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
$g\_1 q\_1 + ... + g\_m q\_m = 1$.



Common zero of invariants of finite groups
Venkataramana's answer really leaves little to add. But let me nevertheless add that the assertion is true for all reductive groups. It is known as the "surjectivity of the categorical quotient." This can be found in probably all textbooks on invariant theory. Out of personal bias, let me quote one here: Derksen and Kemper: Computational Invariant Theory, Springer. You can find the result and some context in Chapter 2.


