# Noether Normalization Theorem

Throughout this article, *k* is a field, not necessarily algebraically closed.

Definition.Let A be a finitely generated k-algebra which is an integral domain.

We say are

algebraically independentover k if they are so as elements of .The

transcendence degreeof 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 .

**Note**

Hence Frac(*A*) is algebraic over so .

**Proof**

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

Claim.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. ♦

**Exercise**

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.

Corollary 1.Let A be a finitely generated k-algebra. If A is a field, then it is a finite extension of k.

**Proof**

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 .

Corollary 2.If is a proper ideal then .

**Proof**

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 . ♦

Theorem (Nullstellensatz).If is a radical ideal, then

.

**Proof**

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.

Proposition 1.For any field k, .

**Proof**

(≥) 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 . ♦

Corollary 3.Let A be a finitely generated k-algebra which is also an integral domain. Then

.

**Proof**

By NNT pick which are algebraically independent over *k* such that is finite. Then

♦

Next, the following is consistent with our intuition.

Proposition 2.Now suppose k is algebraically closed.

If A, B are finitely generated k-algebras which are integral domains, then

**Note**

Recall (exercise B here) that if we write and for irreducible affine varieties, then is an integral domain since is also irreducible.

**Proof**

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 . ♦

The equation in case 1 of the proof of the theorem “Case 1 : x_m is transcendental over Frac(B).”

Should it be $latex …. B [x_m] = A ” at the end?

Thanks! Corrected.

Could you please be elaborate in the proof of the claim regarding the statement “Since we see that these values are distinct across all the terms."

It’s basically base-M expansion. Over all tuples satisfying , the values are all distinct.