Noether Normalization Theorem
Throughout this article, k is a field, not necessarily algebraically closed.
Let A be a finitely generated k-algebra which is an integral domain.
We say are algebraically independent over k if they are so as elements of .
The transcendence degree of A over k refers to that of over k.
Immediately we present the main theorem of the day.
Noether Normalization Theorem (NNT).
Let be a finitely generated k-algebra. There exist which are algebraically independent over k such that A is finite over .
Hence Frac(A) is algebraic over so .
Write for some . We will prove the result by induction on m; when m = 0 there is nothing to show so suppose m ≥ 1.
Let so that . By induction hypothesis we can find which are algebraically independent over k and B is finite over . Now we consider whether is algebraic or transcendental over Frac(B).
Case 1 : is transcendental over Frac(B).
Thus are algebraically independent over k. Also
is a finite extension so we are done.
Case 2 : is algebraic over Frac(B).
Since is algebraic over we can write
for some .
Clearing denominators we get
where , not all zero. We reparametrize for , where the will be determined later. Substituting then gives us
There exist such that expanding the above gives
where and .
Proof of Claim.
Pick and let . Expanding , we obtain a sum of terms of the form
The top exponent of upon expansion is . Since , we see that these values are distinct across all the terms. Thus there is a unique term of where D is the maximum over these .
This proves the claim. ♦
Resuming the proof of the theorem, we see that is integral over , so we have finite extensions
and we are done. ♦
Let . For each of the following A, find an algebraically independent sequence such that is finite.
Proof of Nullstellensatz
The NNT allows us to prove the Nullstellensatz quite easily.
Let A be a finitely generated k-algebra. If A is a field, then it is a finite extension of k.
By NNT we can find algebraically independent over k such that is finite. If then is integral over which is absurd since is a UFD and hence normal. ♦
For the rest of this section suppose k is algebraically closed. Recall the correspondence between radical ideals of and closed subsets .
If is a proper ideal then .
Let be any maximal ideal of A containing . Then is a finitely generated k-algebra so by corollary 1, it is isomorphic to k as a k-algebra. From , suppose . Then . Hence . ♦
If is a radical ideal, then
We already know (⊇) holds. For (⊆), pick any . Since A is noetherian has a finite set of generators . Consider the ideal of . Note that . We claim that .
- If not, by corollary 2 there exists a point such that for and . Then but , contradicting .
Hence so for some m > 0. Since is radical this means . ♦
Consequences of NNT
This gives us the following highly non-trivial result.
For any field k, .
(≥) Take the prime chain .
(≤) The proof is by induction on n : when n = 0 this is clear so let n > 0. Suppose is any non-zero prime ideal. Now contains some irreducible f so for , by exercise B here, we have
By NNT, we can find which are algebraically independent such that A is finite over where . By induction hypothesis,
and since A is finite over it we have too. Since for all non-zero prime we have . ♦
Let A be a finitely generated k-algebra which is also an integral domain. Then
By NNT pick which are algebraically independent over k such that is finite. Then
Next, the following is consistent with our intuition.
Now suppose k is algebraically closed.
If A, B are finitely generated k-algebras which are integral domains, then
Recall (exercise B here) that if we write and for irreducible affine varieties, then is an integral domain since is also irreducible.
Pick (resp. ) which are algebraically independent over k such that
are finite extensions. Now we have an injection
which is also a finite extension. Thus . ♦