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 "k", and any finitely generated commutative "k"-algebra "A", there exists a non-negative integer "d" and algebraically independent elements "y"1, "y"2, ..., "y""d" in "A" such that "A" is a finitely generated module over the polynomial ring "S" = "k"["y"1, "y"2, ..., "y""d"].
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 formula_1, 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 formula_2. 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.
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.
Proof.
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, formula_3. We shall induct on "m". If formula_4, then the assertion is trivial. Assume now formula_5. It is enough to show that there is a subring "S" of "A" that is generated by formula_6 elements, such that "A" is finite over "S." Indeed, by the inductive hypothesis, we can find algebraically independent elements formula_7 of "S" such that "S" is finite over formula_8. 
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 formula_12 is a monomial appearing in formula_13, with coefficient formula_14, the highest term in formula_15 after expanding the product looks like
Whenever the above exponent agrees with the highest formula_15 exponent produced by some other monomial, it is possible that the highest term in formula_15 of formula_19 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 formula_20 encodes a unique base "r" number, so this does not occur. Thus formula_15 is integral over formula_22. Since formula_23 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 formula_24 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 formula_25 being trivial. Since formula_26 is a chain of prime ideals, the dimension is at least "d". To get the reverse estimate, let formula_27 be a chain of prime ideals. Let formula_28. We apply the noether normalization and get formula_29 (in the normalization process, we're free to choose the first variable) such that "S" is integral over "T". By the inductive hypothesis, formula_30 has dimension "d" - 1. By incomparability, formula_31 is a chain of length formula_32 and then, in formula_33, it becomes a chain of length formula_6. Since formula_35, we have formula_36. Hence, formula_37.
Refinement.
The following refinement appears in Eisenbud's book, which builds on Nagata's idea:
Geometrically speaking, the last part of the theorem says that for formula_38 any general linear projection formula_39 induces a finite morphism formula_40 (cf. the lede); besides Eisenbud, see also .
Illustrative application: generic freeness.
The proof of generic freeness (the statement later) illustrates a typical yet nontrivial application of the normalization lemma. The generic freeness says: let formula_41 be rings such that formula_42 is a Noetherian integral domain and suppose there is a ring homomorphism formula_43 that exhibits formula_44 as a finitely generated algebra over formula_42. Then there is some formula_46 such that formula_47 is a free formula_48-module.
Let formula_49 be the fraction field of formula_42. We argue by induction on the Krull dimension of formula_51. The basic case is when the Krull dimension is formula_52; i.e., formula_53. This is to say there is some formula_46 such that formula_55 and so formula_47 is free as an formula_48-module. For the inductive step, note formula_51 is a finitely generated formula_49-algebra. Hence, by the Noether normalization lemma, formula_51 contains algebraically independent elements formula_61 such that formula_51 is finite over the polynomial ring formula_63. Multiplying each formula_64 by elements of formula_42, we can assume formula_64 are in formula_44. We now consider:
It need not be the case that formula_44 is finite over formula_70. But that will be the case after inverting a single element, as follows. If formula_71 is an element of formula_44, then, as an element of formula_51, it is integral over formula_63; i.e., formula_75 for some formula_76 in formula_63. Thus, some formula_46 kills all the denominators of the coefficients of formula_76 and so formula_71 is integral over formula_81. Choosing some finitely many generators of formula_44 as an formula_70-algebra and applying this observation to each generator, we find some formula_46 such that formula_47 is integral (thus finite) over formula_81. Replace formula_87 by formula_88 and then we can assume formula_44 is finite over formula_90.
To finish, consider a finite filtration formula_91 by formula_70-submodules such that formula_93 for prime ideals formula_94 (such a filtration exists by the theory of associated primes). For each "i", if formula_95, by inductive hypothesis, we can choose some formula_96 in formula_42 such that formula_98 is free as an formula_99-module, while formula_70 is a polynomial ring and thus free. Hence, with formula_101, formula_47 is a free module over formula_48. formula_104