Ideal generated by homogeneous polynomials

516 Views Asked by At

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

There are 1 best solutions below

0
On BEST ANSWER

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 $IR\subseteq J$.