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

## 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.

**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 $2 \le i \le n$. $\square$

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

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

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

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

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. $\square$

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

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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #5420 by Brian Shih on

Comment #5647 by Johan on