Noether Normalizations

Section 14.2 Noether Normalizations

“I think normalization is a good thing.”
―Jamie Dimon

Lemma 14.17. Making a Pure-Power Leading Term.

Let $A$ be a domain, $R=A[x_1,\dots,x_n]\text{,}$ and let $f\in R$ be a polynomial of degree at most $N\text{.}$ The $A$-algebra automorphism of $R$ given by

\begin{equation*} \phi(x_i)=x_i + x_n^{N^{n-i}} \textrm{ for } i<n \quad \textrm{ and } \quad \phi(x_n)=x_n \end{equation*}

maps $f$ to a polynomial that, viewed as a polynomial in $x_n$ with coefficients in $A[x_1,\dots,x_{n-1}]\text{,}$ has leading term $d x_n^a$ for some $d\in A$ and $a\in \N\text{.}$

Proof.

The map $\phi$ sends a monomial term $d x_1^{a_1} \cdots x_n^{a_n}$ to a polynomial with unique highest degree term $d x_n^{a_1 N^{n-1} + a_2 N^{n-2} + \cdots + a_{n-1} N + a_n}\text{.}$ For each of the monomials $d x_1^{a_1} \cdots x_n^{a_n}$ in $f$ with nonzero coefficient $d \neq 0\text{,}$ we must have each $a_i \leqslant N\text{,}$ so the map

\begin{equation*} (a_1,\dots,a_n) \mapsto a_1 N^{n-1} + a_2 N^{n-2} + \cdots + a_{n-1} N + a_n \end{equation*}

is injective when restricted to the set of exponent tuples of $f\text{.}$ Therefore, none of the terms can cancel. We find that the leading term is of the promised form.

Lemma 14.18. Making a Pure-Power Leading Term (Graded Version).

Let $k$ be an infinite field, and let $R=k[x_1,\dots,x_n]$ be standard graded, meaning $\deg(x_i)=1\text{.}$ Let $f\in R$ be a homogeneous polynomial of degree $N\text{.}$ There is a degree-preserving $k$-algebra automorphism of $R$ given by $\phi(x_i) = x_i + a_i x_n$ for $i<n$ and $\phi(x_n)=x_n$ that maps $f$ to a polynomial that viewed as a polynomial in $x_n$ with coefficients in $k[x_1,\dots,x_{n-1}]\text{,}$ has leading term $a x_n^N$ for some (nonzero) $a\in k\text{.}$

Proof.

Given , we just need to show that the $x^N$ coefficient of $\phi(f)$ is nonzero for some choice of $a_i\text{.}$ One can check that the coefficient of the $x^N$ term is $f(-a_1,\dots,-a_{n-1},1)\text{.}$ But $f(-a_1,\dots,-a_{n-1},1)\text{,}$ when thought of as a polynomial in the $a_i\text{,}$ is identically zero, then $f$ must be the zero polynomial.

Theorem 14.19. Noether Normalization.

Let $A$ be a domain, and $R$ be a finitely generated $A$-algebra. There is some nonzero $a\in A$ and $x_1,\dots,x_t\in R$ algebraically independent over $A$ such that $R_a$ is module-finite over $A_a[x_1,\dots,x_t]\text{.}$

In particular, if $k$ is a field and $R$ is a finitely generated $k$-algebra, then there exist $x_1, \ldots, x_t \in R$ algebraically independent over $k$ such that $k[x_1,\ldots,x_t] \subseteq R$ is module-finite.

Proof.

We proceed by induction on the number of algebra generators $n$ of $R$ over $A\text{.}$ There is nothing to prove in the case when $n=0\text{.}$

Now suppose that we know the result holds for $A$-algebras generated by at most $n-1$ elements, and let $R=A[r_1,\dots,r_n]\text{.}$ If $r_1,\dots,r_n$ are algebraically independent over $A\text{,}$ we are done: $R$ is module-finite over $R = A[r_1,\dots,r_n]\text{.}$ If not, there is some algebraic relation among the generators $r_1, \ldots, r_n\text{,}$ meaning there exists $f(x_1,\dots,x_n) \in A[x_1,\dots,x_n]$ such that $f(r_1,\dots,r_n)=0\text{.}$ After possibly applying to change our choice of algebra generators, we can assume that $f$ has leading term $a x_n^N$ for some $a\text{.}$ Then $f$ is monic in $x_n$ after inverting $a\text{,}$ so $r_n$ is integral over $A_a[r_1,\dots,r_{n-1}]\text{,}$ and thus $R_a$ is module-finite over $A_a[r_1,\dots,r_{n-1}]$ by . By hypothesis, $A_{ab}[r_1,\dots,r_{n-1}]$ is module-finite over $A_{ab}[x_1,\dots,x_s]$ for some $b\in A$ and $x_1,\dots,x_s$ that are algebraically independent over $A\text{.}$ Since $R_{ab}$ is module-finite over $A_{ab}[r_1,\dots,r_{n-1}]\text{,}$ then by $R_{ab}$ must also be module-finite over $A_{ab}[x_1,\dots,x_s]\text{,}$ and we are done.

Definition 14.20. Noether Normalization.

Let $R$ be a finitely generated algebra for some field $k\text{.}$ Then a noether normalization of $R$ is a polynomial ring $A=k[x_1,\ldots, x_t]\subseteq R$ such that $x_1,\ldots, x_t$ are algebraically independent over $k$ and $R$ is module-finite over $A\text{.}$

Theorem 14.21. Graded Noether Normalization.

Let $k$ be an infinite field, and $R$ be a finitely generated $\N$-graded $k$-algebra with $R_0=k$ and $R=k[R_1]\text{.}$ There are homogeneous elements $x_1,\dots,x_t\in R_1$ algebraically independent over $k$ such that $R$ is module-finite over $k[x_1,\dots,x_t]\text{.}$

Proof.

We repeat the proof of but use in place of .

Remark 14.22.

There also exist Noether normalizations for quotients of power series rings over fields: the key change is that after a change of coordinates, one can rewrite any nonzero power series in $k [[ x_1,\dots, x_n \Rbracket$ as a series of the form $u (x_n^d + a_{d-1} x_n^{d-1} + \cdots + a_0)$ for a unit $u$ and $a_0,\dots,a_{d-1}\in k [[ x_1,\dots,x_{n-1}\Rbracket\text{.}$ This is called . The rest of the proof of the Noether normalization theorem proceeds in essentially the same way. Thus, given $k[[ x_1,\dots, x_n \Rbracket / I\text{,}$ we have some module-finite inclusion of another power series ring $k[[ z_1,\dots, z_d \Rbracket \subseteq k[[ x_1,\dots, x_n \Rbracket/I\text{.}$

Theorem 14.23. Noether Normalization Height and Dimension.

Let $R$ be a domain that is a finitely generated algebra over a field $k\text{,}$ or a quotient of a power series ring over a field. Let $k[z_1,\dots,z_d]$ be any noether normalization for $R\text{.}$ For any maximal ideal $\fm$ of $R\text{,}$ the length of any saturated chain of primes from $0$ to $\fm$ is $d\text{.}$ In particular, $\dim(R) = d\text{.}$

Proof.

We will show the proof in the case when $R$ is a finitely generated domain over a field $k\text{;}$ the power series case is similar, and left as an exercise. We will prove by induction on~$d$ that for any finitely generated domain with a Noether normalization with $d$ algebraically independent elements, any saturated chain of primes ending in a maximal ideal has length~$d\text{.}$

When $d=0\text{,}$ $R$ is a domain that is integral over a field, hence $R$ is a field by . Now suppose the statement holds for $d-1\text{,}$ and let $R$ be a finitely generated domain over some field $k$ with Noether normalization $A = k[z_1, \ldots, z_d]\text{.}$ Consider a maximal ideal $\fm$ of $R$ and a saturated chain

\begin{equation*} 0 \subsetneq Q_1 \subsetneq \cdots \subsetneq Q_s = \fm. \end{equation*}

By , its contraction to $A=k[z_1,\dots,z_d]$ is a chain of $s$ distinct primes in $R\text{:}$

\begin{equation*} 0 \subsetneq P_1 := Q_1 \cap A \subsetneq \cdots \subsetneq P_s := Q_s \cap A. \end{equation*}

Our assumption that the original chain is saturated implies that $Q_1$ has height 1. Suppose that $P_1$ had height $2$ or more. By , we would be able to construct a chain up to $Q_1 $ of length $2$ or more, but $Q_1$ has height $1\text{.}$ Thus $P_1$ has height $1\text{.}$ Since $k[z_1,\dots,z_d]$ is a UFD, $P_1=(f)$ for some prime element $f\text{,}$ by . After a change of variables, as in , we can assume that $f$ is monic in $z_d$ with coefficients in $k[z_1,\dots,z_{d-1}]\text{.}$ So $k[z_1,\dots,z_{d-1}] \subseteq A/(f) \subseteq R/Q_1$ are module-finite extensions, and the induction hypothesis applies to $R/Q_1\text{.}$ Now

\begin{equation*} 0 = Q_1/Q_1 \subsetneq Q_2/Q_1 \subsetneq \cdots\subsetneq Q_s/Q_1 = \fm/Q_1 \end{equation*}

is a saturated chain in the domain $R/Q_1$ going up to the maximal ideal $\fm/Q_1\text{.}$ The induction hypothesis then says that this chain has length $d-1\text{,}$ so $k-1=d-1\text{,}$ and $k=d\text{.}$

Corollary 14.24. Dimension of Polynomial Ring.

The dimension of the polynomial ring $k[x_1,\dots,x_d]$ is $d\text{.}$

Proof.

The polynomial ring $k[x_1,\dots,x_d]$ is a Noether normalization of itself, so says that it must have dimension $d\text{.}$

Corollary 14.25. Dimension of Algebra.

If $R$ is a $k$-algebra, the dimension of $R$ is less than or equal to the minimal size of an algebra generating set for $R$ over $k\text{.}$ If $R = k[f_1, \ldots, f_d]$ and $\dim(R) = d\text{,}$ then $R$ is isomorphic to a polynomial ring over $k\text{,}$ and the generators $f_i$ are algebraically independent.

Proof.

The first statement is trivial unless $R$ is finitely generated, in which case we can write $R=k[f_1,\dots,f_s] \cong k[x_1,\dots,x_s]/I$ for some ideal $I\text{,}$ so

\begin{equation*} \dim(R) \leqslant \dim(k[x_1,\dots,x_s]) = d. \end{equation*}

Suppose we chose $s$ to be minimal. If $I\neq 0\text{,}$ then $\dim(R)< s\text{,}$ since the zero ideal is not contained in $I\text{.}$

Corollary 14.26. F.G. Algebra Catenary Domain.

Let $R$ be a finitely generated algebra or a quotient of a power series ring over a field. (a) $R$ is catenary. If additionally $R$ is a domain, then (b) $R$ is equidimensional, and (c) $\ht(I)=\dim(R)-\dim(R/I)$ for all ideals $I\text{.}$

Proof.

Let $P \subseteq Q$ be primes in $R\text{.}$ After quotient out by $P\text{,}$ we can assume that $R$ is a domain and $P=0\text{.}$ Fix a saturated chain $C$ from $Q$ to a maximal ideal $\fm\text{.}$ Given two saturated chains $C'\text{,}$ $C''$ from $0$ to $Q\text{,}$ the concatenations $C'|C$ and $C''|C$ are saturated chains from $0$ to $\fm\text{,}$ so by they must have the same length. It follows that $C'$ and $C''$ have the same length.

Equidimensionality is immediate from .

We have

\begin{equation*} \ht(I) = \min\{\ht(P) \mid P\in \Min(I)\} \end{equation*}

and

\begin{equation*} \dim(R/I) = \max\{\dim(R/P) \mid P \in \Min(I)\}. \end{equation*}

Therefore, it suffices to show the equality for prime ideals, since if $P \in \Min(I)$ attains $\ht(I)\text{,}$ which is the minimal value of $\ht(Q)$ for $Q \in \Min(I)\text{,}$ then it also attains the maximal value of $\dim(R/Q)$ for $Q \in \Min(I)\text{.}$ Now, take a saturated chain of primes $C$ from $0$ to $P\text{,}$ and a saturated chain $C'$ from $P$ to a maximal ideal $\fm\text{.}$ Since $R$ is catenary, $C$ has length $\ht(P)\text{.}$ Moreover, $C'$ has length $\dim(R/P)$ by , and $C|C'$ has length $\dim(R)$ by .

Definition 14.27. Transcendence Basis.

Let $K\subseteq L$ be an extension of fields. A transcendence basis for $L$ over $K$ is a maximal algebraically independent subset of $L$ over $K\text{.}$

Lemma 14.28. Transcendence Basis Well-Defined 1.

Let $\{x_1,\dots,x_m\}$ and $\{y_1,\dots,y_n\}$ be two transcendence bases for $L$ over $K\text{.}$ Then, there is some $i$ such that $\{x_i,y_2,\dots,y_n\}$ is a transcendence basis.

Proof.

Since $L$ is algebraic over $K(y_1,\dots,y_n)\text{,}$ for each $i$ there is some $p_i(t)\in K(y_1,\dots,y_n)[t]$ such that $p_i(x_i)=0\text{.}$ We can clear denominators to assume without loss of generality that $p_i(x_i)\in K[y_1,\dots,y_n][t]\text{.}$

We claim that there is some $i$ such that $p_i(t) \notin K[y_2,\dots,y_n][t]\text{.}$ If not, then

\begin{equation*} p_i(t) \in K[y_2,\dots,y_n][t] \end{equation*}

for all $i\text{,}$ and thus that each $x_i$ is algebraic over $K(y_2,\dots,y_n)\text{.}$ Thus, $K(x_1,\dots,x_m)$ is algebraic over $K(y_2,\dots,y_n)\text{,}$ and since $L$ is algebraic over $K(x_1,\dots,x_m)\text{,}$ $y$ is algebraic over $K(y_2,\dots,y_n)\text{,}$ which contradicts that $\{y_1,\dots,y_n\}$ is a transcendence basis. This shows the claim.

Now, we claim that for such $i\text{,}$ $\{x_i,y_2,\dots,y_n\}$ is a transcendence basis. Thinking of the equation $p_i(x_i)=0$ as a polynomial expression in $K[x_i,y_2,\dots,y_n][y_1]\text{,}$ $y_1$ is algebraic over $K(x_i,y_2,\dots,y_n)\text{,}$ hence $K(y_1,\dots,y_n)$ is algebraic over $K(x_i,y_2,\dots,y_n)\text{,}$ and $L$ as well.

If $\{x_i,y_2,\dots,y_n\}$ were algebraically dependent, then there is some polynomial equation $p(x_i,y_2,\dots,y_n)=0\text{.}$ This equation must involve $x_i\text{,}$ since $y_2,\dots,y_n$ are algebraically independent. We would then have $K(x_i,y_2,\dots,y_n)$ is algebraic over $K(y_2,\dots,y_n)\text{.}$ But since $y_1$ is algebraic over $K(x_i,y_2,\dots,y_n)\text{,}$ we would have that $K(y_1,\dots,y_n)$ is algebraic over $K(y_2,\dots,y_n)\text{,}$ which would contradict that $y_1,\dots,y_n$ is a transcendence basis.

Lemma 14.29. Transcendence Basis Well-Defined 2.

If $\{x_1,\dots,x_m\}$ and $\{y_1,\dots,y_n\}$ are two transcendence bases for $L$ over $K\text{,}$ then $m=n\text{.}$

Proof.

Say that $m\leqslant n\text{.}$ If the intersection has $s<m$ elements, then without loss of generality $y_1\notin \{x_1,\dots,x_m\}\text{.}$ Then, for some $i\text{,}$ $\{x_i, y_2\dots,y_n\}$ is a transcendence basis, and $\{x_1,\dots,x_m\} \cap \{x_i, y_2\dots,y_n\}$ has $s+1$ elements. Replacing $\{y_1,\dots,y_n\}$ with $\{x_i, y_2\dots,y_n\}$ and repeating this process, we obtain a transcendence basis with $n$ elements such that $\{x_1,\dots,x_m\} \subseteq \{y_1,\dots,y_n\}\text{.}$ But we must then have that these two transcendence bases are equal, so $m=n\text{.}$

EMPTY

Corollary 14.30. Dimension in F.G. Domain.

If $R$ is a finitely generated domain over a field $k\text{,}$ then

\begin{equation*} \dim(R) = \mathrm{trdeg}_k(\mathrm{frac}(R)). \end{equation*}

Proof.

If $S \subseteq R$ is module-finite, then by $\mathrm{frac}(S)\subseteq \mathrm{frac}(R)$ is integral, or equivalently algebraic. Hence $\mathrm{frac}(S)$ and $\mathrm{frac}(R)$ have the same transcendence degree over~$k\text{.}$ In particular, if $A = k[z_1,\dots,z_d]$ is a Noether normalization for $R\text{,}$

\begin{equation*} \mathrm{trdeg}_k(\mathrm{frac}(R))= \mathrm{trdeg}_k(\mathrm{frac}(A)) = \mathrm{trdeg}_k(k(z_1,\dots,z_d)) = d = \dim(A) = \dim(R). \end{equation*}

Example 14.31. .

Let $R = k[xu,xv,yu,yv] \subseteq k[x,y,u,v]\text{,}$ where $k$ is an algebraically closed field and $x,y,u,v$ are indeterminates. Then

\begin{equation*} \mathrm{frac}(R) = k(xu,xv,yu,yv) = k \left( xu, \frac{v}{u},\frac{y}{x},\frac{yv}{xu} \right) = k\left(xu,\frac{v}{u},\frac{y}{x}\right), \end{equation*}

and these last three are algebraically independent over $k\text{.}$ Thus, $\dim(R)=3\text{.}$

Example 14.32. .

Let us use our dimension theorems to give two different proofs that over any field $k\text{,}$ $\displaystyle R=k[x,y,z]/(y^2-xz)$ has dimension 2.

First, we claim that $A = k[x,z]$ is a Noether normalization of $R\text{.}$ First, note that $x$ and $z$ are algebraically independent over $k\text{.}$ Moreover, $R = A[y]$ and $y$ satisfies the monic polynomial $t^2-xz \in A[t]\text{,}$ so $A \subseteq R = A[y]$ is integral, and thus module-finite by . Therefore, $\dim(R) = 2\text{.}$

Alternatively, one can show that $y^2-xz$ is irreducible, e.g., by thinking of it as a polynomial in $y$ and applying Eisenstein’s criterion. Then $(y^2-xz)$ is a prime of height one, so by the dimension of $R$ is

\begin{equation*} \dim(R) = \dim(k[x,y,z])-\ht((y^2-xz))=3-1=2. \end{equation*}

Example 14.33. .

Let us compute the dimension of the ring $R = k[a,b,c,d]/I\text{,}$ where

\begin{equation*} I=(b^2-ac,c^2-bd,bc-ad). \end{equation*}

We claim that $A = k[a,d]\subseteq R$ is a Noether normalization. The inclusion is integral, since

\begin{equation*} b^2=ac \implies b^4=a^2 c^2 = a^2bd, \end{equation*}

so $b$ satisfies $t^4 - a^2d t =0\text{.}$ Similarly, one can show that $c$ satisfies $t^4-ad^2t\text{.}$

\begin{equation*} c^2=bd \implies c^4 = b^2 d^2 = acd^2, \end{equation*}

so $c$ satisfies $t^4-ad^2t\text{.}$

We also need to show that $a,d$ are algebraically independent over $k\text{,}$ or equivalently that the map from the polynomial ring $\psi\! : k[u,v] \to R$ with $\psi(u)=a\text{,}$ $\psi(v)=d$ is injective.

First assume that $k=\overline{k}$ is algebraically closed. Observe that the map #empty

is surjective: given $a,d \in k\text{,}$ write $a=\alpha^3, d=\delta^3\text{,}$ and note that $(\alpha^3, \alpha^2 \delta, \alpha \delta^2, \delta^3)$ is an element of $X=\cZ(I)$ that maps to $(a,d)\text{.}$ Thus, the kernel of the induced map on coordinate rings $k[\mathbb{A}^2] \to k[X]$ is $\cI(\mathbb{A}^2) = 0\text{;}$ i.e, the map is injective. By the Nullstellensatz, $\mathcal{I}(\cZ(I))= \sqrt{I}\text{,}$ and our induced map on coordinate rings is the map #empty Since this map is injective, and this factors as #empty the first map is injective.

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