Noether normalization lemma

<h2 id="proof">Proof</h2>
Theorem. (Noether Normalization Lemma) Let k be a field and 
 
 
 
 A
 =
 k
 [
 
 y
 
 1
 
 ′
 
 ,
 .
 .
 .
 ,
 
 y
 
 m
 
 ′
 
 ]
 
 
 {\displaystyle A=k[y_{1}',...,y_{m}']}
 
 be a finitely generated k-algebra. Then for some integer d, 
 
 
 
 0
 ≤
 d
 ≤
 m
 
 
 {\displaystyle 0\leq d\leq m}
 
, there exist 
 
 
 
 
 y
 
 1
 
 
 ,
 …
 ,
 
 y
 
 d
 
 
 ∈
 A
 
 
 {\displaystyle 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
 
 
 ,
 …
 ,
 
 y
 
 d
 
 
 ]
 
 
 {\displaystyle 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 <a href="/facts/Masayoshi_Nagata/9F5JmrEN">Nagata</a> and appears in <a href="/facts/David_Mumford/WJoJ8K2u">Mumford</a>'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
 
 
 {\displaystyle m=0}
 
 is 
 
 
 
 k
 =
 A
 
 
 {\displaystyle k=A}
 
 and there is nothing to prove. Assume 
 
 
 
 m
 =
 1
 
 
 {\displaystyle m=1}
 
. Then 
 
 
 
 A
 ≅
 k
 [
 y
 ]
 
 /
 
 I
 
 
 {\displaystyle A\cong k[y]/I}
 
 as k-algebras, where 
 
 
 
 I
 ⊂
 k
 [
 y
 ]
 
 
 {\displaystyle I\subset k[y]}
 
 is some ideal. Since 
 
 
 
 k
 [
 y
 ]
 
 
 {\displaystyle k[y]}
 
 is a PID (it is a Euclidean domain), 
 
 
 
 I
 =
 (
 f
 )
 
 
 {\displaystyle I=(f)}
 
. If 
 
 
 
 f
 =
 0
 
 
 {\displaystyle f=0}
 
 we are done, so assume 
 
 
 
 f
 ≠
 0
 
 
 {\displaystyle f\neq 0}
 
. Let e be the degree of f. Then A is generated, as a k-vector space, by 
 
 
 
 1
 ,
 y
 ,
 
 y
 
 2
 
 
 ,
 …
 ,
 
 y
 
 e
 −
 1
 
 
 
 
 {\displaystyle 1,y,y^{2},\dots ,y^{e-1}}
 
. Thus A is finite over k. Assume now 
 
 
 
 m
 ≥
 2
 
 
 {\displaystyle m\geq 2}
 
. If the 
 
 
 
 
 y
 
 i
 
 ′
 
 
 
 {\displaystyle y_{i}'}
 
 are algebraically independent, then by setting 
 
 
 
 
 y
 
 i
 
 
 =
 
 y
 
 i
 
 ′
 
 
 
 {\displaystyle 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
 
 
 {\displaystyle m-1}
 
 elements, such that A is finite over S. Indeed, by the inductive hypothesis, we can find, for some integer d, 
 
 
 
 0
 ≤
 d
 ≤
 m
 −
 1
 
 
 {\displaystyle 0\leq d\leq m-1}
 
, algebraically independent elements 
 
 
 
 
 y
 
 1
 
 
 ,
 .
 .
 .
 ,
 
 y
 
 d
 
 
 
 
 {\displaystyle y_{1},...,y_{d}}
 
 of S such that S is finite over 
 
 
 
 k
 [
 
 y
 
 1
 
 
 ,
 .
 .
 .
 ,
 
 y
 
 d
 
 
 ]
 
 
 {\displaystyle k[y_{1},...,y_{d}]}
 
. Since A is finite over S, and S is finite over 
 
 
 
 k
 [
 
 y
 
 1
 
 
 ,
 .
 .
 .
 ,
 
 y
 
 d
 
 
 ]
 
 
 {\displaystyle k[y_{1},...,y_{d}]}
 
, we obtain the desired conclusion that A is finite over 
 
 
 
 k
 [
 
 y
 
 1
 
 
 ,
 .
 .
 .
 ,
 
 y
 
 d
 
 
 ]
 
 
 {\displaystyle k[y_{1},...,y_{d}]}
 
. 
To prove the claim, we assume by hypothesis that the 
 
 
 
 
 y
 
 i
 
 ′
 
 
 
 {\displaystyle y_{i}'}
 
 are not algebraically independent, so that there is a nonzero polynomial f in m variables over k such that

f
 (
 
 y
 
 1
 
 ′
 
 ,
 …
 ,
 
 y
 
 m
 
 ′
 
 )
 =
 0
 
 
 {\displaystyle 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
              
            
          
        
        ,
        
        2
        ≤
        i
        ≤
        m
        ,
      
    
    {\displaystyle z_{i}=y_{i}'-(y_{1}')^{r^{i-1}},\quad 2\leq i\leq m,}

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

Then the preceding reads:

f
        (
        
          
            
              y
              ~
            
          
        
        ,
        
          z
          
            2
          
        
        +
        
          
            
              
                y
                ~
              
            
          
          
            r
          
        
        ,
        
          z
          
            3
          
        
        +
        
          
            
              
                y
                ~
              
            
          
          
            
              r
              
                2
              
            
          
        
        ,
        …
        ,
        
          z
          
            m
          
        
        +
        
          
            
              
                y
                ~
              
            
          
          
            
              r
              
                m
                −
                1
              
            
          
        
        )
        =
        0.
         
         
         
        (
        ∗
        )
      
    
    {\displaystyle 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
 
 
 
 
 y
 ~
 
 
 
 
 
 α
 
 1
 
 
 
 
 
 ∏
 
 2
 
 
 m
 
 
 (
 
 z
 
 i
 
 
 +
 
 
 
 
 y
 ~
 
 
 
 
 
 r
 
 i
 −
 1
 
 
 
 
 
 )
 
 
 α
 
 i
 
 
 
 
 
 
 {\displaystyle 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
 ∈
 k
 
 
 {\displaystyle a\in k}
 
, the highest term in 
 
 
 
 
 
 
 y
 ~
 
 
 
 
 
 {\displaystyle {\tilde {y}}}
 
 after expanding the product looks like

a
 
 
 
 
 y
 ~
 
 
 
 
 
 α
 
 1
 
 
 +
 
 α
 
 2
 
 
 r
 +
 ⋯
 +
 
 α
 
 m
 
 
 
 r
 
 m
 −
 1
 
 
 
 
 
 
 {\displaystyle a{\tilde {y}}^{\alpha _{1}+\alpha _{2}r+\cdots +\alpha _{m}r^{m-1}}}
 
.
Whenever the above exponent agrees with the highest 
 
 
 
 
 
 
 y
 ~
 
 
 
 
 
 {\displaystyle {\tilde {y}}}
 
 exponent produced by some other monomial, it is possible that the highest term in 
 
 
 
 
 
 
 y
 ~
 
 
 
 
 
 {\displaystyle {\tilde {y}}}
 
 of 
 
 
 
 f
 (
 
 
 
 y
 ~
 
 
 
 ,
 
 z
 
 2
 
 
 +
 
 
 
 
 y
 ~
 
 
 
 
 r
 
 
 ,
 
 z
 
 3
 
 
 +
 
 
 
 
 y
 ~
 
 
 
 
 
 r
 
 2
 
 
 
 
 ,
 .
 .
 .
 ,
 
 z
 
 m
 
 
 +
 
 
 
 
 y
 ~
 
 
 
 
 
 r
 
 m
 −
 1
 
 
 
 
 )
 
 
 {\displaystyle 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
 
 
 {\displaystyle r=1+\deg f}
 
), then each 
 
 
 
 
 α
 
 1
 
 
 +
 
 α
 
 2
 
 
 r
 +
 ⋯
 +
 
 α
 
 m
 
 
 
 r
 
 m
 −
 1
 
 
 
 
 {\displaystyle \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
 ∈
 k
 
 
 {\displaystyle c\in k}
 
 be the coefficient of the unique monomial of f of multidegree 
 
 
 
 (
 
 α
 
 1
 
 
 ,
 …
 ,
 
 α
 
 m
 
 
 )
 
 
 {\displaystyle (\alpha _{1},\dots ,\alpha _{m})}
 
 for which the quantity 
 
 
 
 
 α
 
 1
 
 
 +
 
 α
 
 2
 
 
 r
 +
 ⋯
 +
 
 α
 
 m
 
 
 
 r
 
 m
 −
 1
 
 
 
 
 {\displaystyle \alpha _{1}+\alpha _{2}r+\cdots +\alpha _{m}r^{m-1}}
 
 is maximal. Multiplication of 
 
 
 
 (
 ∗
 )
 
 
 {\displaystyle (*)}
 
 by 
 
 
 
 1
 
 /
 
 c
 
 
 {\displaystyle 1/c}
 
 gives an integral dependence equation of 
 
 
 
 
 
 
 y
 ~
 
 
 
 
 
 {\displaystyle {\tilde {y}}}
 
 over 
 
 
 
 S
 =
 k
 [
 
 z
 
 2
 
 
 ,
 .
 .
 .
 ,
 
 z
 
 m
 
 
 ]
 
 
 {\displaystyle S=k[z_{2},...,z_{m}]}
 
, i.e., 
 
 
 
 
 y
 
 1
 
 ′
 
 (
 =
 
 
 
 y
 ~
 
 
 
 )
 
 
 {\displaystyle y_{1}'(={\tilde {y}})}
 
 is integral over S. Moreover, because 
 
 
 
 A
 =
 S
 [
 
 y
 
 1
 
 ′
 
 ]
 
 
 {\displaystyle 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
 
 
 ]
 
 
 {\displaystyle 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 <a href="/facts/Dimension_theory_(algebra)/qDqrj97u">dimension theory</a>.) We induct on d, with the case 
 
 
 
 d
 =
 0
 
 
 {\displaystyle d=0}
 
 being trivial. Since 
 
 
 
 0
 ⊊
 (
 
 y
 
 1
 
 
 )
 ⊊
 (
 
 y
 
 1
 
 
 ,
 
 y
 
 2
 
 
 )
 ⊊
 ⋯
 ⊊
 (
 
 y
 
 1
 
 
 ,
 …
 ,
 
 y
 
 d
 
 
 )
 
 
 {\displaystyle 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
 ⊊
 
 
 
 p
 
 
 
 1
 
 
 ⊊
 ⋯
 ⊊
 
 
 
 p
 
 
 
 m
 
 
 
 
 {\displaystyle 0\subsetneq {\mathfrak {p}}_{1}\subsetneq \cdots \subsetneq {\mathfrak {p}}_{m}}
 
 be a chain of prime ideals. Let 
 
 
 
 0
 ≠
 u
 ∈
 
 
 
 p
 
 
 
 1
 
 
 
 
 {\displaystyle 0\neq u\in {\mathfrak {p}}_{1}}
 
. We apply the Noether normalization and get 
 
 
 
 T
 =
 k
 [
 u
 ,
 
 z
 
 2
 
 
 ,
 …
 ,
 
 z
 
 d
 
 
 ]
 
 
 {\displaystyle 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
 )
 
 
 {\displaystyle T/(u)}
 
 has dimension 
 
 
 
 d
 −
 1
 
 
 {\displaystyle d-1}
 
. By <a href="/facts/Incomparability_property_(commutative_algebra)/oJCRHFIa">incomparability</a>, 
 
 
 
 
 
 
 p
 
 
 
 i
 
 
 ∩
 T
 
 
 {\displaystyle {\mathfrak {p}}_{i}\cap T}
 
 is a chain of length 
 
 
 
 m
 
 
 {\displaystyle m}
 
 and then, in 
 
 
 
 T
 
 /
 
 (
 
 
 
 p
 
 
 
 1
 
 
 ∩
 T
 )
 
 
 {\displaystyle T/({\mathfrak {p}}_{1}\cap T)}
 
, it becomes a chain of length 
 
 
 
 m
 −
 1
 
 
 {\displaystyle m-1}
 
. Since 
 
 
 
 dim
 ⁡
 T
 
 /
 
 (
 
 
 
 p
 
 
 
 1
 
 
 ∩
 T
 )
 ≤
 dim
 ⁡
 T
 
 /
 
 (
 u
 )
 
 
 {\displaystyle \operatorname {dim} T/({\mathfrak {p}}_{1}\cap T)\leq \operatorname {dim} T/(u)}
 
, we have 
 
 
 
 m
 −
 1
 ≤
 d
 −
 1
 
 
 {\displaystyle m-1\leq d-1}
 
. Hence, 
 
 
 
 dim
 ⁡
 S
 ≤
 d
 
 
 {\displaystyle \dim S\leq d}
 
. 
 
 
 
 ◻
 
 
 {\displaystyle \square }

<h2 id="refinement">Refinement</h2>
The following refinement appears in Eisenbud's book, which builds on Nagata's idea:<a class="footnote-ref" id="fnref:3" href="#fn:3">3</a>

Theorem—Let A be a finitely generated algebra over a field k, and 
 
 
 
 
 I
 
 1
 
 
 ⊂
 ⋯
 ⊂
 
 I
 
 m
 
 
 
 
 {\displaystyle I_{1}\subset \dots \subset I_{m}}
 
 be a chain of ideals such that 
 
 
 
 dim
 ⁡
 (
 A
 
 /
 
 
 I
 
 i
 
 
 )
 =
 
 d
 
 i
 
 
 >
 
 d
 
 i
 +
 1
 
 
 .
 
 
 {\displaystyle \operatorname {dim} (A/I_{i})=d_{i}>d_{i+1}.}
 
 Then there exists algebraically independent elements y1, ..., yd in A such that

<ol><li>A is a finitely generated module over the polynomial subring S = k[y1, ..., yd].</li>
<li>
 
 
 
 
 I
 
 i
 
 
 ∩
 S
 =
 (
 
 y
 
 
 d
 
 i
 
 
 +
 1
 
 
 ,
 …
 ,
 
 y
 
 d
 
 
 )
 
 
 {\displaystyle I_{i}\cap S=(y_{d_{i}+1},\dots ,y_{d})}
 
.</li>
<li>If the 
 
 
 
 
 I
 
 i
 
 
 
 
 {\displaystyle I_{i}}
 
's are homogeneous, then yi's may be taken to be homogeneous.</li></ol>
Moreover, if k is an infinite field, then any sufficiently general choice of yI'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
 =
 Spec
 ⁡
 A
 ⊂
 
 
 A
 
 
 m
 
 
 
 
 {\displaystyle X=\operatorname {Spec} A\subset \mathbf {A} ^{m}}
 
 any general linear projection 
 
 
 
 
 
 A
 
 
 m
 
 
 →
 
 
 A
 
 
 d
 
 
 
 
 {\displaystyle \mathbf {A} ^{m}\to \mathbf {A} ^{d}}
 
 induces a <a href="/facts/Finite_morphism/djrynAmJ">finite morphism</a> 
 
 
 
 X
 →
 
 
 A
 
 
 d
 
 
 
 
 {\displaystyle X\to \mathbf {A} ^{d}}
 
 (cf. the lede); besides Eisenbud, see also <a href="https://arxiv.org/abs/1209.5993">[1]</a>.

Corollary—Let A be an integral domain that is a finitely generated algebra over a field. If 
 
 
 
 
 
 p
 
 
 
 
 {\displaystyle {\mathfrak {p}}}
 
 is a prime ideal of A, then

dim
 ⁡
 A
 =
 height
 ⁡
 
 
 p
 
 
 +
 dim
 ⁡
 A
 
 /
 
 
 
 p
 
 
 
 
 {\displaystyle \dim A=\operatorname {height} {\mathfrak {p}}+\dim A/{\mathfrak {p}}}
 
.
In particular, the Krull dimension of the localization of A at any maximal ideal is dim A.

Corollary—Let 
 
 
 
 A
 ⊂
 B
 
 
 {\displaystyle A\subset B}
 
 be integral domains that are finitely generated algebras over a field. Then

dim
        ⁡
        B
        =
        dim
        ⁡
        A
        +
        
          
            t
            r
            .
            d
            e
            g
          
          
            Q
            (
            A
            )
          
        
        ⁡
        Q
        (
        B
        )
      
    
    {\displaystyle \dim B=\dim A+\operatorname {tr.deg} _{Q(A)}Q(B)}

(the special case of <a href="/facts/Nagata%2527s_altitude_formula/qDqrj97u">Nagata's altitude formula</a>).

<h2 id="illustrative-application-generic-freeness">Illustrative application: generic freeness</h2>
A typical nontrivial application of the normalization lemma is the <a href="/facts/Generic_freeness/sd0em6cC">generic freeness</a> theorem: Let 
 
 
 
 A
 ,
 B
 
 
 {\displaystyle A,B}
 
 be rings such that 
 
 
 
 A
 
 
 {\displaystyle A}
 
 is a Noetherian integral domain and suppose there is a ring homomorphism 
 
 
 
 A
 →
 B
 
 
 {\displaystyle A\to B}
 
 that exhibits 
 
 
 
 B
 
 
 {\displaystyle B}
 
 as a finitely generated algebra over 
 
 
 
 A
 
 
 {\displaystyle A}
 
. Then there is some 
 
 
 
 0
 ≠
 g
 ∈
 A
 
 
 {\displaystyle 0\neq g\in A}
 
 such that 
 
 
 
 B
 [
 
 g
 
 −
 1
 
 
 ]
 
 
 {\displaystyle B[g^{-1}]}
 
 is a free 
 
 
 
 A
 [
 
 g
 
 −
 1
 
 
 ]
 
 
 {\displaystyle A[g^{-1}]}
 
-module.
To prove this, let 
 
 
 
 F
 
 
 {\displaystyle F}
 
 be the <a href="/facts/Fraction_field/cmlJrDSS">fraction field</a> of 
 
 
 
 A
 
 
 {\displaystyle A}
 
. We argue by induction on the Krull dimension of 
 
 
 
 F
 
 ⊗
 
 A
 
 
 B
 
 
 {\displaystyle F\otimes _{A}B}
 
. The base case is when the Krull dimension is 
 
 
 
 −
 ∞
 
 
 {\displaystyle -\infty }
 
; i.e., 
 
 
 
 F
 
 ⊗
 
 A
 
 
 B
 =
 0
 
 
 {\displaystyle F\otimes _{A}B=0}
 
; that is, when there is some 
 
 
 
 0
 ≠
 g
 ∈
 A
 
 
 {\displaystyle 0\neq g\in A}
 
 such that 
 
 
 
 g
 B
 =
 0
 
 
 {\displaystyle gB=0}
 
 , so that 
 
 
 
 B
 [
 
 g
 
 −
 1
 
 
 ]
 
 
 {\displaystyle B[g^{-1}]}
 
 is free as an 
 
 
 
 A
 [
 
 g
 
 −
 1
 
 
 ]
 
 
 {\displaystyle A[g^{-1}]}
 
-module. For the inductive step, note that 
 
 
 
 F
 
 ⊗
 
 A
 
 
 B
 
 
 {\displaystyle F\otimes _{A}B}
 
 is a finitely generated 
 
 
 
 F
 
 
 {\displaystyle F}
 
-algebra. Hence by the Noether normalization lemma, 
 
 
 
 F
 
 ⊗
 
 A
 
 
 B
 
 
 {\displaystyle F\otimes _{A}B}
 
 contains algebraically independent elements 
 
 
 
 
 x
 
 1
 
 
 ,
 …
 ,
 
 x
 
 d
 
 
 
 
 {\displaystyle x_{1},\dots ,x_{d}}
 
 such that 
 
 
 
 F
 
 ⊗
 
 A
 
 
 B
 
 
 {\displaystyle F\otimes _{A}B}
 
 is finite over the polynomial ring 
 
 
 
 F
 [
 
 x
 
 1
 
 
 ,
 …
 ,
 
 x
 
 d
 
 
 ]
 
 
 {\displaystyle F[x_{1},\dots ,x_{d}]}
 
. Multiplying each 
 
 
 
 
 x
 
 i
 
 
 
 
 {\displaystyle x_{i}}
 
 by elements of 
 
 
 
 A
 
 
 {\displaystyle A}
 
, we can assume 
 
 
 
 
 x
 
 i
 
 
 
 
 {\displaystyle x_{i}}
 
 are in 
 
 
 
 B
 
 
 {\displaystyle B}
 
. We now consider:

A
          ′
        
        :=
        A
        [
        
          x
          
            1
          
        
        ,
        …
        ,
        
          x
          
            d
          
        
        ]
        →
        B
        .
      
    
    {\displaystyle A':=A[x_{1},\dots ,x_{d}]\to B.}

Now 
 
 
 
 B
 
 
 {\displaystyle B}
 
 may not be finite over 
 
 
 
 
 A
 ′
 
 
 
 {\displaystyle A'}
 
, but it will become finite after inverting a single element as follows. If 
 
 
 
 b
 
 
 {\displaystyle b}
 
 is an element of 
 
 
 
 B
 
 
 {\displaystyle B}
 
, then, as an element of 
 
 
 
 F
 
 ⊗
 
 A
 
 
 B
 
 
 {\displaystyle F\otimes _{A}B}
 
, it is integral over 
 
 
 
 F
 [
 
 x
 
 1
 
 
 ,
 …
 ,
 
 x
 
 d
 
 
 ]
 
 
 {\displaystyle F[x_{1},\dots ,x_{d}]}
 
; i.e., 
 
 
 
 
 b
 
 n
 
 
 +
 
 a
 
 1
 
 
 
 b
 
 n
 −
 1
 
 
 +
 ⋯
 +
 
 a
 
 n
 
 
 =
 0
 
 
 {\displaystyle b^{n}+a_{1}b^{n-1}+\dots +a_{n}=0}
 
 for some 
 
 
 
 
 a
 
 i
 
 
 
 
 {\displaystyle a_{i}}
 
 in 
 
 
 
 F
 [
 
 x
 
 1
 
 
 ,
 …
 ,
 
 x
 
 d
 
 
 ]
 
 
 {\displaystyle F[x_{1},\dots ,x_{d}]}
 
. Thus, some 
 
 
 
 0
 ≠
 g
 ∈
 A
 
 
 {\displaystyle 0\neq g\in A}
 
 kills all the denominators of the coefficients of 
 
 
 
 
 a
 
 i
 
 
 
 
 {\displaystyle a_{i}}
 
 and so 
 
 
 
 b
 
 
 {\displaystyle b}
 
 is integral over 
 
 
 
 
 A
 ′
 
 [
 
 g
 
 −
 1
 
 
 ]
 
 
 {\displaystyle A'[g^{-1}]}
 
. Choosing some finitely many generators of 
 
 
 
 B
 
 
 {\displaystyle B}
 
 as an 
 
 
 
 
 A
 ′
 
 
 
 {\displaystyle A'}
 
-algebra and applying this observation to each generator, we find some 
 
 
 
 0
 ≠
 g
 ∈
 A
 
 
 {\displaystyle 0\neq g\in A}
 
 such that 
 
 
 
 B
 [
 
 g
 
 −
 1
 
 
 ]
 
 
 {\displaystyle B[g^{-1}]}
 
 is integral (thus finite) over 
 
 
 
 
 A
 ′
 
 [
 
 g
 
 −
 1
 
 
 ]
 
 
 {\displaystyle A'[g^{-1}]}
 
. Replace 
 
 
 
 B
 ,
 A
 
 
 {\displaystyle B,A}
 
 by 
 
 
 
 B
 [
 
 g
 
 −
 1
 
 
 ]
 ,
 A
 [
 
 g
 
 −
 1
 
 
 ]
 
 
 {\displaystyle B[g^{-1}],A[g^{-1}]}
 
 and then we can assume 
 
 
 
 B
 
 
 {\displaystyle B}
 
 is finite over 
 
 
 
 
 A
 ′
 
 :=
 A
 [
 
 x
 
 1
 
 
 ,
 …
 ,
 
 x
 
 d
 
 
 ]
 
 
 {\displaystyle A':=A[x_{1},\dots ,x_{d}]}
 
.
To finish, consider a finite filtration 
 
 
 
 B
 =
 
 B
 
 0
 
 
 ⊃
 
 B
 
 1
 
 
 ⊃
 
 B
 
 2
 
 
 ⊃
 ⋯
 ⊃
 
 B
 
 r
 
 
 
 
 {\displaystyle B=B_{0}\supset B_{1}\supset B_{2}\supset \cdots \supset B_{r}}
 
 by 
 
 
 
 
 A
 ′
 
 
 
 {\displaystyle A'}
 
-submodules such that 
 
 
 
 
 B
 
 i
 
 
 
 /
 
 
 B
 
 i
 +
 1
 
 
 ≃
 
 A
 ′
 
 
 /
 
 
 
 
 p
 
 
 
 i
 
 
 
 
 {\displaystyle B_{i}/B_{i+1}\simeq A'/{\mathfrak {p}}_{i}}
 
 for prime ideals 
 
 
 
 
 
 
 p
 
 
 
 i
 
 
 
 
 {\displaystyle {\mathfrak {p}}_{i}}
 
 (such a filtration exists by the theory of <a href="/facts/Associated_prime/PtXW8lal">associated primes</a>). For each i, if 
 
 
 
 
 
 
 p
 
 
 
 i
 
 
 ≠
 0
 
 
 {\displaystyle {\mathfrak {p}}_{i}\neq 0}
 
, by inductive hypothesis, we can choose some 
 
 
 
 
 g
 
 i
 
 
 ≠
 0
 
 
 {\displaystyle g_{i}\neq 0}
 
 in 
 
 
 
 A
 
 
 {\displaystyle A}
 
 such that 
 
 
 
 
 A
 ′
 
 
 /
 
 
 
 
 p
 
 
 
 i
 
 
 [
 
 g
 
 i
 
 
 −
 1
 
 
 ]
 
 
 {\displaystyle A'/{\mathfrak {p}}_{i}[g_{i}^{-1}]}
 
 is free as an 
 
 
 
 A
 [
 
 g
 
 i
 
 
 −
 1
 
 
 ]
 
 
 {\displaystyle A[g_{i}^{-1}]}
 
-module, while 
 
 
 
 
 A
 ′
 
 
 
 {\displaystyle A'}
 
 is a polynomial ring and thus free. Hence, with 
 
 
 
 g
 =
 
 g
 
 0
 
 
 ⋯
 
 g
 
 r
 
 
 
 
 {\displaystyle g=g_{0}\cdots g_{r}}
 
, 
 
 
 
 B
 [
 
 g
 
 −
 1
 
 
 ]
 
 
 {\displaystyle B[g^{-1}]}
 
 is a free module over 
 
 
 
 A
 [
 
 g
 
 −
 1
 
 
 ]
 
 
 {\displaystyle A[g^{-1}]}
 
. 
 
 
 
 ◻
 
 
 {\displaystyle \square }

<h2 id="notes">Notes</h2>

<ul><li><a href="/facts/David_Eisenbud/hw2yEnnx">Eisenbud, David</a> (1995), Commutative algebra. With a view toward algebraic geometry, <a href="/facts/Graduate_Texts_in_Mathematics/VTIfY29j">Graduate Texts in Mathematics</a>, vol. 150, Berlin, New York: <a href="/facts/Springer-Verlag/nAesf6nT">Springer-Verlag</a>, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 3-540-94268-8, <a href="/facts/MR_(identifier)/uP137L11">MR</a> <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=1322960">1322960</a>, <a href="/facts/Zbl_(identifier)/P6rFxKKx">Zbl</a> <a href="https://zbmath.org/?format=complete&q=an:0819.13001">0819.13001</a></li>
<li><a href="/facts/David_Mumford/WJoJ8K2u">Mumford, David</a> (1988), The Red Book of Varieties and Schemes, <a href="/facts/Lecture_Notes_in_Mathematics/5DmSFMhJ">Lecture Notes in Mathematics</a>, Berlin, Heidelberg: <a href="/facts/Springer-Verlag/nAesf6nT">Springer-Verlag</a>, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-3-662-21581-4</li>
<li><a href="https://www.encyclopediaofmath.org/index.php?title=Noether_theorem">"Noether theorem"</a>, <a href="/facts/Encyclopedia_of_Mathematics/WC6mGtPm">Encyclopedia of Mathematics</a>, <a href="/facts/European_Mathematical_Society/B3h7b672">EMS Press</a>, 2001 [1994]. NB the lemma is in the updating comments.</li>
<li><a href="/facts/Emmy_Noether/agcYrVLt">Noether, Emmy</a> (1926), <a href="https://web.archive.org/web/20130308102929/http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=63971">"Der Endlichkeitsatz der Invarianten endlicher linearer Gruppen der Charakteristik p"</a>, Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen: 28–35, archived from <a href="http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=63971">the original</a> on March 8, 2013</li></ul>
<h2 id="further-reading">Further reading</h2>
<ul><li>Robertz, D.: Noether normalization guided by monomial cone decompositions. <a href="/facts/Journal_of_Symbolic_Computation/hEnNr3L1">Journal of Symbolic Computation</a> 44(10), 1359–1373 (2009)</li></ul>

<h2 id="references">References</h2>

<ol>
<li id="fn:1">Noether 1926 - Noether, Emmy (1926), "Der Endlichkeitsatz der Invarianten endlicher linearer Gruppen der Charakteristik p", Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen: 28–35, archived from the original on March 8, 2013 <a href="https://web.archive.org/web/20130308102929/http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=63971" target="_blank">https://web.archive.org/web/20130308102929/http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=63971</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></li>
<li id="fn:2">Eisenbud 1995, Theorem 13.3 - Eisenbud, David (1995), Commutative algebra. With a view toward algebraic geometry, Graduate Texts in Mathematics, vol. 150, Berlin, New York: Springer-Verlag, ISBN 3-540-94268-8, MR 1322960, Zbl 0819.13001 <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=1322960" target="_blank">https://mathscinet.ams.org/mathscinet-getitem?mr=1322960</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></li>
<li id="fn:3">Eisenbud 1995, Theorem 13.3 - Eisenbud, David (1995), Commutative algebra. With a view toward algebraic geometry, Graduate Texts in Mathematics, vol. 150, Berlin, New York: Springer-Verlag, ISBN 3-540-94268-8, MR 1322960, Zbl 0819.13001 <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=1322960" target="_blank">https://mathscinet.ams.org/mathscinet-getitem?mr=1322960</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></li>
</ol>

Noether normalization lemma open-in-new

Noether normalization lemma