Noether normalization lemma

Theorem. (Noether Normalization Lemma) Let k be a field and $A=k[y\_1\text{'},...,y\_m\text{'}]$ be a finitely generated k-algebra. Then for some integer d, $0\backslash leq\; d\backslash leq\; m$, there exist $y\_1,\backslash ldots,y\_d\backslash in\; A$ algebraically independent over k such that A is finite (i.e., finitely generated as a module) over $k[y\_1,\backslash ldots,y\_d]$ (the integer d is then equal to the Krull dimension of A). If A is an integral domain, then d is also the transcendence degree of the field of fractions of A over k.

The following proof is due to Nagata and appears in Mumford's red book. A more geometric proof is given on page 176 of the red book.

Proof: We shall induct on m. Case $m\; =\; 0$ is $k=A$ and there is nothing to prove. Assume $m=1$. Then $A\backslash cong\; k[y]/I$ as k-algebras, where $I\backslash subset\; k[y]$ is some ideal. Since $k[y]$ is a PID (it is a Euclidean domain), $I=(f)$. If $f=0$ we are done, so assume $f\backslash neq\; 0$. Let e be the degree of f. Then A is generated, as a k-vector space, by $1,y,y^2,\backslash dots,y^\{e-1\}$. Thus A is finite over k. Assume now $m\; \backslash geq\; 2$. If the $y\_i\text{'}$ are algebraically independent, then by setting $y\_i=y\_i\text{'}$, we are done. If not, it is enough to prove the claim 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, for some integer d, $0\backslash leq\; d\backslash leq\; m-1$, algebraically independent elements $y\_1,\; ...,\; y\_\{d\}$ of S such that S is finite over $k[y\_1,\; ...,\; y\_d]$. Since A is finite over S, and S is finite over $k[y\_1,\; ...,\; y\_d]$, we obtain the desired conclusion that A is finite over $k[y\_1,\; ...,\; y\_d]$.

To prove the claim, we assume by hypothesis that the $y\_i\text{'}$ are not algebraically independent, so that there is a nonzero polynomial f in m variables over k such that

:$f(y\_1\text{'},\; \backslash ldots,\; y\_m\text{'})\; =\; 0$.

Given an integer r which is determined later, set

:$z\_i\; =\; y\_i\text{'}\; -\; (y\_1\text{'})^\{r^\{i-1\}\},\; \backslash quad\; 2\; \backslash le\; i\; \backslash le\; m,$

and, for simplification of notation, write $\backslash tilde\{y\}=y\_1\text{'}.$

Then the preceding reads:

:$f(\backslash tilde\{y\},\; z\_2\; +\; \backslash tilde\{y\}^r,\; z\_3\; +\; \backslash tilde\{y\}^\{r^2\},\; \backslash ldots,\; z\_m\; +\; \{\backslash tilde\{y\}\}^\{r^\{m-1\}\})\; =\; 0.\backslash \; \backslash \; \backslash \; (*)$

Now, if $a\; \backslash tilde\{y\}^\{\backslash alpha\_1\}\; \backslash prod\_2^m\; (z\_i\; +\; \{\backslash tilde\{y\}\}^\{r^\{i-1\}\})^\{\backslash alpha\_i\}$ is a monomial appearing in the left-hand side of the above equation, with coefficient $a\; \backslash in\; k$, the highest term in $\backslash tilde\{y\}$ after expanding the product looks like

:$a\; \{\backslash tilde\{y\}\}^\{\backslash alpha\_1\; +\; \backslash alpha\_2\; r\; +\; \backslash cdots\; +\; \backslash alpha\_m\; r^\{m-1\}\}$.

Whenever the above exponent agrees with the highest $\backslash tilde\{y\}$ exponent produced by some other monomial, it is possible that the highest term in $\backslash tilde\{y\}$ of $f(\backslash tilde\{y\},\; z\_2\; +\; \backslash tilde\{y\}^r,\; z\_3\; +\; \backslash tilde\{y\}^\{r^2\},\; ...,\; z\_m\; +\; \{\backslash tilde\{y\}\}^\{r^\{m-1\}\})$ will not be of the above form, because it may be affected by cancellation. However, if r is large enough (e.g., we can set $r\; =\; 1\; +\; \backslash deg\; f$), then each $\backslash alpha\_1\; +\; \backslash alpha\_2\; r\; +\; \backslash cdots\; +\; \backslash alpha\_m\; r^\{m-1\}$ encodes a unique base r number, so this does not occur. For such an r, let $c\backslash in\; k$ be the coefficient of the unique monomial of f of multidegree $(\backslash alpha\_1,\backslash dots,\backslash alpha\_m)$ for which the quantity $\backslash alpha\_1\; +\; \backslash alpha\_2\; r\; +\; \backslash cdots\; +\; \backslash alpha\_m\; r^\{m-1\}$ is maximal. Multiplication of $(*)$ by $1/c$ gives an integral dependence equation of $\backslash tilde\{y\}$ over $S\; =\; k[z\_2,\; ...,\; z\_m]$, i.e., $y\_1\text{'}(=\backslash tilde\{y\})$ is integral over S. Moreover, because $A=S[y\_1\text{'}]$, A is in fact finite over S. This completes the proof of the claim, so we are done with the first part.

Moreover, if A is an integral domain, then d is the transcendence degree of its field of fractions. Indeed, A and the polynomial ring $S\; =\; k[y\_1,\; ...,\; y\_d]$ 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 S is d. (This is also a consequence of dimension theory.) We induct on d, with the case $d=0$ being trivial. Since $0\; \backslash subsetneq\; (y\_1)\; \backslash subsetneq\; (y\_1,\; y\_2)\; \backslash subsetneq\; \backslash cdots\; \backslash subsetneq\; (y\_1,\; \backslash dots,\; y\_d)$ is a chain of prime ideals, the dimension is at least d. To get the reverse estimate, let $0\; \backslash subsetneq\; \backslash mathfrak\{p\}\_1\; \backslash subsetneq\; \backslash cdots\; \backslash subsetneq\; \backslash mathfrak\{p\}\_m$ be a chain of prime ideals. Let $0\; \backslash ne\; u\; \backslash in\; \backslash mathfrak\{p\}\_1$. We apply the Noether normalization and get $T\; =\; k[u,\; z\_2,\; \backslash dots,\; z\_d]$ (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, $\backslash mathfrak\{p\}\_i\; \backslash cap\; T$ is a chain of length $m$ and then, in $T/(\backslash mathfrak\{p\}\_1\; \backslash cap\; T)$, it becomes a chain of length $m-1$. Since $\backslash operatorname\{dim\}\; T/(\backslash mathfrak\{p\}\_1\; \backslash cap\; T)\; \backslash le\; \backslash operatorname\{dim\}\; T/(u)$, we have $m\; -\; 1\; \backslash le\; d\; -\; 1$. Hence, $\backslash dim\; S\; \backslash le\; d$. $\backslash square$

The following refinement appears in Eisenbud's book, which builds on Nagata's idea:

{{math_theorem

|Let A be a finitely generated algebra over a field k, and $I\_1\; \backslash subset\; \backslash dots\; \backslash subset\; I\_m$ be a chain of ideals such that $\backslash operatorname\{dim\}\; (A/I\_i)\; =\; d\_i\; >\; d\_\{i+1\}.$ Then there exists algebraically independent elements y₁, ..., y_d in A such that

1. A is a finitely generated module over the polynomial subring S = k[y₁, ..., y_d].
2. $I\_i\; \backslash cap\; S\; =\; (y\_\{d\_i+1\},\; \backslash dots,\; y\_d)$.
3. If the $I\_i$'s are homogeneous, then y_i's may be taken to be homogeneous.

Moreover, if k is an infinite field, then any sufficiently general choice of y_I's has Property 1 above ("sufficiently general" is made precise in the proof).}}

Geometrically speaking, the last part of the theorem says that for $X\; =\; \backslash operatorname\{Spec\}\; A\; \backslash subset\; \backslash mathbf\{A\}^m$ any general linear projection $\backslash mathbf\{A\}^m\; \backslash to\; \backslash mathbf\{A\}^d$ induces a finite morphism $X\; \backslash to\; \backslash mathbf\{A\}^d$ (cf. the lede); besides Eisenbud, see also [https://arxiv.org/abs/1209.5993].

{{math_theorem|name=Corollary|Let A be an integral domain that is a finitely generated algebra over a field. If $\backslash mathfrak\{p\}$ is a prime ideal of A, then

:$\backslash dim\; A\; =\; \backslash operatorname\{height\}\; \backslash mathfrak\{p\}\; +\; \backslash dim\; A/\backslash mathfrak\{p\}$.

In particular, the Krull dimension of the localization of A at any maximal ideal is dim A.}}

{{math_theorem|name=Corollary|Let $A\; \backslash subset\; B$ be integral domains that are finitely generated algebras over a field. Then

:$\backslash dim\; B\; =\; \backslash dim\; A\; +\; \backslash operatorname\{tr.deg\}\_\{Q(A)\}\; Q(B)$

(the special case of Nagata's altitude formula).}}

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 $A\; \backslash to\; B$ that exhibits $B$ as a finitely generated algebra over $A$. Then there is some $0\; \backslash ne\; g\; \backslash in\; A$ 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 $F\; \backslash otimes\_A\; B$. The base case is when the Krull dimension is $-\backslash infty$; i.e., $F\; \backslash otimes\_A\; B\; =\; 0$; that is, when there is some $0\; \backslash ne\; g\; \backslash in\; A$ such that $g\; B\; =\; 0$ , so that $B[g^\{-1\}]$ is free as an $A[g^\{-1\}]$-module. For the inductive step, note that $F\; \backslash otimes\_A\; B$ is a finitely generated $F$-algebra. Hence by the Noether normalization lemma, $F\; \backslash otimes\_A\; B$ contains algebraically independent elements $x\_1,\; \backslash dots,\; x\_d$ such that $F\; \backslash otimes\_A\; B$ is finite over the polynomial ring $F[x\_1,\; \backslash dots,\; x\_d]$. Multiplying each $x\_i$ by elements of $A$, we can assume $x\_i$ are in $B$. We now consider:

:$A\text{'}\; :=\; A[x\_1,\; \backslash dots,\; x\_d]\; \backslash to\; B.$

Now $B$ may not be finite over $A\text{'}$, but it will become finite after inverting a single element as follows. If $b$ is an element of $B$, then, as an element of $F\; \backslash otimes\_A\; B$, it is integral over $F[x\_1,\; \backslash dots,\; x\_d]$; i.e., $b^n\; +\; a\_1\; b^\{n-1\}\; +\; \backslash dots\; +\; a\_n\; =\; 0$ for some $a\_i$ in $F[x\_1,\; \backslash dots,\; x\_d]$. Thus, some $0\; \backslash ne\; g\; \backslash in\; A$ kills all the denominators of the coefficients of $a\_i$ and so $b$ is integral over $A\text{'}[g^\{-1\}]$. Choosing some finitely many generators of $B$ as an $A\text{'}$-algebra and applying this observation to each generator, we find some $0\; \backslash ne\; g\; \backslash in\; A$ such that $B[g^\{-1\}]$ is integral (thus finite) over $A\text{'}[g^\{-1\}]$. Replace $B,\; A$ by $B[g^\{-1\}],\; A[g^\{-1\}]$ and then we can assume $B$ is finite over $A\text{'}\; :=\; A[x\_1,\; \backslash dots,\; x\_d]$.

To finish, consider a finite filtration $B\; =\; B\_0\; \backslash supset\; B\_1\; \backslash supset\; B\_2\; \backslash supset\; \backslash cdots\; \backslash supset\; B\_r$ by $A\text{'}$-submodules such that $B\_i\; /\; B\_\{i+1\}\; \backslash simeq\; A\text{'}/\backslash mathfrak\{p\}\_i$ for prime ideals $\backslash mathfrak\{p\}\_i$ (such a filtration exists by the theory of associated primes). For each i, if $\backslash mathfrak\{p\}\_i\; \backslash ne\; 0$, by inductive hypothesis, we can choose some $g\_i\; \backslash ne\; 0$ in $A$ such that $A\text{'}/\backslash mathfrak\{p\}\_i[g\_i^\{-1\}]$ is free as an $A[g\_i^\{-1\}]$-module, while $A\text{'}$ is a polynomial ring and thus free. Hence, with $g\; =\; g\_0\; \backslash cdots\; g\_r$, $B[g^\{-1\}]$ is a free module over $A[g^\{-1\}]$. $\backslash square$

{{reflist}}

• {{Citation | last1=Eisenbud | first1=David | author1-link=David Eisenbud | title=Commutative algebra. With a view toward algebraic geometry | publisher=Springer-Verlag | location=Berlin, New York | series=Graduate Texts in Mathematics | isbn=3-540-94268-8 | mr=1322960 | year=1995 | volume=150 | zbl=0819.13001 }}
• {{Citation |last1=Mumford |first1=David |author1-link=David Mumford |title=The Red Book of Varieties and Schemes |publisher=Springer-Verlag |location=Berlin, Heidelberg |series=Lecture Notes in Mathematics |isbn=978-3-662-21581-4 |mr= |year=1988 |volume= |zbl=}}
• {{Springer|id=n/n066790|title=Noether theorem}}. 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

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|archiveurl = https://web.archive.org/web/20130308102929/http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=63971

|archivedate = 2013-03-08

}}

• Robertz, D.: Noether normalization guided by monomial cone decompositions. Journal of Symbolic Computation 44(10), 1359–1373 (2009)

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