Let $R=k[x_1,x_2,...,x_n]$, and $S\subset R$ be the ring of invariant polynomials under the action of a finite subset $G\subset \operatorname{GL}(n,k)$. Let $M$ be the ideal of $S$ generated by all homogeneous elements of $S$ of positive degree.

The question is: show we can find homogeneous $f_1,f_2,...,f_k\in S$ such that these $f_i$ generate the ideal $MR\subset R$.

My book leaves this part of the proof out as something trivial... However I'm lost why this should be obvious (or even why this could be done for the matter). My guess is that this shouldn't be a constructive proof. Any ideas to prove this claim?

  • 1
    $\begingroup$ It would suffice to show that $M$ is finitely generated. $\endgroup$
    – Servaes
    Mar 12, 2019 at 23:58
  • $\begingroup$ @Servaes that makes sense. Thanks $\endgroup$
    – gws
    Mar 13, 2019 at 0:17
  • $\begingroup$ Perhaps, applying the Hilbert basis theorem to your ideal $MR$ first. Then averaging the generators with respect to $\langle G \rangle$, at least when the order of $\langle G \rangle$ is finite and invertible in $k$. The homogeneous parts of the resulting polynomials would be even a finite generating set of $M$ in $S$. $\endgroup$
    – Orat
    Mar 13, 2019 at 3:04
  • $\begingroup$ Is "show we can find" asking for a constructive proof? I am pretty sure there is one, at least when $G$ is a group and $k$ has characteristic $0$ (see Noether's bound). $\endgroup$ Mar 13, 2019 at 16:01

1 Answer 1


Let $S$ be a subring of a noetherian ring $R$ and let $B$ be a set of generators of the ideal $I$ of $S$. Then the ideal $IR$ is generated by finitely many elements from $B$.

Because: by assumption $IR$ is finitely generated. Let $c_1,\ldots ,c_r\in IR$ be a set of generators. Then

$c_k=\sum\limits_{i=1}^{m_k}r_ib_{ki},\; r_i\in R, b_{ki}\in B.$

Then $IR$ is generated by the finite set $B_0$ consisting of the elements $b_{ki}$, since obviously $J\subseteq IR$ for the ideal $J$ of $R$ generated by $B_0$. On the other hand $c_k\in J$ for all $k$, hence $IR\subseteq J$.


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