No I don't think this is right. For example the extension $\Bbb{Q}(\sqrt{3},\sqrt{2})$ is a Galois extension of $\Bbb{Q}$ because it is the splitting field of $(x^2 - 3)(x^2-2)$. This polynomial has exactly 4 distinct roots in $\Bbb{Q}(\sqrt{3},\sqrt{2})$ and it can easily be shown that its Galois group is the Klein 4-group $V_4$. But $V_4$ has order 4 that is not equal to $4!=24$.
Now you are thinking that the Galois group has to be $S_4$. Let me tell you why this is not possible. Let us call $\sqrt{3}$ root #1, $\sqrt{2}$ root #2, $-\sqrt{3}$ root #3, $-\sqrt{2}$ root #4. Given a cycle in $S_4$ let that cycle act on the roots simply by permuting the numbers $1,2,3$ and $4$. For example, the cycle $(12)$ exchanges $\sqrt{3}$ and $\sqrt{2}$ and keeps the negative guys fixed. But then this cannot possibly be a valid element of the Galois group because:
Elements of the Galois group must send for example $\sqrt{2}$ to another root of the minimal polynomial of $\sqrt{2}$ over $\Bbb{Q}$. The minimal polynomial of $\sqrt{2}$ when viewed as an element of $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is $x^2 -2$. Therefore the only possibility for where $\sqrt{2}$ can be sent to is $-\sqrt{2}$, because this is the only other root of this polynomial in any splitting field. So therefore the cycle $(12)$ above cannot be a valid element of the Galois group (if we view the Galois group as sitting inside of $S_4$).
It follows that the Galois group in this case can only be viewed as a proper subgroup of $S_4$ and hence cannot have order $4!=24$.
Edit: Since you seem to be having some trouble understanding the Galois group, let me explain a bit more here. Now I assume that you know what an $F$ - algebra is (otherwise how would you understand field extensions?)
The following is the start of how one describes the Galois group:
Let $A = F[x]$ where $F$ is a field. Let $\iota_A$ denote inclusion of $F$ into $A$. Then any $F$ - algebra homomorphism from $F[x]$ to some other $F$ - algebra $B$ (where the homomorphism in question for $B$ is just the inclusion map $\iota_B : F \to B$) is completely determined by specifying the image of $x$ in $B$. This is because if we have an $F$ - algebra homomorphism $\varphi$ from $F[x] \rightarrow B$, we must have that
$$\iota_B = \varphi \circ \iota_A.$$
In particular this means that $\varphi$ must be the identity on the coefficients of a polynomial in $F[x]$. This explains why $\varphi$ is completely determined by its action on $x$. Now we claim that we have a bijection of sets
$$\Big\{\operatorname{Hom}_{\text{$F$ -algebra}} \big(F[x],B\big)\Big\} \longleftrightarrow B $$
where the bijection is given by $f$ that maps $\varphi$ on the left to $\varphi(x)$ with inverse $g$ that maps an element in $b \in B$ to the homomorphism $\varphi_b$ which is evaluation at $b$. Viz. $\varphi_b$ is just the homomorphism that sends $x$ to $b$. You can check that $f$ and $g$ are mutual inverses.
Now a corollary of this is that we have a bijection
$$\Big\{\operatorname{Hom}_{\text{$F$ -algebra}} \big(F[x]/(f(x)),B\big)\Big\} \longleftrightarrow \Big\{b\in B : \varphi_b(f(x)) = 0 \Big\}. $$
I will leave you to work out the details of how this comes from the fact I stated before. Essentially it is due to the universal property of quotient rings that says given a unique ring homomorphism $\varphi$ from $F[x]$ to $B$ we have unique ring homomorphism from the quotient of $F[x]/(\ker \varphi)$to $B$ such that $\varphi$ factorises through the quotient. I can edit my post to explain this more if you wish.
This is the start of how one gets a description of the Galois group because giving a homomorphism from some field say $F(\alpha)$ to itself (which is automatically an automorphism by the Rank - Nullity Theorem) is by my description above equivalent equivalent to specifying a root of the minimal polynomial of $\alpha$ over $F$ in $B$. But then our $B$ here is exactly what we started with, that is $F(\alpha)$ so that $\alpha$ must be sent to another root of its minimal polynomial over $F$.
Does this help to explain more to you?
We define a Galois extension $L/K$ to be an extension of fields that is
- Normal: if $x\in L$ has minimal polynomial $f(X) \in K[X]$, and $y$ is another root of $f$, then $y\in L$.
- Separable: if $x\in L$ has minimal polynomial $f(X) \in K[X]$, then $f$ has distinct roots in its splitting field.
When $L/K$ is a finite extension, these conditions are equivalent to $L$ being the splitting field of a separable polynomial $f(X) \in K[X]$ - i.e. your condition $1$. This is a fact which is proven in any course in Galois theory. See for example Theorem 3.10 in these lecture notes.
Your condition $2$ is certainly false: for example $\mathbb Q(\sqrt2)/\mathbb Q$ is a Galois extension, but is not the splitting field of $X^5+3X+2$ or of any other (irreducible) polynomial other than $X^2-2$.
Best Answer
First just to point out that this is when $E$ is a finite separable normal extension of $F$. At the steps you ask about $\alpha$ and $\beta$ have the same min poly over $F$ and so an earlier result shows that $F(\alpha)$ is isomorphic to $F(\beta)$ and there is an isomorphism $\psi_{j}$ say that acts as the identity on $F$ and maps $\alpha$ to $\beta$. But now the splitting field $E$ sits on top of both these fields and (essentially) using the result that proves splitting fields are isomorphic you can construct an isomorphism $\phi_{j}$ say from $E$ to $E$ such that its restriction to $F(\alpha )$ is $\psi_{j}$. Now there is a $\phi_{j}$ for each root and there are $r$ of these and $\phi_{j}$ is in Gal($E$/$F$). Let $\theta_{s}$ be the elements of Gal($E$/$F(\alpha))$ and so there are $n/r$ such $\theta_{s}$. Now look at the maps $\phi_{j}\theta_{s}$. They are distinct, there are $n$ of them and you show they exhaust Gal($E$/$F$).
Hope this helps