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.
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, the highest term in of looks
Thus, if r is larger than any exponent appearing in f, then the highest term of in also has the form as above. In other words, 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 obviously 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, 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 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 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 invertible one 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 .