# Common zero of invariants of finite groups

Let $G$ be a finite group $n = |G|$. Let $\sigma : G \rightarrow GL(n,\mathbb{C})$ be the regular representation. Hence every element of $G$ can be seen as a permutation matrix. Let $\mathbb{Q}[x_1,...,x_n]^G = \mathbb{Q}[f_1,...,f_m]$ be the invariant ring. Let $s_1,...,s_r$ (this might be empty) be the algebraich dependencys of the $f_i$, that is $s_j(f_1,...,f_m) = 0$ for $j=1,..,r$. Suppose we choose rational numbers $a_1,..,a_m$ which fullfill the algebraic dependecys, that is $s_j(a_1,...,a_m)=0$. Does there exists $\alpha_1,...,\alpha_n \in \mathbb{C}$ such that $f_i(\alpha_1,...,\alpha_n) = a_i$? My first thought is to use Hilbert's Nullstellensatz to prove this by contradiction, but I am unsure if it is always true. I would consider the polynomials $g_i = f_i - a_i$. Then by Hilberts Nullstellensatz exactely one thing is true:

1) The $g_i$ have a common zero $\alpha \in \mathbb{C}^n$ hence $f_i(\alpha) = a_i$

2) There exist polynomials $q_1,...,q_m$ such that $g_1 q_1 + ... + g_m q_m = 1$.

• yes, it is true. You can extend a maximal ideal in the ring of invariants to a maximal ideal in the whole polynomial ring. Otherwise, the polynomial span of the smaller maximal ideal generates the unit ideal (nullstellensatz) , and by taking invariants under the finite group you get that the maximal ideal contains $1$. This argument holds for any finite integral extension Jan 2, 2016 at 14:24
• Thanks! Do you have a reference where I could read this up and fill in the details of the proof you just gave?
– user6671
Jan 2, 2016 at 14:27
• Atiyah Mcdonald (commutative algebra) have a section on finite integral extension a Jan 2, 2016 at 15:13