# Is a polynomial which is invariant in the roots of some separable polynomial also invariant in the usual sense?

Let $\alpha_1,\cdots,\alpha_n \in \mathbb{C}$ be the roots of a separable polynomial with rational coefficients. Let $K := \mathbb{Q}(\alpha_1,\cdots,\alpha_n)$. Then the field extension $K/\mathbb{Q}$ is galois and since the galois group $G:=Gal(K/\mathbb{Q})$ permutes the $\alpha_i$, we have an injective group homomorphism

$\pi : Gal(K/\mathbb{Q}) \rightarrow S_n$.

Let $f(x_1,\cdots,x_n) \in \mathbb{Q}[x_1,\cdots,x_n]$ such that the following holds:

$f(\alpha_1,\cdots,\alpha_n) = f(\alpha_{\pi(\sigma)(1)}, \cdots,\alpha_{\pi(\sigma)(n)})$ for all $\sigma \in G$.

Does it follow that:

$f(x_1,\cdots,x_n) = f(x_{\pi(\sigma)(1)}, \cdots,x_{\pi(\sigma)(n)})$ for all $\sigma \in G$?

• Hmm I guess it's hard to be precise in the title and concise at the same time, so never mind my comment then. – user21820 Jan 3 '16 at 12:40

## 1 Answer

$P$ is the separable polynomial $x^2+1$ its roots are $i,-i$. the Galois group is $S_2$ $f(x,y)=x^2y+y$ $f(i,-i) =i^2(-i)-i=0, f(-i,i)=(-i)^2i+i=0$, but $f(x,y)\neq f(y,x)$.