Commutative Algebra 42

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 \alpha_1, \ldots, \alpha_n \in A are algebraically independent over k if they are so as elements of \mathrm{Frac} A.

The transcendence degree of A over k refers to that of \mathrm{Frac} A over k.

Immediately we present the main theorem of the day.

Noether Normalization Theorem (NNT).

Let A be a finitely generated k-algebra. There exist \alpha_1, \ldots, \alpha_n \in A which are algebraically independent over k such that A is finite over k[\alpha_1, \ldots, \alpha_n].


Hence Frac(A) is algebraic over k(\alpha_1, \ldots, \alpha_n) so \mathrm{trdeg} A/k = n.


Write A = k[x_1, \ldots, x_m] for some x_1, \ldots, x_m \in A. We will prove the result by induction on m; when m = 0 there is nothing to show so suppose m ≥ 1.

Let B:=k[x_1, \ldots, x_{m-1}] so that k\subseteq B \subseteq A. By induction hypothesis we can find \alpha_1, \ldots, \alpha_n \in B which are algebraically independent over k and B is finite over k[\alpha_1, \ldots, \alpha_n]. Now we consider whether x := x_m is algebraic or transcendental over Frac(B).

Case 1 : x_m is transcendental over Frac(B).

Thus \alpha_1, \ldots, \alpha_n, x_m are algebraically independent over k. Also

k[\alpha_1, \ldots, \alpha_n, x_m] \subseteq B[x_m] = A

is a finite extension so we are done.

Case 2 : x_m is algebraic over Frac(B).

Since x_m is algebraic over k(\alpha_1, \ldots, \alpha_n) we can write

x_m^d + c_{d-1} x_m^{d-1} + \ldots + c_1 x_m + c_0 = 0 for some c_0, \ldots, c_{d-1} \in k(\alpha_1, \ldots, \alpha_n).

Clearing denominators we get

p_d(\alpha_1, \ldots, \alpha_n) x_m^d + p_{d-1}(\alpha_1, \ldots, \alpha_n) x_m^{d-1} + \ldots + p_0(\alpha_1, \ldots, \alpha_n) = 0

where p_i \in k[X_1, \ldots, X_n], not all zero. We reparametrize \beta_i = \alpha_i - x_m^{d_i} for 1\le i\le n, where the d_i \ge 0 will be determined later. Substituting then gives us

\sum_{j=0}^d p_j(\beta_1 + x_m^{d_1}, \beta_2 + x_m^{d_2}, \ldots, \beta_n + x_m^{d_n}) x_m^j = 0.


There exist d_1, \ldots, d_n \ge 0 such that expanding the above gives

\sum_{j=0}^D q_j(\beta_1, \ldots, \beta_n) x_m^j = 0 where q_j \in k[X_1, \ldots, X_n] and q_D \in k-\{0\}.

Proof of Claim.

Pick M > \max(d, \deg p_0, \ldots, \deg p_d) and let d_i = M^i. Expanding p_j(\beta_1 + x_m^{d_1}, \beta_2 + x_m^{d_2}, \ldots, \beta_n + x_m^{d_n}) x_m^j, we obtain a sum of terms of the form

(\beta_1 + x_m^{d_1})^{e_1} (\beta_2 + x_m^{d_2})^{e_2} \ldots (\beta_n + x_m^{d_n})^{e_n}\cdot x_m^j, \quad e_1, \ldots, e_n \ge 0.

The top exponent of x_m upon expansion is d_1 e_1 + \ldots + d_n e_n + j = M e_1 + \ldots + M^n e_n + j. Since 0\le e_1, \ldots, e_n, j < M, we see that these values are distinct across all the terms. Thus there is a unique term of x_m^D where D is the maximum over these j + \sum_i d_i e_i.

This proves the claim. ♦

Resuming the proof of the theorem, we see that x_m is integral over k[\beta_1, \ldots, \beta_n], so we have finite extensions

k[\beta_1, \ldots, \beta_n] \subseteq k[\beta_1, \ldots, \beta_n, x_m] = k[\alpha_1, \ldots, \alpha_n, x_m] \subseteq B[x_m] = A

and we are done. ♦


Let k = \mathbb C. For each of the following A, find an algebraically independent sequence \alpha_1, \ldots, \alpha_n \in A such that \mathbb C[\alpha_1, \ldots, \alpha_n] \subseteq A is finite.

\mathbb C[X, Y]/(XY - 1), \quad \mathbb C[X, Y, Z]/(XY + YZ + ZX - 1).


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.


By NNT we can find algebraically independent \alpha_1, \ldots, \alpha_n \in A over k such that k[\alpha_1, \ldots, \alpha_n] \subseteq A is finite. If n\ge 1 then \frac 1 {\alpha_1} \in A is integral over k[\alpha_1, \ldots, \alpha_n] which is absurd since k[\alpha_1, \ldots, \alpha_n] is a UFD and hence normal. ♦

For the rest of this section suppose k is algebraically closed. Recall the correspondence between radical ideals of A := k[X_1, \ldots, X_n] and closed subsets V\subseteq \mathbb A^n(k).

Corollary 2.

If \mathrm a \subseteq A is a proper ideal then V(\mathfrak a) \ne\emptyset.


Let \mathfrak m be any maximal ideal of A containing \mathfrak a. Then A/\mathfrak m is a finitely generated k-algebra so by corollary 1, it is isomorphic to k as a k-algebra. From A/\mathfrak m \cong k, suppose X_i \mapsto v_i \in k. Then \mathfrak m = (X_1 - v_1, \ldots, X_n - v_n). Hence (v_1, \ldots, v_n) \in V(\mathfrak a). ♦

Theorem (Nullstellensatz).

If \mathfrak a \subseteq A is a radical ideal, then

IV(\mathfrak a) = \mathfrak a.


We already know (⊇) holds. For (⊆), pick any f\in IV(\mathfrak a). Since A is noetherian \mathfrak a has a finite set of generators (f_1, \ldots, f_k). Consider the ideal \mathfrak b := (f_1, \ldots, f_k, Y\cdot f - 1) of B = k[X_1, \ldots, X_n, Y]. Note that B/\mathfrak b \cong (A/\mathfrak a)_f. We claim that \mathfrak b = B.

  • If not, by corollary 2 there exists a point (v_1, \ldots, v_n, w) \in \mathbb A^{n+1} such that f_i (v_1, \ldots, v_n) = 0 for 1\le i\le k and f(v_1, \ldots, v_n)w = 1. Then (v_1, \ldots, v_n) \in V(\mathfrak a) but f(v_1,\ldots, v_n) \ne 0, contradicting f\in IV(\mathfrak a).

Hence (A/\mathfrak a)_f = 0 so f^m \in \mathfrak a for some m > 0. Since \mathfrak a is radical this means f\in \mathfrak a. ♦


Consequences of NNT

This gives us the following highly non-trivial result.

Proposition 1.

For any field k, \dim k[X_1, \ldots, X_n] = n.


(≥) Take the prime chain 0 \subset (X_1) \subset (X_1, X_2) \subset \ldots \subset (X_1, \ldots, X_n).

(≤) The proof is by induction on n : when n = 0 this is clear so let n > 0. Suppose \mathfrak p \subset k[X_1, \ldots, X_n] is any non-zero prime ideal. Now \mathfrak p contains some irreducible f so for A = k[X_1, \ldots, X_n]/\mathfrak p, by exercise B here, we have

\mathrm{trdeg} A/k \le n-1.

By NNT, we can find \alpha_1, \ldots, \alpha_m \in A which are algebraically independent such that A is finite over k[\alpha_1, \ldots, \alpha_m] where m = \mathrm{trdeg} A/k \le n-1. By induction hypothesis,

\dim k[\alpha_1, \ldots, \alpha_m] = m

and since A is finite over it we have \dim A = m \le n-1 too. Since \dim A/\mathfrak p \le n-1 for all non-zero prime \mathfrak p we have \dim A \le n. ♦

Corollary 3.

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

\dim A = \mathrm{trdeg} A/k.


By NNT pick \alpha_1, \ldots, \alpha_n \in A which are algebraically independent over k such that k[\alpha_1, \ldots, \alpha_n] \subseteq A is finite. Then

\dim A = \dim k[\alpha_1, \ldots, \alpha_n] = n.

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

\dim (A\otimes_k B) = \dim A + \dim B.


Recall (exercise B here) that if we write A\cong k[V] and B \cong k[W] for irreducible affine varieties, then A\otimes_k B \cong k[V\times W] is an integral domain since V\times W is also irreducible.


Pick \alpha_1, \ldots, \alpha_n \in A (resp. \beta_1, \ldots, \beta_m \in B) which are algebraically independent over k such that

k[\alpha_1, \ldots, \alpha_n] \subseteq A, \quad k[\beta_1, \ldots, \beta_m] \subseteq B

are finite extensions. Now we have an injection

k[X_1, \ldots, X_n, Y_1, \ldots, Y_m] \cong k[\alpha_1, \ldots, \alpha_n] \otimes_k k[\beta_1, \ldots, \beta_m] \hookrightarrow A\otimes_k B

which is also a finite extension. Thus \dim (A\otimes_k B) = m + n. ♦


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4 Responses to Commutative Algebra 42

  1. Vanya says:

    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?

  2. Vanya says:

    Could you please be elaborate in the proof of the claim regarding the statement “Since 0\le e_1, \ldots, e_n, j < M, we see that these values are distinct across all the terms."

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