The Stacks project

10.115 Noether normalization

In this section we prove variants of the Noether normalization lemma. The key ingredient we will use is contained in the following two lemmas.

Lemma 10.115.1. Let $n \in \mathbf{N}$. Let $N$ be a finite nonempty set of multi-indices $\nu = (\nu _1, \ldots , \nu _ n)$. Given $e = (e_1, \ldots , e_ n)$ we set $e \cdot \nu = \sum e_ i\nu _ i$. Then for $e_1 \gg e_2 \gg \ldots \gg e_{n-1} \gg e_ n$ we have: If $\nu , \nu ' \in N$ then

\[ (e \cdot \nu = e \cdot \nu ') \Leftrightarrow (\nu = \nu ') \]

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 $n \ge 2$,

\[ e_1(\nu _1 - \nu '_1) = \sum \nolimits _{i = 2}^ n e_ i(\nu '_ i - \nu _ i). \]

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

\[ e_1(\nu _1 - \nu '_1) \ge e_1 > A_2e_2 + \ldots + A_ ne_ n \ge \sum \nolimits _{i = 2}^ n e_ i|\nu '_ i - \nu _ i| \ge \sum \nolimits _{i = 2}^ n e_ i(\nu '_ i - \nu _ i). \]

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

Lemma 10.115.2. Let $R$ be a ring. Let $g \in R[x_1, \ldots , x_ n]$ be an element which is nonconstant, i.e., $g \not\in R$. For $e_1 \gg e_2 \gg \ldots \gg e_{n-1} \gg e_ n = 1$ the polynomial

\[ g(x_1 + x_ n^{e_1}, x_2 + x_ n^{e_2}, \ldots , x_{n - 1} + x_ n^{e_{n - 1}}, x_ n) = ax_ n^ d + \text{lower order terms in }x_ n \]

where $d > 0$ and $a \in R$ is one of the nonzero coefficients of $g$.

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

\[ (x_1 + x_ n^{e_1})^{\nu _1} \ldots (x_{n-1} + x_ n^{e_{n-1}})^{\nu _{n-1}} x_ n^{\nu _ n} \quad \text{is}\quad x_ n^{e_1\nu _1 + \ldots + e_{n-1}\nu _{n-1} + \nu _ n}. \]

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. $\square$

Lemma 10.115.3. Let $k$ be a field. Let $S = k[x_1, \ldots , x_ n]/I$ for some proper ideal $I$. If $I \not= 0$, then there exist $y_1, \ldots , y_{n-1} \in k[x_1, \ldots , x_ n]$ such that $S$ is finite over $k[y_1, \ldots , y_{n-1}]$. Moreover we may choose $y_ i$ to be in the $\mathbf{Z}$-subalgebra of $k[x_1, \ldots , x_ n]$ generated by $x_1, \ldots , x_ n$.

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

\[ f(y_1 + x_ n^{e_1}, \ldots , y_{n-1} + x_ n^{e_{n-1}}, x_ n) = 0. \]

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. $\square$


Lemma 10.115.4. Let $k$ be a field. Let $S = k[x_1, \ldots , x_ n]/I$ for some ideal $I$. If $I \neq (1)$, there exist $r\geq 0$, and $y_1, \ldots , y_ r \in k[x_1, \ldots , x_ n]$ such that (a) the map $k[y_1, \ldots , y_ r] \to S$ is injective, and (b) the map $k[y_1, \ldots , y_ r] \to S$ is finite. In this case the integer $r$ is the dimension of $S$. Moreover we may choose $y_ i$ to be in the $\mathbf{Z}$-subalgebra of $k[x_1, \ldots , x_ n]$ generated by $x_1, \ldots , x_ n$.

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. $\square$

Lemma 10.115.5. Let $k$ be a field. Let $S$ be a finite type $k$ algebra and denote $X = \mathop{\mathrm{Spec}}(S)$. Let $\mathfrak q$ be a prime of $S$, and let $x \in X$ be the corresponding point. There exists a $g \in S$, $g \not\in \mathfrak q$ such that $\dim (S_ g) = \dim _ x(X) =: d$ and such that there exists a finite injective map $k[y_1, \ldots , y_ d] \to S_ g$.

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. $\square$

Lemma 10.115.6. Let $k$ be a field. Let $\mathfrak q \subset k[x_1, \ldots , x_ n]$ be a prime ideal. Set $r = \text{trdeg}_ k\ \kappa (\mathfrak q)$. Then there exists a finite ring map $\varphi : k[y_1, \ldots , y_ n] \to k[x_1, \ldots , x_ n]$ such that $\varphi ^{-1}(\mathfrak q) = (y_{r + 1}, \ldots , y_ n)$.

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

\[ k[x_1, \ldots , x_ n] \longrightarrow k[x_1, \ldots , x_ n], \quad x_ n \mapsto x_ n, \quad x_ i \mapsto x_ i + x_ n^{e_ i}\ (i < n) \]

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

\[ k[y_1, \ldots , y_ n] \longrightarrow k[x_1, \ldots , x_ n], \quad y_ n \mapsto f, \quad y_ i \mapsto x_ i\ (i < n) \]

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

\[ k[z_1, \ldots , z_ n] \to k[y_1, \ldots , y_ n] \to k[x_1, \ldots , x_ n] \]

solves the problem. $\square$

Lemma 10.115.7. Let $R \to S$ be an injective finite type ring map. Assume $R$ is a domain. Then there exists an integer $d$ and a factorization

\[ R \to R[y_1, \ldots , y_ d] \to S' \to S \]

by injective maps such that $S'$ is finite over $R[y_1, \ldots , y_ d]$ and such that $S'_ f \cong S_ f$ for some nonzero $f \in R$.

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. $\square$

Comments (2)

Comment #5420 by Brian Shih on

What does the notation >> mean?

Comment #5647 by on

If we say then that translates into " is much larger than " in the English language. Does that help?

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