Noether normalization

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 exists a non-negative integer ''d'' and algebraically independent elements ''y''₁, ''y''₂, ..., ''y''_''d'' in ''A'' such that ''A'' is a finitely generated module over the polynomial ring ''S'' = ''k'' 'y''₁, ''y''₂, ..., ''y''_''d''

The integer ''d'' above is uniquely determined; it is the Krull dimension of the ring ''A''. When ''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 integral. Let ''S'' be the coordinate ring of the ''d''-dimensional affine space $\mathbb A^d_k$, and let ''A'' be the coordinate ring of some other ''d''-dimensional affine variety ''X''. Then the inclusion map ''S'' → ''A'' induces a surjective finite morphism of affine varieties $X\to \mathbb A^d_k$. The conclusion is that 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 follows: every affine ''k''-scheme (of finite type) ''X'' is finite morphism, 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 appropriate dimensions.

The form of the Noether normalization lemma stated above can be used as an important step in proving Hilbert's Nullstellensatz. This gives it further geometric importance, at least formally, as the Nullstellensatz underlies the development of much of classical algebraic geometry. The theorem is also an important tool in establishing the notions of Krull dimension for ''k''-algebras.

Proof

The following proof is due to Nagata and is taken from Mumford's red book. A proof in the geometric flavor is also given in the page 127 of the red book an
this mathoverflow thread

The ring ''A'' in the lemma is generated as a ''k''-algebra by elements, say, $y_1, ..., y_m$. We shall induct on ''m''. If $m = 0$, then the assertion is trivial. Assume now $m > 0$. It is enough to show that there is a subring ''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
:$z_i = y_i - y_1^, \quad 2 \le i \le m.$
Then the preceding reads:
:$f(y_1, z_2 + y_1^r, z_3 + y_1^, \ldots, z_m + y_1^) = 0$.
Now, if $a y_1^ \prod_2^m (z_i + y_1^)^$ 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
:$a y_1^.$
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 $f(y_1, z_2 + y_1^r, z_3 + y_1^, ..., z_m + y_1^)$ 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 $\alpha_1 + r \alpha_2 + \cdots + \alpha_m r^$ encodes a unique base ''r'' number, so this does not occur. Thus $y_1$ is integral over $S = k[z_2, ..., z_m]$. Since $y_i = z_i + y_1^$ 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 (algebra), 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 $0 \subsetneq \mathfrak_1 \subsetneq \cdots \subsetneq \mathfrak_m$ be a chain of prime ideals. Let $0 \ne u \in \mathfrak_1$. 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 property (commutative algebra), incomparability, $\mathfrak_i \cap T$ is a chain of length $m$ and then, in $T/(\mathfrak_1 \cap T)$, it becomes a chain of length $m-1$. Since $\operatorname T/(\mathfrak_1 \cap T) \le \operatorname T/(u)$, 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 $X = \operatorname A \subset \mathbf^m$ any general linear projection $\mathbf^m \to \mathbf^d$ induces a finite morphism $X \to \mathbf^d$ (cf. the lede); besides Eisenbud, see als

Illustrative application: generic freeness

The proof of generic freeness (the statement later) illustrates a typical yet nontrivial application of the normalization lemma. The generic freeness says: 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 $B[g^]$ is a free $A[g^]$-module.

Let $F$ be the fraction field of $A$. We argue by induction on the Krull dimension of $F \otimes_A B$. The basic case is when the Krull dimension is $-\infty$; i.e., $F \otimes_A B = 0$. This is to say there is some $0 \ne g \in A$ such that $g B = 0$ and so $B[g^]$ is free as an $A[g^]$-module. For the inductive step, note $F \otimes_A B$ 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:
:$A' := A[x_1, \dots, x_d] \to B.$
It need not be the case that $B$ is finite over $A'$. But that will be the case 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., $b^n + a_1 b^ + \dots + a_n = 0$ 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 $A'[g^]$. 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 $B[g^]$ is integral (thus finite) over $A'[g^]$. Replace $B, A$ by $B[g^], A[g^]$ 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 $B_i / B_ \simeq A'/\mathfrak_i$ for prime ideals $\mathfrak_i$ (such a filtration exists by the theory of associated primes). For each ''i'', if $\mathfrak_i \ne 0$, by inductive hypothesis, we can choose some $g_i \ne 0$ in $A$ such that $A'/\mathfrak_i[g_i^]$ is free as an $A[g_i^]$-module, while $A'$ is a polynomial ring and thus free. Hence, with $g = g_0 \cdots g_r$, $B[g^]$ is a free module over $A[g^]$. $\square$

Notes

References

*
*. NB the lemma is in the updating comments.
*{{citation
, last = Noether
, first = Emmy
, authorlink = Emmy Noether
, year = 1926
, title = Der Endlichkeitsatz der Invarianten endlicher linearer Gruppen der Charakteristik ''p''
, url = http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=63971
, journal = Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen
, pages = 28–35
, url-status = dead
, archiveurl = https://web.archive.org/web/20130308102929/http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=63971
, archivedate = 2013-03-08

Further reading

*Robertz, D.: Noether normalization guided by monomial cone decompositions. J. Symbolic Comput. 44(10), 1359–1373 (2009)

Commutative algebra
Algebraic varieties
Lemmas in algebra
Algebraic geometry