Noether normalization lemma

If $A$ is a finitely generated algebra over a field $k$, then there are algebraically independent elements $y_1,\dots ,y_d\in A$ such that $A$ is finite, hence integral, over $k[y_1,\dots ,y_d]$.

To normalize $k[a_1,\dots ,a_n]$ over an arbitrary field, choose a nonzero polynomial relation and a base $M$ larger than all its exponents. The substitutions $b_i=a_i-a_n^{M^i}$ give distinct top weights to its monomials, so its highest power of $a_n$ has a nonzero scalar coefficient. It becomes a monic polynomial equation, making the algebra finite over $k[b_1,\dots ,b_{n-1}]$. Induction proves the Noether normalization lemma, including over finite fields.

For a nonconstant polynomial over any field, use triangular substitutions $x_i=y_i+y_n^{B^i}$, $x_n=y_n$, where $B$ exceeds every exponent in its support. Base-$B$ weights distinguish all original monomials, so the transformed polynomial has a unique highest $y_n$ power with coefficient in $k^{\times }$. Rescale to make it monic. The monic polynomial quotient is finite free over $k[y_1,\dots ,y_{n-1}]$, yielding a finite flat projection. The nonlinear substitution works over finite fields as well as infinite ones; a generic linear-direction argument alone would not do so.

Letover a field in which $3$ is invertible. Put $u=x-z$ and $v=y-z$. ThensoDivision by this monic quadratic makes $T$ a free $k[u,v]$-module with basis $1,z$. Thus $k[u,v]$ is a polynomial algebra and $T$ is integral over it.