↑ arvau

Wikipedia / Noether normalization lemma

In mathematics, the Noether normalization lemma is a result of commutative algebra, introduced by Emmy Noether in 1926. It states that for any field, and any finitely generated commutative k-algebra that are algebraically independent is a finitely generated module over the polynomial ring is equal to the Krull dimension of the ring is an integral domain is also the transcendence degree of the field of fractions 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 schemess, 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 idealss of R (equivalently, closed subsets of X) that are finite over the affine coordinate subspaces of the corresponding dimensions.

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.

Statement and proof

Theorem. (Noether Normalization Lemma) Let k be a finitely generated k-algebra. Then for some integer d algebraically independent over k such that A (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 as k is a PID (it is a Euclidean domain. Let e be the degree of f. Then A is generated, as a k-vector space. Thus A is finite over k, we are done. If not, it is enough to prove the claim that there is a k-subalgebra S of A elements, such that A is finite over S. Indeed, by the inductive hypothesis, we can find, for some integer d of S such that S. Since A is finite over S, and S, we obtain the desired conclusion that A.

are not algebraically independent, so that there is a nonzero polynomial f in m variables over k such that

  1. .

Given an integer r which is determined later, set

Then the preceding reads:

after expanding the product looks like

  1. .

will not be of the above form, because it may be affected by cancellation. However, if r encodes a unique base r number, so this does not occur. For such an r be the coefficient of the unique monomial of f is integral over S, 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 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 is a chain of prime ideals, the dimension is at least d (in the normalization process, we're free to choose the first variable) such that S is integral over T. By incomparability

Refinement

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

induces a finite morphism (cf. the lede); besides Eisenbud, see also https://arxiv.org/abs/1209.5993.

Illustrative application: generic freeness

A typical nontrivial application of the normalization lemma is the generic freeness is a Noetherian integral domain and suppose there is a ring homomorphism-module.

be the fraction field. We now consider:

(such a filtration exists by the theory of associated primes). For each i is a free module

Notes

References

  • . NB the lemma is in the updating comments.

Further reading