TITLE: Correct my intuition: every Galois group is $S_n$, and other obviously incorrect statements
QUESTION [12 upvotes]: (I hope that this question is acceptable and within the rules of math.stackexchange. If not, mods should edit at will and let me know if this question must be broken into several different questions. I ask these all together at once because they seem crucially tied together insofar as answers would correct my misunderstandings.)
I am currently studying Galois Theory, but I am unable to get a handle on the subject. My intuition leads to me conclusions which are obviously incorrect, so I will ask a brief series of questions which I think will help me correct my course.
Let $Q$ be the rationals. Let $f$ be an irreducible polynomial (hence separable) over $Q$ with degree $n$. Let $F$ be the splitting field of $f$. So $F/Q$ is Galois. Let $G$ be the Galois group of $F/Q$. Let $a_1,\dots ,a_n$ be the distinct roots of $f$. My understanding is that $F=Q(a_1,\dots ,a_n)$.
Question 1: Since every automorphism of $F/Q$ permutes $a_1,\dots ,a_n$, it is clear that $G$ can be embedded into $S_n$ as a subgroup. Why is it not the case that $G$ is automatically all of $S_n$? Surely any permutation of the roots gives an automorphism of $F$ preserving $Q$? If not, what might be an instructive minimal example?
Question 2: Conceptually speaking, what exactly prevents certain permutations from being acceptable automorphisms of $F/Q$?
Question 3:  Alternatively, if $f$ is not irreducible, then $F$ is the splitting field of some polynomial which is not irreducible. In this case, I believe that roots from different irreducible components cannot jump to roots of other irreducible components. Why is this the case?
Question 4: Again we assume that $f$ is irreducible. What must be unique about the situation in order for $G$ to really be all of $S_n$?
Question 5: Now set $n=4$. I know that $A_{4}$ is the only subgroup of $S_4$ with order $12$. Suppose that $G=S_4$. Suppose $K=Q(a_1)$. Why is it not the case that $F/K$ has order $12$ and hence has Galois group $A_4$? It seems that the Galois group of $F/K$ could include all permutations of $a_1,\dots ,a_4$ that map $a_1$ to itself.
Question 6: Suppose we are in the case of Question 5. Why is it the case that the Galois group of $F/K$ has a transposition?

REPLY [2 votes]: Question 1 and 2:
A $\mathbb Q$-homomorphism does not only permute the roots of some polynomial. It's also a field automorphism, and it has to obey the axioms of a field automorphism. This may restrict the choices of permutations if the roots can be written as polynomials of each other, since homomorphisms have very strong restrictions when applied to polynomial expressions. For instance, if $\alpha$ is one root of $f$, and the others are $\alpha^2,\dots,\alpha^n$, then the image of that one single root $\alpha$ under a $\mathbb Q$-homomorphism $\sigma$ already determines the images of all the other roots as well, since $\sigma(\alpha^k)=\sigma(\alpha)^k$, removing a lot of choice. An example of such a polynomial is $X^4+X^3+X^2+X+1$, whose roots are $\zeta:=e^{2\pi i/5}$ and $\zeta^2,\dots,\zeta^4$, the primitive 5th roots of unity.
Question 3:
A $\mathbb Q$-automorphism maps a root of any polynomial to another root of the same polynomial, you already got that. If a polynomial can be factorized, let's say $f=gh$, and $\alpha$ is a root of $f$, then $\alpha$ must also be a root of one of the factors (say $g$), which is where the restrictions come from: all polynomial relationships have to be preserved by $\mathbb Q$-automorphisms, not just an arbitrarily chosen one. So such an automorphism preserves both $f(\alpha)=0$ and $g(\alpha)=0$. So $\alpha$ has to be mapped to a root of both $f$ and $g$ (so essentially a root of $g$).
Question 4:
Essentially, the roots of the polynomial must not admit too many polynomial relationships between the roots (as in, one root is the image of another under a polynomial), because those would put the restrictions discussed above on the choice of permutations.
Question 5 and 6:
Why would the order be 12? If the minimal polynomial of $F/\mathbb Q$ is $(X-a_1)\dots(X-a_4)$, then the minimal polynomial of $F/\mathbb Q(a_1)$ is $(X-a_2)\dots(X-a_4)$, just drop the factor with $a_1$. The Galois group of this polynomial is at most $S_3$. In fact, taking only the subgroup of $S_4$ whose permutations fix $a_1$ is essentially the same as taking the group of permutations of only $a_2,\dots,a_4$. That's $S_3$, realized as a subgroup of $S_4$. And this realization of $S_3$ as a subgroup contains a transposition.