Encyclopedia

Noether normalization lemma

Result of commutative algebra

Wikipedia contributorsAdapted for Fathomly · View authors

6 min read
Source updated September 8, 2026

In mathematics, the Noether normalization lemma (sometimes referred to as a theorem rather than a 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 $y_{1},y_{2},\ldots ,y_{d}$ in $A$ that are algebraically independent over $k$ and such that $A$ is a finitely generated module over the polynomial ring $S=k[y_{1},y_{2},\ldots ,y_{d}]$ . 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 $\mathbb {A} _{k}^{d}$ . Then the inclusion map $S\hookrightarrow A$ induces a surjective finite morphism of affine varieties $X\to \mathbb {A} _{k}^{d}$ : 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 yields Zariski's lemma as a corollary. In turn, Zariski's lemma can be used to give a straightforward proof of Hilbert's Nullstellensatz, one of the most fundamental results of classical algebraic geometry. Noether normalization is also an important tool in establishing the notion of Krull dimension for k-algebras.

01Statement and proof

Theorem. (Noether Normalization Lemma) Let k be a field and $A=k[y_{1}',...,y_{m}']$ be a finitely generated k-algebra. Then for some integer d, $0\leq d\leq m$ , there exist $y_{1},\ldots ,y_{d}\in A$ algebraically independent over k such that A is finite (i.e., finitely generated as a module) over $k[y_{1},\ldots ,y_{d}]$ (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 second, more geometric proof is given on page 176 of the same text.

Proof: 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 $m\geq 2$ . If the $y_{i}'$ are algebraically independent, then by setting $y_{i}=y_{i}'$ , 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, $0\leq d\leq m-1$ , 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 $y_{i}'$ are not algebraically independent, so 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

$z_{i}=y_{i}'-(y_{1}')^{r^{i-1}},\quad 2\leq i\leq m,$

and, for simplification of notation, write ${\tilde {y}}=y_{1}'.$

Then the preceding reads:

$f({\tilde {y}},z_{2}+{\tilde {y}}^{r},z_{3}+{\tilde {y}}^{r^{2}},\ldots ,z_{m}+{\tilde {y}}^{r^{m-1}})=0.\ \ \ (*)$

Now, if $a{\tilde {y}}^{\alpha _{1}}\prod _{2}^{m}(z_{i}+{\tilde {y}}^{r^{i-1}})^{\alpha _{i}}$ is a monomial appearing in the left-hand side of the above equation, with coefficient $a\in k$ , the highest term in ${\tilde {y}}$ after expanding the product looks like

$a{\tilde {y}}^{\alpha _{1}+\alpha _{2}r+\cdots +\alpha _{m}r^{m-1}}$ .

Whenever the above exponent agrees with the highest ${\tilde {y}}$ exponent produced by some other monomial, it is possible that the highest term in ${\tilde {y}}$ of $f({\tilde {y}},z_{2}+{\tilde {y}}^{r},z_{3}+{\tilde {y}}^{r^{2}},...,z_{m}+{\tilde {y}}^{r^{m-1}})$ 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 $r=1+\deg f$ ), then each $\alpha _{1}+\alpha _{2}r+\cdots +\alpha _{m}r^{m-1}$ 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 $\alpha _{1}+\alpha _{2}r+\cdots +\alpha _{m}r^{m-1}$ is maximal. Multiplication of $(*)$ by $1/c$ gives an integral dependence equation of ${\tilde {y}}$ over $S=k[z_{2},...,z_{m}]$ , i.e., $y_{1}'(={\tilde {y}})$ is integral over S. Moreover, because $A=S[y_{1}']$ , 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 $0\subsetneq (y_{1})\subsetneq (y_{1},y_{2})\subsetneq \cdots \subsetneq (y_{1},\dots ,y_{d})$ 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 $0\neq u\in {\mathfrak {p}}_{1}$ . 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 $\operatorname {dim} T/({\mathfrak {p}}_{1}\cap T)\leq \operatorname {dim} T/(u)$ , we have $m-1\leq d-1$ . Hence, $\dim S\leq d$ . $\square$

03Illustrative 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 $0\neq g\in A$ 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 $0\neq g\in A$ such that $gB=0$ , 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 $0\neq g\in A$ 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 $0\neq g\in A$ 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 ${\mathfrak {p}}_{i}\neq 0$ , by inductive hypothesis, we can choose some $g_{i}\neq 0$ 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 $A[g^{-1}]$ . $\square$

Watch videos about Noether normalization lemmaExplainers and documentaries on YouTube (opens in a new tab)

Sources and credits

This article is adapted from the Wikipedia article “Noether normalization lemma”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

Fathomly is not affiliated with or endorsed by the Wikimedia Foundation. Spotted a problem? Tell us.

Noether normalization lemma

01Statement and proof

02Refinement

03Illustrative application: generic freeness

Sources and credits