Polynomial Rings and Finitely Generated Modules

Question

I've been trying to solve the following question:

Show that the polynomial ring $\mathbb{Z}[x,y,z]$ is a finitely generated module over its subring generated by the following three elements:$$x+y+z,\ xy+xz+yz,\ xyz.$$

Could you give me a hint? Say $R$ is the ring generated by the above three elements. My idea was to show that $x$ is integral over $R$, i.e. that the subring $R[x]$ of $\mathbb{Z}[x,y,z]$ is finitely generated as an R-submodule of $\mathbb{Z}[x,y,z]$.

Alternatively, I tried finding the monic polynomial in $R[w]$ that $x$ satisfies, but neither method worked (or maybe I just missed something).

I am quite confused and feel like I'm doing something completely wrong.

Answer 1

Set $a = x + y + z$, $b = xy + xz + yz$, and $c = xyz$. Then $$ (t-x)(t-y)(t-z) = t^3 - at^2 + bt - c \in R[t] $$ so $x$, $y$, and $z$ are all integral over $R$. There is a standard property relating integrality to finitely generated modules: when $B/A$ is a ring extension and $b \in B$, the following properties are equivalent:

(i) $b$ is integral over $A$,

(ii) $A[b]$ is a finitely generated $A$-module,

(iii) $b$ is contained in a subring $S$ of $B$ such that $A \subset S \subset B$ and $S$ is a finitely generated $A$-module.

In the proof of this equivalence, it is the direction from (iii) to (i) that is the most technical.

Anyway, this equivalence implies that the sum and product of elements in $B$ integral over $A$ is integral over $A$, so the set of elements in $B$ integral over $A$ is a subring of $B$.

The proof that (i) implies (ii) shows that when $b$ and $b'$ in $B$ are integral over $A$, $A[b,b']$ is a finitely generated $A$-module, and the same is true for the subring of $B$ generated over $A$ by any finite set of elements in $B$ integral over $A$.