Are you an EPFL student looking for a semester project?
Work with us on data science and visualisation projects, and deploy your project as an app on top of Graph Search.
Concept
Noether normalization lemma
Summary
In mathematics, the Noether normalization 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 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 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.
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. 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.
Official source
About this result
This page is automatically generated and may contain information that is not correct, complete, up-to-date, or relevant to your search query. The same applies to every other page on this website. Please make sure to verify the information with EPFL's official sources.