Noether normalization lemma

A finitely generated algebra over a field is integral over a polynomial subalgebra.
Noether normalization lemma

Noether normalization is the foundational structural theorem for finitely generated algebras over a field: after choosing suitable “coordinates,” the algebra becomes an of a polynomial ring. It is one of the key inputs behind dimension theory (via ) and the algebra–geometry dictionary (via the ).

Statement

Let kk be a , and let AA be a finitely generated kk-algebra. Then there exist elements

y1,,ydA y_1,\dots,y_d \in A

that are algebraically independent over kk such that AA is integral over the kk-subalgebra k[y1,,yd]k[y_1,\dots,y_d].

Equivalently, there exists an injective kk-algebra homomorphism

k[t1,,td]A k[t_1,\dots,t_d]\hookrightarrow A

whose image is a polynomial subalgebra, and such that AA is a finitely generated module over k[t1,,td]k[t_1,\dots,t_d] (i.e. AA is module-finite over that subring). In the language of , this says every element of AA is integral over the subring k[y1,,yd]k[y_1,\dots,y_d].

Moreover, one can choose d=dim(A)d=\dim(A) in the sense of .

Examples

  1. Polynomial rings normalize themselves.
    If A=k[x1,,xn]A=k[x_1,\dots,x_n], take yi=xiy_i=x_i and d=nd=n. Then A=k[y1,,yn]A=k[y_1,\dots,y_n], so AA is integral over the chosen polynomial subalgebra in the strongest possible way (equality).

  2. A plane curve coordinate ring.
    Let

    A=k[x,y]/(y2x3x). A = k[x,y]/(y^2-x^3-x).

    Let xˉ,yˉ\bar x,\bar y be the residue classes of x,yx,y in AA. Then yˉ\bar y satisfies a monic polynomial over k[xˉ]k[\bar x]:

    yˉ2xˉ3xˉ=0. \bar y^2 - \bar x^3 - \bar x = 0.

    Hence yˉ\bar y is integral over k[xˉ]k[\bar x], and AA is integral over the polynomial subalgebra k[xˉ]k[t]k[\bar x]\cong k[t] (so here d=1d=1).

  3. A reducible example: k[x,y]/(xy)k[x,y]/(xy).
    Let

    A=k[x,y]/(xy). A = k[x,y]/(xy).

    Set u=xˉ+yˉAu=\bar x+\bar y\in A (bars denote residue classes). Then xˉ\bar x satisfies the monic equation

    T2uT=0 T^2-uT=0

    in A[T]A[T] (since xˉ2uxˉ=xˉ(xˉ(xˉ+yˉ))=xˉyˉ=0\bar x^2-u\bar x=\bar x(\bar x-(\bar x+\bar y))=-\bar x\bar y=0), so xˉ\bar x is integral over k[u]k[u]. Similarly, yˉ\bar y is integral over k[u]k[u]. Therefore AA is integral over the polynomial subalgebra k[u]k[t]k[u]\cong k[t].

This lemma is frequently combined with (to choose “good” linear combinations) and with localization techniques such as when passing to local statements.