Noether normalization of the quadratic surface xy plus yz plus zx
= Noether normalization of the quadratic surface xy plus yz plus zx
{c}
Let
$$
T=k[x,y,z]/(xy+yz+zx)
$$
over a field in which $3$ is invertible. Put $u=x-z$ and $v=y-z$. Then
$$
xy+yz+zx=uv+2(u+v)z+3z^2,
$$
so
$$
T\cong k[u,v][z]\big/\left(z^2+\frac23(u+v)z+\frac13uv\right).
$$
Division 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.