Noether Normalization Lemma

Statement

Let $k$ be a field and let $A$ be a finitely generated commutative $k$-algebra. Then there exist elements

that are algebraically independent over $k$ such that the natural inclusion

makes $A$ an integral extension , i.e. every element of $A$ is integral over $k[y_1,\dots,y_d]$. Equivalently, $A$ is a finite (module-finite) $k[y_1,\dots,y_d]$-module.

Consequences/interpretation:

• $k[y_1,\dots,y_d]$ is a polynomial ring in $d$ variables.
• The integer $d$ equals $\dim(A)$, the Krull dimension of $A$ (in particular when $A$ is a domain, $d$ is the transcendence degree of $\mathrm{Frac}(A)$ over $k$).
• Since $A$ is finite over a Noetherian ring, $A$ is Noetherian (compare Hilbert basis theorem , Hilbert basis corollary ).

Examples

1. Polynomial ring itself.
   If $A=k[x_1,\dots,x_n]$, take $y_i=x_i$. Then $A=k[y_1,\dots,y_n]$, so $A$ is (trivially) integral over the polynomial subalgebra, with $d=n$.

2. A parabola (finite map to $\mathbb A^1$).
   Let

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

   Set $t=y\in A$. Then $x$ satisfies the monic polynomial

   $$X^2 - t = 0 \quad\text{in } A,$$

   so $x$ is integral over $k[t]$. Hence $A$ is integral (indeed finite) over $k[t]$, and $d=1$.

3. Union of coordinate axes.
   Let

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

   Set $t=x+y\in A$. Then $x$ satisfies

   $$X^2 - tX = 0$$

   because $x^2 - (x+y)x = x^2 - x^2 - xy = 0$ in $A$. This is monic in $X$, so $x$ is integral over $k[t]$, and then $y=t-x$ is also integral. Thus $A$ is integral over $k[t]$, again with $d=1$.