Noether normalization lemma

Short description: Result of commutative algebra

In mathematics, the Noether normalization lemma (sometimes referred to as a theorem rather than a lemma) is a result of commutative algebra, introduced by Emmy Noether in 1926. It states that for any field $k$, and any finitely generated commutative k-algebra $A$, there exist elements $y_1,y_2,\dots ,y_d$ in $A$ that are algebraically independent over $k$ and such that $A$ is a finitely generated module over the polynomial ring $S=k[y_1,y_2,\dots ,y_d]$. The integer $d$ is equal to the Krull dimension of the ring $A$; and if $A$ is an integral domain, $d$ is also the transcendence degree of the field of fractions of $A$ over k.

The theorem has a geometric interpretation. Suppose A is the coordinate ring of an affine variety X, and consider S as the coordinate ring of a d-dimensional affine space ${\mathbb{𝔸}}_k^d$. Then the inclusion map $S\hookrightarrow A$ induces a surjective finite morphism of affine varieties $X\to {\mathbb{𝔸}}_k^d$: that is, any affine variety is a branched covering of affine space. When k is infinite, such a branched covering map can be constructed by taking a general projection from an affine space containing X to a d-dimensional subspace.

More generally, in the language of schemes, the theorem can equivalently be stated as: every affine k-scheme (of finite type) X is finite over an affine n-dimensional space. The theorem can be refined to include a chain of ideals of R (equivalently, closed subsets of X) that are finite over the affine coordinate subspaces of the corresponding dimensions.

The Noether normalization lemma yields Zariski's lemma as a corollary. In turn, Zariski's lemma can be used to give a straightforward proof of Hilbert's Nullstellensatz, one of the most fundamental results of classical algebraic geometry. Noether normalization is also an important tool in establishing the notion of Krull dimension for k-algebras.

Statement and proof

Theorem. (Noether Normalization Lemma) Let k be a field and $A=k[{y_1}^{\prime },...,{y_m}^{\prime }]$ be a finitely generated k-algebra. Then for some integer d, $0\le d\le m$, there exist $y_1,\dots ,y_d\in A$ algebraically independent over k such that A is finite (i.e., finitely generated as a module) over $k[y_1,\dots ,y_d]$ (the integer d is then equal to the Krull dimension of A). If A is an integral domain, then d is also the transcendence degree of the field of fractions of A over k.

The following proof is due to Nagata and appears in Mumford's Red Book. A second, more geometric proof is given on page 176 of the same text.

Proof: We shall induct on m. Case $m=0$ is $k=A$ and there is nothing to prove. Assume $m=1$. Then $A\cong k[y]/I$ as k-algebras, where $I\subset k[y]$ is some ideal. Since $k[y]$ is a PID (it is a Euclidean domain), $I=(f)$. If $f=0$ we are done, so assume $f\ne 0$. Let e be the degree of f. Then A is generated, as a k-vector space, by $1,y,y^2,\dots ,y^{e-1}$. Thus A is finite over k. Assume now $m\ge 2$. If the ${y_i}^{\prime }$ are algebraically independent, then by setting $y_i={y_i}^{\prime }$, we are done. If not, it is enough to prove the claim that there is a k-subalgebra S of A that is generated by $m-1$ elements, such that A is finite over S. Indeed, by the inductive hypothesis, we can find, for some integer d, $0\le d\le m-1$, algebraically independent elements $y_1,...,y_d$ of S such that S is finite over $k[y_1,...,y_d]$. Since A is finite over S, and S is finite over $k[y_1,...,y_d]$, we obtain the desired conclusion that A is finite over $k[y_1,...,y_d]$.

To prove the claim, we assume by hypothesis that the ${y_i}^{\prime }$ are not algebraically independent, so that there is a nonzero polynomial f in m variables over k such that

$f({y_1}^{\prime },\dots ,{y_m}^{\prime })=0$.

Given an integer r which is determined later, set

$z_i={y_i}^{\prime }-({y_1}^{\prime }{)}^{r^{i-1}},2\le i\le m,$

and, for simplification of notation, write $\tilde{y}={y_1}^{\prime }.$

Then the preceding reads:

$f(\tilde{y},z_2+{\tilde{y}}^r,z_3+{\tilde{y}}^{r^2},\dots ,z_m+{\tilde{y}}^{r^{m-1}})=0.(*)$

Now, if $a{\tilde{y}}^{{\alpha }_1}\prod _2^m(z_i+{\tilde{y}}^{r^{i-1}}{)}^{{\alpha }_i}$ is a monomial appearing in the left-hand side of the above equation, with coefficient $a\in k$, the highest term in $\tilde{y}$ after expanding the product looks like

$a{\tilde{y}}^{{\alpha }_1+{\alpha }_{2}r+\cdots +{\alpha }_m{r}^{m-1}}$.

Whenever the above exponent agrees with the highest $\tilde{y}$ exponent produced by some other monomial, it is possible that the highest term in $\tilde{y}$ of $f(\tilde{y},z_2+{\tilde{y}}^r,z_3+{\tilde{y}}^{r^2},...,z_m+{\tilde{y}}^{r^{m-1}})$ will not be of the above form, because it may be affected by cancellation. However, if r is large enough (e.g., we can set $r=1+\mathrm{deg}f$), then each ${\alpha }_1+{\alpha }_{2}r+\cdots +{\alpha }_m{r}^{m-1}$ encodes a unique base r number, so this does not occur. For such an r, let $c\in k$ be the coefficient of the unique monomial of f of multidegree $({\alpha }_1,\dots ,{\alpha }_m)$ for which the quantity ${\alpha }_1+{\alpha }_{2}r+\cdots +{\alpha }_m{r}^{m-1}$ is maximal. Multiplication of $(*)$ by $1/c$ gives an integral dependence equation of $\tilde{y}$ over $S=k[z_2,...,z_m]$, i.e., ${y_1}^{\prime }(=\tilde{y})$ is integral over S. Moreover, because $A=S[{y_1}^{\prime }]$, A is in fact finite over S. This completes the proof of the claim, so we are done with the first part.

Moreover, if A is an integral domain, then d is the transcendence degree of its field of fractions. Indeed, A and the polynomial ring $S=k[y_1,...,y_d]$ have the same transcendence degree (i.e., the degree of the field of fractions) since the field of fractions of A is algebraic over that of S (as A is integral over S) and S has transcendence degree d. Thus, it remains to show the Krull dimension of S is d. (This is also a consequence of dimension theory.) We induct on d, with the case $d=0$ being trivial. Since $0⊊(y_1)⊊(y_1,y_2)⊊\cdots ⊊(y_1,\dots ,y_d)$ is a chain of prime ideals, the dimension is at least d. To get the reverse estimate, let $0⊊{\mathfrak{𝔭}}_1⊊\cdots ⊊{\mathfrak{𝔭}}_m$ be a chain of prime ideals. Let $0\ne u\in {\mathfrak{𝔭}}_1$. We apply the Noether normalization and get $T=k[u,z_2,\dots ,z_d]$ (in the normalization process, we're free to choose the first variable) such that S is integral over T. By the inductive hypothesis, $T/(u)$ has dimension $d-1$. By incomparability, ${\mathfrak{𝔭}}_i\cap T$ is a chain of length $m$ and then, in $T/({\mathfrak{𝔭}}_1\cap T)$, it becomes a chain of length $m-1$. Since $\dim T/({\mathfrak{𝔭}}_1\cap T)\le \dim T/(u)$, we have $m-1\le d-1$. Hence, $\dim S\le d$. $◻$

Refinement

The following refinement appears in Eisenbud's book, which builds on Nagata's idea:

Theorem — Let A be a finitely generated algebra over a field k, and $I_1\subset \dots \subset I_m$ be a chain of ideals such that $\dim (A/I_i)=d_i>d_{i+1}.$ Then there exists algebraically independent elements y₁, ..., y_d in A such that

1. A is a finitely generated module over the polynomial subring S = k[y₁, ..., y_d].
2. $I_i\cap S=(y_{d_i+1},\dots ,y_d)$.
3. If the $I_i$'s are homogeneous, then y_i's may be taken to be homogeneous.

Moreover, if k is an infinite field, then any sufficiently general choice of y_I's has Property 1 above ("sufficiently general" is made precise in the proof).

Geometrically speaking, the last part of the theorem says that for $X=\mathrm{Spec}A\subset {\mathbf{𝐀}}^m$ any general linear projection ${\mathbf{𝐀}}^m\to {\mathbf{𝐀}}^d$ induces a finite morphism $X\to {\mathbf{𝐀}}^d$ (cf. the lede); besides Eisenbud, see also [1].

Corollary — Let A be an integral domain that is a finitely generated algebra over a field. If $\mathfrak{𝔭}$ is a prime ideal of A, then

$\dim A=\mathrm{height}\mathfrak{𝔭}+\dim A/\mathfrak{𝔭}$.

In particular, the Krull dimension of the localization of A at any maximal ideal is dim A.

Corollary — Let $A\subset B$ be integral domains that are finitely generated algebras over a field. Then

$\dim B=\dim A+{\mathrm{tr.deg}}_{Q(A)}Q(B)$

(the special case of Nagata's altitude formula).

Illustrative application: generic freeness

A typical nontrivial application of the normalization lemma is the generic freeness theorem: Let $A,B$ be rings such that $A$ is a Noetherian integral domain and suppose there is a ring homomorphism $A\to B$ that exhibits $B$ as a finitely generated algebra over $A$. Then there is some $0\ne g\in A$ such that $B[g^{-1}]$ is a free $A[g^{-1}]$-module.

To prove this, let $F$ be the fraction field of $A$. We argue by induction on the Krull dimension of $F{\otimes }_{A}B$. The base case is when the Krull dimension is $-\mathrm{\infty }$; i.e., $F{\otimes }_{A}B=0$; that is, when there is some $0\ne g\in A$ such that $gB=0$ , so that $B[g^{-1}]$ is free as an $A[g^{-1}]$-module. For the inductive step, note that $F{\otimes }_{A}B$ is a finitely generated $F$-algebra. Hence by the Noether normalization lemma, $F{\otimes }_{A}B$ contains algebraically independent elements $x_1,\dots ,x_d$ such that $F{\otimes }_{A}B$ is finite over the polynomial ring $F[x_1,\dots ,x_d]$. Multiplying each $x_i$ by elements of $A$, we can assume $x_i$ are in $B$. We now consider:

$A^{\prime }:=A[x_1,\dots ,x_d]\to B.$

Now $B$ may not be finite over $A^{\prime }$, but it will become finite after inverting a single element as follows. If $b$ is an element of $B$, then, as an element of $F{\otimes }_{A}B$, it is integral over $F[x_1,\dots ,x_d]$; i.e., $b^n+a_1{b}^{n-1}+\dots +a_n=0$ for some $a_i$ in $F[x_1,\dots ,x_d]$. Thus, some $0\ne g\in A$ kills all the denominators of the coefficients of $a_i$ and so $b$ is integral over $A^{\prime }[g^{-1}]$. Choosing some finitely many generators of $B$ as an $A^{\prime }$-algebra and applying this observation to each generator, we find some $0\ne g\in A$ such that $B[g^{-1}]$ is integral (thus finite) over $A^{\prime }[g^{-1}]$. Replace $B,A$ by $B[g^{-1}],A[g^{-1}]$ and then we can assume $B$ is finite over $A^{\prime }:=A[x_1,\dots ,x_d]$. To finish, consider a finite filtration $B=B_0\supset B_1\supset B_2\supset \cdots \supset B_r$ by $A^{\prime }$-submodules such that $B_i/B_{i+1}\simeq A^{\prime }/{\mathfrak{𝔭}}_i$ for prime ideals ${\mathfrak{𝔭}}_i$ (such a filtration exists by the theory of associated primes). For each i, if ${\mathfrak{𝔭}}_i\ne 0$, by inductive hypothesis, we can choose some $g_i\ne 0$ in $A$ such that $A^{\prime }/{\mathfrak{𝔭}}_i[g_i^{-1}]$ is free as an $A[g_i^{-1}]$-module, while $A^{\prime }$ is a polynomial ring and thus free. Hence, with $g=g_0\cdots g_r$, $B[g^{-1}]$ is a free module over $A[g^{-1}]$. $◻$

References

• Eisenbud, David (1995), Commutative algebra. With a view toward algebraic geometry, Graduate Texts in Mathematics, 150, Berlin, New York: Springer-Verlag, ISBN 3-540-94268-8 
• Mumford, David (1988), The Red Book of Varieties and Schemes, Lecture Notes in Mathematics, Berlin, Heidelberg: Springer-Verlag, ISBN 978-3-662-21581-4 
• Hazewinkel, Michiel, ed. (2001), "Noether theorem", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=n/n066790 . NB the lemma is in the updating comments.
• Noether, Emmy (1926), "Der Endlichkeitsatz der Invarianten endlicher linearer Gruppen der Charakteristik p", Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen: 28–35, archived from the original on 2013-03-08, https://web.archive.org/web/20130308102929/http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=63971 

Further reading

• Robertz, D.: Noether normalization guided by monomial cone decompositions. Journal of Symbolic Computation 44(10), 1359–1373 (2009)