Correct my intuition: every Galois group is $S_n$, and other obviously incorrect statements

(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?

Solution 1:

Question 1: Consider $f(x) = x^4 + 1$, which is irreducible over $\mathbb{Q}$. Then we may write all the roots of $f(x)$ as $\zeta_8^i$ where $i = 1, 3, 5, 7$. In particular $[F : \mathbb{Q}] = [\mathbb{Q}(\zeta_8) : \mathbb{Q}] = 4$. Thus the galois group cannot be all of $S_4$ - if you write out the automorphisms you will see that saying where $\zeta_8$ goes is the same as saying where all the $\zeta_8^i$ go.

Question 2: This was answered above, sometimes there is dependencies between the roots, so because each element of the galois group is a field automorphism, some will be disallowed.

Question 3: If $F$ is the splitting field of $f(x)$ and $f(x) = g(x)h(x)$ then if $\sigma \in Gal(F / \mathbb{Q})$ we have $\sigma(g(x)) = g(\sigma(x))$. In particular if $\alpha$ is a root of $g(x)$ then so is $\sigma(\alpha)$.

Question 4: This is a somewhat difficult question to answer. There are many criteria for when a subgroup of $S_n$ might be the symmetric group (e.g., if it is a transitive subgroup containing an $(n-1)$-cycle and a transposition). In general if I were trying to answer this in some situation I would do some algebraic number theory and look mod $p$.

Question 5: The degree $[\mathbb{Q}(a_i): \mathbb{Q}] = 4$ ($a_i$ a root of an irreducible polynomial of degree $4$, in particular $[F : \mathbb{Q}(a_i)] = 24/4 = 6$ by the tower law. You are correct, the elements are the permutations that fix $a_i$, but that is just $S_3$!.

Question 6: The element fixing $a_1$ and $a_2$ but permuting $a_3$ and $a_4$ is in $Gal( F / \mathbb{Q}(a_1))$ and is clearly a transposition.

Solution 2:

It's important that the automorphism do more than merely move the roots around, they must be algebraically indistinguishable. Say I start with the base field of the rationals and I extend to include the solutions to $x^2+1$ which has the complex roots $\pm i$ and we can now write all numbers in this field in the form $a + bi$ in $\mathbb{Q}[i]$. I could however set $j=-i$ and then written all the numbers in the form $a + bj$ without changing anything. Complex multiplication works exactly the same way for $j$ because $j^2=-1$ just like it does for $i$.

Now compare this to field $\mathbb{Q}[\sqrt{2},\sqrt{3}]$. Lets use $j= \sqrt{2}$ and $k=\sqrt{3}$ for this example so that $j^2=2$ and $k^2=3$. If I take numbers of the form $a + bj$ and $a + bk$ then the multiplcation doesn't work the same. To see this compare $$(a+bj)(c+dj)=ac+2bd +(bc+ad)j \\ (a+bk)(c+dk)=ac+3bd +(bc+ad)k$$ and so we can see the that if I replace $j$ with $k$ the multiplication behaves differently and so this permutation is not an automorhphism. There are two automorhphisms related to sign switching as in the previous example. I can't tell $\sqrt{2}$ and $-\sqrt{2}$ apart in the above equations in the same way as in the first example and similarly for $\sqrt{3}$ and $-\sqrt{3}$. This means we have identified two non-trivial elements and they have order two. The group must have at least one other element as $2$ does not divide $3$ and these will be related to $jk = \pm \sqrt{6}$ being indistinguishable. Number in this field are all of the form $a + bj + ck + d(jk)$ and any other permutation will change the multiplication so the Galois group is the Klein four group. Permuting some of the roots gave us automorphisms but others changed the multiplication so they weren't automorphisms.