Noether normalization lemma

Remarks

The proof is by induction on the number of generators of $A$. Suppose the generators $x_1, \ldots, x_n$ satisfy a nonzero polynomial relation. The substitution $x_i' = x_i - x_n^{r^i}$ for $i < n$, with $r$ large, turns the relation into one that is monic in $x_n$ up to a nonzero constant. Then $A$ is finite over $k[x_1', \ldots, x_{n-1}']$, and one continues with this smaller algebra. When $k$ is infinite, linear substitutions $x_i' = x_i - c_i x_n$ with suitable $c_i \in k$ suffice.

The number $d$ is the Krull dimension of $A$, by going-up and incomparability. When $A$ is a domain, $d$ is also the transcendence degree of its field of fractions over $k$.

Example: let $A = k[x, y]/(xy - 1)$ and $t = x + y$. In $A$, $x(t - x) = xy = 1$, so $x^2 - t x + 1 = 0$ and likewise for $y$; hence $A$ is generated by $1$ and $x$ as a $k[t]$-module. The element $t$ is transcendental over $k$, since $A$ is not finite-dimensional over $k$, so $d = 1$. Note that $x$ alone does not work: $A = k[x, x^{-1}]$ is not finite over $k[x]$.

Going deeper

For a student

Noether normalization is the main tool to compute the dimension of finitely generated algebras over a field and to prove that it equals the transcendence degree. It gives a short proof of Zariski's lemma and hence of the weak Nullstellensatz. If $A$ is a field, then $k[y_1, \ldots, y_d]$ is a field by integral extensions of fields, so $d = 0$. It is also used to show that finitely generated domains over a field are catenary.