Let us consider the finitely generated $k$-algebra $M=k[x,y,z]/(xy,yz,zx)$. Its Noether normalisation is $R=k[w]$, where $w=x+y+z$. Now I need to show that $M$ is not a flat $R$-module. For this it is enough to show that $M$ is torsion as $R$ is PID.
M is generated by $(1,x,y,z)$ as an $R$-module. So I need to find a torsion element in $M$. How do I find a torsion element in $M$? Is there a procedure?
Thank you.
In the comments, I showed that $M$ is actually generated by $1,x,y$.
Note that this a linear independent set of generators, i.e. a basis:
Take a linear combination, which is zero:
$$0=f(w) \cdot 1 + g(w) \cdot x + h(w) \cdot y = f(x+y+z) + xg(x)+yh(y).$$
Assume $f \neq 0$. Note that $f$ cannot be constant, because the other summands do not admit constant terms to cancel the constant. Let $d > 0$ be the degree of $f$, then $f(x+y+z)$ admits the term $z^d$, which cannot be canceled by the other terms. Hence $f=0$ and from there it is trivial to see that $g=h=0$.
Summarizing, $M$ is a free $k[w]$-module of rank $3$.