# Noether Normalization Theorem

Definition.

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 $k[\alpha_1, \ldots, \alpha_n]$.

Note

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

Proof

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

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

Claim.

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. ♦

Exercise

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.

Proof

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 $V(\mathfrak a) \ne\emptyset$.

Proof

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

$IV(\mathfrak a) = \mathfrak a$.

Proof

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, $\dim k[X_1, \ldots, X_n] = n$.

Proof

(≥) 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

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

Proof

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

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

Note

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.

Proof

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?

• limsup says:

Thanks! Corrected.

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."

• limsup says:

It’s basically base-M expansion. Over all tuples $e_1, \ldots, e_n, j$ satisfying $0 \le e_1, \ldots, e_n, j < M$, the values $j + M e_1 + \ldots + M^n e_n$ are all distinct.