Let $F$ be a field, and let $f(x) ∈ F[x]$ be a polynomial with only simple roots. Let $E$ be a splitting field for $f(x)$. Consider the Galois group $G_f$ of the polynomial $f(x)$.
Now I need to prove $G_f$ is the set of permutations $σ ∈ S_n$ that take every $F$-algebraic relation between the roots of $f(x)$ to another $F$-algebraic relation between the roots.
That is: I need to show that:
A permutation $σ ∈ S_n$ comes from an automorphism in $Gal(E/F)$ $\Leftrightarrow$ $(\forall g(x_1,...,x_n) ∈ F[x_1,...,x_n] )(g(α_1,...,α_n) = 0\Rightarrow g(α_{σ(1)},...,α_{σ(n)}) = 0)$.
(Here $\alpha_1,...,\alpha_n$ are all the roots of $f(x)$).
I am considering the following track for one direction:
We have a unique $F$-algebra homomorphism $ψ: F[x_1,...,x_n] → E$ defined by: $x_i \mapsto α_i$.
For $σ ∈ S_n$. Let $σ'$ denote the unique automorphism of the $F$-algebra $F[x_1,...,x_n]$ defined by $x_i \mapsto x_{σ(i)}$.
Take the composition $ψ ◦ σ'$. It is a homomorphism $ϕ: F[x_1,...,x_n] → E$. Using the universal property of quotients, may be we can prove that there is an automorphism $σ_0 ∈ Gal(E/F)$ such that $ψ ◦ σ' = σ_0 ◦ ψ$ and we also have that the permutation of the roots of $f(x)$ induced by $σ_0$ is actually the same as our original permutation $σ$.
And then I am a bit confused. So may I please ask for a way if we can acutally do it? And also how to deal with the other direction? Thanks a lot!
Let $K =F(\alpha_1,\ldots,\alpha_n)$ be the splitting field of an irreducible polynomial $f \in F[x]$ whose roots are $\alpha_1,\ldots,\alpha_n$. We have an isomorphism $$\varphi : F[x_1,\ldots,x_n]/I \to K, \qquad \varphi(p(x_1,\ldots,x_n)+I)= p(\alpha_1,\ldots,\alpha_n)$$ where $I$ is the ideal $$I = \{ p \in F[x_1,\ldots,x_n], p(\alpha_1,\ldots,\alpha_n) = 0\}$$
If and only if $\sigma \in Gal(K/F)$, we have that $\sigma$ induces an automorphism of $K$ permutating the $\alpha_i$, so that $\varphi^{-1}\circ\sigma \circ\varphi$ induces an automorphism of $F[x_1,\ldots,x_n]/I$ permuting the $x_i$.