Noether normalization lemma

In mathematics, the Noether normalization lemma is a result of commutative algebra, introduced by Emmy Noether in 1926.[1] It states that for any field k, and any finitely generated commutative k-algebra A, there exists a non-negative integer d and algebraically independent elements y1, y2, ..., yd in A such that A is a finitely generated module over the polynomial ring S = k [y1, y2, ..., yd].

The integer d above is uniquely determined; it is the Krull dimension of the ring A. When 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 integral. Let S be the coordinate ring of the d-dimensional affine space , and let A be the coordinate ring of some other d-dimensional affine variety X. Then the inclusion map S → A induces a surjective finite morphism of affine varieties . The conclusion is that 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 follows: 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 appropriate dimensions.[2]

The form of the Noether normalization lemma stated above can be used as an important step in proving Hilbert's Nullstellensatz. This gives it further geometric importance, at least formally, as the Nullstellensatz underlies the development of much of classical algebraic geometry. The theorem is also an important tool in establishing the notions of Krull dimension for k-algebras.


The following proof is due to Nagata and is taken from Mumford's red book. A proof in the geometric flavor is also given in the page 127 of the red book and this mathoverflow thread.

The ring A in the lemma is generated as a k-algebra by elements, say,  . We shall induct on m. If  , then the assertion is trivial. Assume now  . It is enough to show that there is a subring S of A that is generated by   elements, such that A is finite over S. Indeed, by the inductive hypothesis, we can find algebraically independent elements   of S such that S is finite over  .

Since otherwise there would be nothing to prove, we can also assume that there is a nonzero polynomial f in m variables over k such that


Given an integer r which is determined later, set


Then the preceding reads:


Now, if   is a monomial appearing in the left-hand side of the above equation, with coefficient  , the highest term in   after expanding the product looks like


Whenever the above exponent agrees with the highest   exponent produced by some other monomial, it is possible that the highest term in   of   will not be of the above form, because it may be affected by cancellation. However, if r is larger than any exponent appearing in f, then each   encodes a unique base r number, so this does not occur. Thus   is integral over  . Since   are also integral over that ring, A is integral over S. It follows A is finite over S, and since S is generated by m-1 elements, by the inductive hypothesis we are done.

If A is an integral domain, then d is the transcendence degree of its field of fractions. Indeed, A and   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 the polynomial ring S is d. (This is also a consequence of dimension theory.) We induct on d, with the case   being trivial. Since   is a chain of prime ideals, the dimension is at least d. To get the reverse estimate, let   be a chain of prime ideals. Let  . We apply the noether normalization and get   (in the normalization process, we're free to choose the first variable) such that S is integral over T. By the inductive hypothesis,   has dimension d - 1. By incomparability,   is a chain of length   and then, in  , it becomes a chain of length  . Since  , we have  . Hence,  .


The following refinement appears in Eisenbud's book, which builds on Nagata's idea:[2]

Theorem — Let A be a finitely generated algebra over a field k, and   be a chain of ideals such that   Then there exists algebraically independent elements y1, ..., yd in A such that

  1. A is a finitely generated module over the polynomial subring S = k[y1, ..., yd].
  2.  .
  3. If the  's are homogeneous, then yi's may be taken to be homogeneous.

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

Geometrically speaking, the last part of the theorem says that for   any general linear projection   induces a finite morphism   (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   is a prime ideal of A, then


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

Corollary — Let   be integral domains that are finitely generated algebras over a field. Then


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

Illustrative application: generic freenessEdit

The proof of generic freeness (the statement later) illustrates a typical yet nontrivial application of the normalization lemma. The generic freeness says: let   be rings such that   is a Noetherian integral domain and suppose there is a ring homomorphism   that exhibits   as a finitely generated algebra over  . Then there is some   such that   is a free  -module.

Let   be the fraction field of  . We argue by induction on the Krull dimension of  . The basic case is when the Krull dimension is  ; i.e.,  . This is to say there is some   such that   and so   is free as an  -module. For the inductive step, note   is a finitely generated  -algebra. Hence, by the Noether normalization lemma,   contains algebraically independent elements   such that   is finite over the polynomial ring  . Multiplying each   by elements of  , we can assume   are in  . We now consider:


It need not be the case that   is finite over  . But that will be the case after inverting a single element, as follows. If   is an element of  , then, as an element of  , it is integral over  ; i.e.,   for some   in  . Thus, some   kills all the denominators of the coefficients of   and so   is integral over  . Choosing some finitely many generators of   as an  -algebra and applying this observation to each generator, we find some   such that   is integral (thus finite) over  . Replace   by   and then we can assume   is finite over  . To finish, consider a finite filtration   by  -submodules such that   for prime ideals   (such a filtration exists by the theory of associated primes). For each i, if  , by inductive hypothesis, we can choose some   in   such that   is free as an  -module, while   is a polynomial ring and thus free. Hence, with  ,   is a free module over  .  


  1. ^ Noether 1926
  2. ^ a b Eisenbud 1995, Theorem 13.3


  • Eisenbud, David (1995), Commutative algebra. With a view toward algebraic geometry, Graduate Texts in Mathematics, vol. 150, Berlin, New York: Springer-Verlag, ISBN 3-540-94268-8, MR 1322960, Zbl 0819.13001
  • "Noether theorem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]. 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 March 8, 2013

Further readingEdit

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