Noether normalization lemma

Equivalently, there exists an injective $k$-algebra homomorphism

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

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

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

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

1. A plane curve coordinate ring. Let
   $$A = k[x,y]/(y^2-x^3-x).$$
   Let $\bar x,\bar y$ be the residue classes of $x,y$ in $A$. Then $\bar y$ satisfies a monic polynomial over $k[\bar x]$:
   $$\bar y^2 - \bar x^3 - \bar x = 0.$$
   Hence $\bar y$ is integral over $k[\bar x]$, and $A$ is integral over the polynomial subalgebra $k[\bar x]\cong k[t]$ (so here $d=1$).

1. A reducible example: $k[x,y]/(xy)$. Let
   $$A = k[x,y]/(xy).$$
   Set $u=\bar x+\bar y\in A$ (bars denote residue classes). Then $\bar x$ satisfies the monic equation
   $$T^2-uT=0$$
   in $A[T]$ (since $\bar x^2-u\bar x=\bar x(\bar x-(\bar x+\bar y))=-\bar x\bar y=0$), so $\bar x$ is integral over $k[u]$. Similarly, $\bar y$ is integral over $k[u]$. Therefore $A$ is integral over the polynomial subalgebra $k[u]\cong k[t]$.