1}}, x_ n)$, i.e., $a = a_\nu $ for the unique $\nu \in N$ such that $e \cdot \nu $ is maximal. $\square$

Proof. Pick $f \in I$, $f\not= 0$. It suffices to show the lemma for $k[x_1, \ldots , x_ n]/(f)$ since $S$ is a quotient of that ring. We will take $y_ i = x_ i - x_ n^{e_ i}$, $i = 1, \ldots , n-1$ for suitable integers $e_ i$. When does this work? It suffices to show that $\overline{x_ n} \in k[x_1, \ldots , x_ n]/(f)$ is integral over the ring $k[y_1, \ldots , y_{n-1}]$. The equation for $\overline{x_ n}$ over this ring is

Hence we are done if we can show there exists integers $e_ i$ such that the leading coefficient with respect to $x_ n$ of the equation above is a nonzero element of $k$. This can be achieved for example by choosing $e_1 \gg e_2 \gg \ldots \gg e_{n-1}$, see Lemma 10.115.2. $\square$

Proof. By induction on $n$, with $n = 0$ being trivial. If $I = 0$, then take $r = n$ and $y_ i = x_ i$. If $I \not= 0$, then choose $y_1, \ldots , y_{n-1}$ as in Lemma 10.115.3. Let $S' \subset S$ be the subring generated by the images of the $y_ i$. By induction we can choose $r$ and $z_1, \ldots , z_ r \in k[y_1, \ldots , y_{n-1}]$ such that (a), (b) hold for $k[z_1, \ldots , z_ r] \to S'$. Since $S' \to S$ is injective and finite we see (a), (b) hold for $k[z_1, \ldots , z_ r] \to S$. The last assertion follows from Lemma 10.112.4. $\square$

Proof. Note that by definition $\dim _ x(X)$ is the minimum of the dimensions of $S_ g$ for $g \in S$, $g \not\in \mathfrak q$, i.e., the minimum is attained. Thus the lemma follows from Lemma 10.115.4. $\square$

Proof. By induction on $n$. The case $n = 0$ is clear. Assume $n > 0$. If $r = n$, then $\mathfrak q = (0)$ and the result is clear. Choose a nonzero $f \in \mathfrak q$. Of course $f$ is nonconstant. After applying an automorphism of the form

we may assume that $f$ is monic in $x_ n$ over $k[x_1, \ldots , x_ n]$, see Lemma 10.115.2. Hence the ring map

is finite. Moreover $y_ n \in \mathfrak q \cap k[y_1, \ldots , y_ n]$ by construction. Thus $\mathfrak q \cap k[y_1, \ldots , y_ n] = \mathfrak pk[y_1, \ldots , y_ n] + (y_ n)$ where $\mathfrak p \subset k[y_1, \ldots , y_{n - 1}]$ is a prime ideal. Note that $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is finite, and hence $r = \text{trdeg}_ k\ \kappa (\mathfrak p)$. Apply the induction hypothesis to the pair $(k[y_1, \ldots , y_{n - 1}], \mathfrak p)$ and we obtain a finite ring map $k[z_1, \ldots , z_{n - 1}] \to k[y_1, \ldots , y_{n - 1}]$ such that $\mathfrak p \cap k[z_1, \ldots , z_{n - 1}] = (z_{r + 1}, \ldots , z_{n - 1})$. We extend the ring map $k[z_1, \ldots , z_{n - 1}] \to k[y_1, \ldots , y_{n - 1}]$ to a ring map $k[z_1, \ldots , z_ n] \to k[y_1, \ldots , y_ n]$ by mapping $z_ n$ to $y_ n$. The composition of the ring maps

solves the problem. $\square$

Proof. Pick $x_1, \ldots , x_ n \in S$ which generate $S$ over $R$. Let $K$ be the fraction field of $R$ and $S_ K = S \otimes _ R K$. By Lemma 10.115.4 we can find $y_1, \ldots , y_ d \in S$ such that $K[y_1, \ldots , y_ d] \to S_ K$ is a finite injective map. Note that $y_ i \in S$ because we may pick the $y_ j$ in the $\mathbf{Z}$-algebra generated by $x_1, \ldots , x_ n$. As a finite ring map is integral (see Lemma 10.36.3) we can find monic $P_ i \in K[y_1, \ldots , y_ d][T]$ such that $P_ i(x_ i) = 0$ in $S_ K$. Let $f \in R$ be a nonzero element such that $fP_ i \in R[y_1, \ldots , y_ d][T]$ for all $i$. Then $fP_ i(x_ i)$ maps to zero in $S_ K$. Hence after replacing $f$ by another nonzero element of $R$ we may also assume $fP_ i(x_ i)$ is zero in $S$. Set $x_ i' = fx_ i$ and let $S' \subset S$ be the $R$-subalgebra generated by $y_1, \ldots , y_ d$ and $x'_1, \ldots , x'_ n$. Note that $x'_ i$ is integral over $R[y_1, \ldots , y_ d]$ as we have $Q_ i(x_ i') = 0$ where $Q_ i = f^{\deg _ T(P_ i)}P_ i(T/f)$ which is a monic polynomial in $T$ with coefficients in $R[y_1, \ldots , y_ d]$ by our choice of $f$. Hence $R[y_1, \ldots , y_ d] \subset S'$ is finite by Lemma 10.36.5. Since $S' \subset S$ we have $S'_ f \subset S_ f$ (localization is exact). On the other hand, the elements $x_ i = x'_ i/f$ in $S'_ f$ generate $S_ f$ over $R_ f$ and hence $S'_ f \to S_ f$ is surjective. Whence $S'_ f \cong S_ f$ and we win. $\square$