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 exist elements y1, y2, ..., yd in A that are algebraically independent over k and such that A is a finitely generated module over the polynomial ring S = k [y1, y2, ..., yd]. 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 . Then the inclusion map induces a surjective finite morphism of affine varieties : 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.[2]
The Noether normalization lemma can be used as an important step in proving Hilbert's Nullstellensatz, one of the most fundamental results of classical algebraic geometry. The normalization theorem is also an important tool in establishing the notions of Krull dimension for k-algebras.
The following proof is due to Nagata, following Mumford's red book. A more geometric proof is given on page 127 of the red book.
The ring A in the lemma is generated as a k-algebra by some elements . We shall induct on m. Case is and there is nothing to prove. Assume . Then as k-algebras, where is some ideal. Since is a PID (it is a Euclidean domain), . If we are done, so assume . Let e be the degree of f. Then A is generated, as a k-vector space, by . Thus A is finite over k. Assume now . It is enough to show that there is a k-subalgebra 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. For such an r, let be the coefficient of the unique monomial of f of multidegree for which the quantity is maximal. Multiplication of the last identity by gives an integral dependence equation of over , i.e., is integral over S. 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
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).
A typical nontrivial application of the normalization lemma is the generic freeness theorem: 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.
To prove this, let be the fraction field of . We argue by induction on the Krull dimension of . The base case is when the Krull dimension is ; i.e., ; that is, when there is some such that , so that is free as an -module. For the inductive step, note that 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:
Now may not be finite over , but it will become finite 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 .