Prove $Aut_F(E)$ is isomorphic to a subgroup of $Perm(X_1)×···×Perm(X_r)$ Let F be a ﬁeld. Let $f(x) ∈ F[x]$ be a nonzero polynomial, let $E$ be a splitting ﬁeld for $f(x)$ over $F$, and let $X$ denote the set of roots of $f(x)$ in E
Let $f(x)\in F[x]$, and suppose $f(x) = f_1(x)···f_r(x)$ is the factorization of $f(x)$ into irreducible polynomials in $F[x]$.
For each irreducible factor $f_i(x)$, let $X_i$ denote the set of roots of $f_i(x)$ in $E$. May I please ask how to prove that $Aut_F(E)$ is isomorphic to a subgroup of $Perm(X_1)×···×Perm(X_r)$.
I have tried induction but stuck, so may I ask if it can be done by induction? Or is there any simpler explaination? May I please ask for an explaination?
 Thanks in advance!
EDIT: May I please ask the assumption that $f_i$'s are irreducible is necessary? May I please ask if why we need it or not.
 A: I shall assume that $E$ is contained in $E'$, the splitting field of $f$. Then $Aut_F (E) $ is a subgroup of $Aut_F E $. So sufficient to assume $E$ is the splitting field of $f $. Then consider the action of $Aut_{F} (E) $ on the Cartesian product of $X_i $. Each $\sigma \in Aut_F (E) $ permits the elements of $X_i$ and only $1_E $ leaves every element of $X_{i} $ fixed. So there exists an injective map from $Aut_F (E) $ to $Perm (X_1)×Perm (X_2)×....×Perm (X_r) $ suppose $\sigma $ sends an element of $X_i$ to an element of $X_{j}$. Then those two elements are conjugate and hence satisfy both $f_i $ and $f_j $ whence we get $f_i=f_j$ and $X_i=X_j $. But then we can drop  $X_j$ from consideration and the previous arguments go through.
A: There are three ideas here:
Call $G$ the automorphism group over $F$ of the splitting field $E$ of $f = f_1 \ldots f_n \in F[x]$. $G = Aut_F(E)$.


*

*Each $f_i \in F[x]$ will be fixed by $G$. This implies that it acts on the roots of $f_i$. (Write $f_i = \Pi (x - \alpha_i)$, which is a true equation in $E$[x], then $g \in G$ preserve the polynomial on the left, because its coefficients are in $F$ and thus fixed by $G$. Also the group action preserves the truth of this equation. So all that can happen is that perhaps the $\alpha_i$ get moved around.)

*If $f_i$ is irreducible, then $G$ will act transivitely on its roots. For this, you have to look at the construction of the splitting field. The feeling is that when you adjoined a root of $f$, you could have adjoined a different one instead, and the result would have been an isomorphic field extension. Then you lift this isomorphism up to the level of the splitting field using the uniqueness of the splitting field. This requires proof and there should be a relevant lemma in your Galois theory text. (This is 18.3 in Isaac's Algebra.)

*$G$ is determined by its action on the roots of $f$. This is because the roots of $f$ will generate $E$ as a field over $F$ (our assumption that $E$ is the splitting field), and the action of elements in $G$ preserves the field operations and leaves $F$ invariant.


Altogether, $1.$ says we get a map into $Perm(X_1) \times \ldots Perm(X_n)$ and $3.$ says it is injective. I guess 2. is actually irrelevant, but it's good to know anyway. (Why did I write it???)
If is $E$ to be contained in the splitting field for$ f$, its automorphism group is a subgroup of $Aut_F (Split(f))$. (Split(f) is temporary notation for the splitting field.)
Let me know if any needs further clarification.
