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 exist elements y1, y2, ..., yd in A that are algebraically independent over k and such that A is a finitely generated module over the polynomial ring S = k [y1, y2, ..., yd]. The integer d is equal to the Krull dimension of the ring A; and if 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 the coordinate ring of an affine variety X, and consider S as the coordinate ring of a d-dimensional affine space $$\mathbb A^d_k$$. Then the inclusion map $$S\hookrightarrow A$$ induces a surjective finite morphism of affine varieties $$X\to \mathbb A^d_k$$: that is, 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: 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 corresponding dimensions.

The Noether normalization lemma can be used as an important step in proving Hilbert's Nullstellensatz, one of the most fundamental results of classical algebraic geometry. The normalization theorem is also an important tool in establishing the notions of Krull dimension for k-algebras.

Proof
The following proof is due to Nagata, following Mumford's red book. A more geometric proof is given on page 127 of the red book.

The ring A in the lemma is generated as a k-algebra by some elements $$y_1, ..., y_m$$. We shall induct on m. Case $$m = 0$$ is $$k=A$$ and there is nothing to prove. Assume $$m=1$$. Then $$A\cong k[y]/I$$ as k-algebras, where $$I\subset k[y]$$ is some ideal. Since $$k[y]$$ is a PID (it is a Euclidean domain), $$I=(f)$$. If $$f=0$$ we are done, so assume $$f\neq 0$$. Let e be the degree of f. Then A is generated, as a k-vector space, by $$1,y,y^2,\dots,y^{e-1}$$. Thus A is finite over k. Assume now $$m \geq 2$$. It is enough to show that there is a k-subalgebra S of A that is generated by $$m-1$$ elements, such that A is finite over S. Indeed, by the inductive hypothesis, we can find algebraically independent elements $$x_1, ..., x_d$$ of S such that S is finite over $$k[x_1, ..., x_d]$$.

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
 * $$f(y_1, \ldots, y_m) = 0$$.

Given an integer r which is determined later, set
 * $$z_i = y_i - y_1^{r^{i-1}}, \quad 2 \le i \le m.$$

Then the preceding reads:
 * $$f(y_1, z_2 + y_1^r, z_3 + y_1^{r^2}, \ldots, z_m + y_1^{r^{m-1}}) = 0$$.

Now, if $$a y_1^{\alpha_1} \prod_2^m (z_i + y_1^{r^{i-1}})^{\alpha_i}$$ is a monomial appearing in the left-hand side of the above equation, with coefficient $$a \in k$$, the highest term in $$y_1$$ after expanding the product looks like
 * $$a y_1^{\alpha_1 + r \alpha_2 + \cdots + \alpha_m r^{m-1}}.$$

Whenever the above exponent agrees with the highest $$y_1$$ exponent produced by some other monomial, it is possible that the highest term in $$y_1$$ of $$f(y_1, z_2 + y_1^r, z_3 + y_1^{r^2}, ..., z_m + y_1^{r^{m-1}})$$ 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 $$\alpha_1 + r \alpha_2 + \cdots + \alpha_m r^{m-1}$$ encodes a unique base r number, so this does not occur. For such an r, let $$c\in k$$ be the coefficient of the unique monomial of f of multidegree $$(\alpha_1,\dots,\alpha_m)$$ for which the quantity $$\alpha_1 + r \alpha_2 + \cdots + \alpha_m r^{m-1}$$ is maximal. Multiplication of the last identity by $$1/c$$ gives an integral dependence equation of $$y_1$$ over $$S = k[z_2, ..., z_m]$$, i.e., $$y_1$$ is integral over S. Since $$y_i = z_i + y_1^{r^{i-1}}$$ 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 $$S = k[y_1, ..., y_d]$$ 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 $$d=0$$ being trivial. Since $$0 \subsetneq (y_1) \subsetneq (y_1, y_2) \subsetneq \cdots \subsetneq (y_1, \dots, y_d)$$ is a chain of prime ideals, the dimension is at least d. To get the reverse estimate, let $$0 \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_m$$ be a chain of prime ideals. Let $$0 \ne u \in \mathfrak{p}_1$$. We apply the noether normalization and get $$T = k[u, z_2, \dots, z_d]$$ (in the normalization process, we're free to choose the first variable) such that S is integral over T. By the inductive hypothesis, $$T/(u)$$ has dimension d - 1. By incomparability, $$\mathfrak{p}_i \cap T$$ is a chain of length $$m$$ and then, in $$T/(\mathfrak{p}_1 \cap T)$$, it becomes a chain of length $$m-1$$. Since $$\operatorname{dim} T/(\mathfrak{p}_1 \cap T) \le \operatorname{dim} T/(u)$$, we have $$m - 1 \le d - 1$$. Hence, $$\dim S \le d$$.

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 $$X = \operatorname{Spec} A \subset \mathbf{A}^m$$ any general linear projection $$\mathbf{A}^m \to \mathbf{A}^d$$ induces a finite morphism $$X \to \mathbf{A}^d$$ (cf. the lede); besides Eisenbud, see also.

Illustrative application: generic freeness
A typical nontrivial application of the normalization lemma is the generic freeness theorem: Let $$A, B$$ be rings such that $$A$$ is a Noetherian integral domain and suppose there is a ring homomorphism $$A \to B$$ that exhibits $$B$$ as a finitely generated algebra over $$A$$. Then there is some $$0 \ne g \in A$$ such that $$B[g^{-1}]$$ is a free $$A[g^{-1}]$$-module.

To prove this, let $$F$$ be the fraction field of $$A$$. We argue by induction on the Krull dimension of $$F \otimes_A B$$. The base case is when the Krull dimension is $$-\infty$$; i.e., $$F \otimes_A B = 0$$; that is, when there is some $$0 \ne g \in A$$ such that $$g B = 0$$, so that $$B[g^{-1}]$$ is free as an $$A[g^{-1}]$$-module. For the inductive step, note that $$F \otimes_A B$$ is a finitely generated $$F$$-algebra. Hence by the Noether normalization lemma, $$F \otimes_A B$$ contains algebraically independent elements $$x_1, \dots, x_d$$ such that $$F \otimes_A B$$ is finite over the polynomial ring $$F[x_1, \dots, x_d]$$. Multiplying each $$x_i$$ by elements of $$A$$, we can assume $$x_i$$ are in $$B$$. We now consider:
 * $$A' := A[x_1, \dots, x_d] \to B.$$

Now $$B$$ may not be finite over $$A'$$, but it will become finite after inverting a single element as follows. If $$b$$ is an element of $$B$$, then, as an element of $$F \otimes_A B$$, it is integral over $$F[x_1, \dots, x_d]$$; i.e., $$b^n + a_1 b^{n-1} + \dots + a_n = 0$$ for some $$a_i$$ in $$F[x_1, \dots, x_d]$$. Thus, some $$0 \ne g \in A$$ kills all the denominators of the coefficients of $$a_i$$ and so $$b$$ is integral over $$A'[g^{-1}]$$. Choosing some finitely many generators of $$B$$ as an $$A'$$-algebra and applying this observation to each generator, we find some $$0 \ne g \in A$$ such that $$B[g^{-1}]$$ is integral (thus finite) over $$A'[g^{-1}]$$. Replace $$B, A$$ by $$B[g^{-1}], A[g^{-1}]$$ and then we can assume $$B$$ is finite over $$A' := A[x_1, \dots, x_d]$$. To finish, consider a finite filtration $$B = B_0 \supset B_1 \supset B_2 \supset \cdots \supset B_r$$ by $$A'$$-submodules such that $$B_i / B_{i+1} \simeq A'/\mathfrak{p}_i$$ for prime ideals $$\mathfrak{p}_i$$ (such a filtration exists by the theory of associated primes). For each i, if $$\mathfrak{p}_i \ne 0$$, by inductive hypothesis, we can choose some $$g_i \ne 0$$ in $$A$$ such that $$A'/\mathfrak{p}_i[g_i^{-1}]$$ is free as an $$A[g_i^{-1}]$$-module, while $$A'$$ is a polynomial ring and thus free. Hence, with $$g = g_0 \cdots g_r$$, $$B[g^{-1}]$$ is a free module over $$A[g^{-1}]$$. $$\square$$