Vrftsjtryk

I am currently working through Joseph Rotman's book "Galois Theory" and am trying to prove the following theorem.

If $f(x) in F[x]$ has $n$ distinct roots in its splitting field $E$, then Gal($E/F$) is isomorphic to a subgroup of the symmetric group $S_n$, thus its order is a divisor of $n!$.

I have struggled to understand the proof in the book and have therefore tried to write my own proof of this fact, I have found some questions close to this on this site, but haven't been able to find exactly what I'm looking for.

My proof (so far) is as follows:

Proof:

Let $X = {alpha_1, ... , alpha_n }$ be the set of roots of $f(x)$, as $E$ is the splitting field of $f$, it may be written as $E = F(alpha_1, ... , alpha_n)$ and noting that $alpha_i neq alpha_j$ for $ineq j$. Let $sigma in$ Gal($E/F$) then $sigma(X) = X$ (a result from an earlier lemma). Now I would like to construct an injective homomorphism from $Gal(E/F)$ to $S_n$, my guess is the map which sends $sigma$ to its restriction in $X$, which I shall denote by $sigma_X$. Let $g$ be this map, then it is clear that $g$ is a homomorphism.

Now I try to show that $g$ is injective, consider $sigma , phi in Gal(E/F)$ where $sigma neq phi$, both of these functions are F-automorphisms of $E/F$, so they "fix" $F$, then we can conclude that $sigma$ and $phi$ can be different if and only if they act differently on $X$, which in the language of mathematics is precisely $sigma neq phi Leftrightarrow sigma_X neq phi_X$. This is the definition of an injective function.

This means that $|Gal(E/F)| leq |S_n| = n!$, however due to having constructed a homomorphism, we have mapped a group to another group, hence we have mapped $Gal(E/F)$ to a subgroup of $S_n$, whose order divides $|S_n|$ and therefore $|Gal(E/F)|$ divides $n!$, completing the proof.

Question

I am wondering whether my proof for this is correct, I think that it is mostly there, with the statement

we can conclude that $sigma$ and $phi$ can be different if and only if they act differently on X

perhaps being the only issue.

Assume $sigma_X = varphi_X$, i.e. $forall i, sigma(alpha_i) = varphi(alpha_i)$. We shall show that $sigma = varphi$, i.e. $forall x in E, sigma(x) = varphi(x)$.

Now note that $E = F(alpha_1, cdots, alpha_n)$, and every $alpha_i$ is algebraic over $F$, so every $x in E$ can be expressed as a polynomial in $(alpha_i)_{i=1}^n$ with coefficients in $F$, say $x = sum f_j alpha_{1}^{v_{j1}} alpha_{2}^{v_{j2}} cdots alpha_{n}^{v_{jn}}$. Then:

$$begin{array}{rcll}
sigma(x) &=& sigma left( sum f_j alpha_{1}^{v_{j1}} alpha_{2}^{v_{j2}} cdots alpha_{n}^{v_{jn}} right) \
&=& sum sigma left( f_j alpha_{1}^{v_{j1}} alpha_{2}^{v_{j2}} cdots alpha_{n}^{v_{jn}} right) & text {$sigma$ preserves addition} \
&=& sum sigma left( f_j right) sigma left( alpha_{1} right)^{v_{j1}} sigma left( alpha_{2} right)^{v_{j2}} cdots sigma left( alpha_{n} right)^{v_{jn}} & text {$sigma$ preserves multiplication} \
&=& sum f_j sigma left( alpha_{1} right)^{v_{j1}} sigma left( alpha_{2} right)^{v_{j2}} cdots sigma left( alpha_{n} right)^{v_{jn}} & text {$sigma$ fixes $F$} \
&=& sum f_j varphi left( alpha_{1} right)^{v_{j1}} varphi left( alpha_{2} right)^{v_{j2}} cdots varphi left( alpha_{n} right)^{v_{jn}} & forall i, sigma(alpha_i) = varphi(alpha_i) \
&=& sum varphi left( f_j right) varphi left( alpha_{1} right)^{v_{j1}} varphi left( alpha_{2} right)^{v_{j2}} cdots varphi left( alpha_{n} right)^{v_{jn}} & text {$varphi$ fixes $F$} \
&=& sum varphi left( f_j alpha_{1}^{v_{j1}} alpha_{2}^{v_{j2}} cdots alpha_{n}^{v_{jn}} right) & text {$varphi$ preserves multiplication} \
&=& varphi left( sum f_j alpha_{1}^{v_{j1}} alpha_{2}^{v_{j2}} cdots alpha_{n}^{v_{jn}} right) & text {$varphi$ preserves addition} \
&=& varphi(x)
end{array}$$

which is what was to be demonstrated.