How exactly does a Galois group behave?

Question

Suppose that $L:K$ is a Galois extension, so that $L$ is the splitting field for a minimal polynomial $f$ over $K$. We can write $f$ as $$(t-\alpha _1)\ldots (t-\alpha _r)$$ so that any $K$-automorphism of $L$ can be thought of as a permutation of the zeroes of $f$. I'm wondering if all such permutations would lead to a $K$-automorphism of $L$, yet I haven't been able to (dis)prove it.

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More generally, for a Galois extension $L:K$, in which $L$ is the splitting field of the polynomial $f$, which can be factored into irreducibles as $$m_1\ldots m_n,$$ if we let $\tau _i$ be an arbitrary permutation of the zeroes of $m_i$, would the product $\tau _1\ldots \tau_n$ be a $K$-automorphism of $L$?

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I would appreciate any help/thoughts!

Answer 1

No, not every permutation need lead to an automorphism.

A cyclotomic field for example, is a field of the form $\mathbb{Q}(\zeta_n)$, where $\zeta_n$ is a primitive $n$th root of unity. It is the splitting field of $x^n-1$, and hence Galois. It is known that the Galois group $\mathrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is always abelian, and so cannot be the full symmetric group if $n>3$ (the degree is $\phi(n)$, Euler’s phi function at $n$).

Now, $x^n-1$ is not irreducible for $n>1$. $\mathbb{Q}(\zeta_n)$ is the splitting field of $\Phi_n(x)$, the $n$th cyclotomic polynomial. For arbitrary $n$ it may not be obvious what the value is, but when $n$ is prime, we have that $$\Phi_p(x) = x^{p-1}+x^{p-2}+\cdots + x + 1$$ (which is $(x^p-1)/(x-1)$). This can be shown to be irreducible over $\mathbb{Q}$ by substituting $x=y+1$ and applying Eisenstein’s Criterion at $p$.

So $[\mathbb{Q}(\zeta_p):\mathbb{Q}] = p-1$, and the Galois group is abelian. If $p>3$, this cannot be the symmetric group on $p-1$ letters.

(In fact, the Galois group of $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$ is isomorphic to the group of units of $\mathbb{Z}/n\mathbb{Z}$, and so for $n$ a prime it will be cyclic of order $p-1$.)

In general, if $f$ is irreducible of degree $n$, then the Galois group is a transitive subgroup of $S_n$; thus, given any two roots $u$ and $v$ of $f$ you can always find an element of the Galois group that sends $u$ to $v$; but in general, you may not even have double transitivity. There may exist two pairs of distinct roots $u_1,u_2$ and $v_1,v_2$ for which no element of the Galois group sends $u_1$ to $v_1$ and $u_2$ to $v_2$.

Answer 2

Let $f \in F[x]$ and $K/F$ a splitting field of $f$. Suppose $f$ is separable so $K/F$ is Galois. Let the roots of $f$ in $K$ be $Z = \{\alpha_1, \dots, \alpha_n\}$. Let $G = Gal(K/F)$. We can define an injective map $G \longrightarrow \Sigma(Z) = S_n$ via $\sigma \mapsto \sigma|_Z$. This is well defined because any element of the Galois group permutes the roots of $f$ (indeed, $\sigma$ fixes $F$ - where the coefficients of $f$ live). It is injective as $K$ is generated over $F$ by the roots of $f$, as it is the splitting field. We can therefore view $G \subseteq S_n$. Your question is whether or not $G = S_n$, and the answer is no.

First of all, if we furthermore assume $f$ to be irreducible, we have that $G$ acts transitively on the roots of $f$. This is a basic fact about Galois theory that I won't prove here. This is the same as saying that $G$ is a transitive subgroup of $S_n$ under this identification, but there are often many of these. For a concrete example, $\langle(1234)\rangle \subseteq S_4$ is a transitive subgroup. To witness this in Galois theory, take $\mathbb F_{p^4}/\mathbb F_p$. Letting $\alpha$ be a generator of $\mathbb F_{p^4}^\times$, $\mathbb F_{p^4} = \mathbb F_p(\alpha)$. Then of course, if $f$ is the minimal polynomial of $\alpha$, $\mathbb F_{p^4}$ is a splitting field. Its Galois group is $\mathbb Z/4$ and cyclically permutes the roots of $f$. Hence, there are permutations in $S_4$ (viewed as permutations of the roots of $f$) which do not induce field automorphisms.

More generally, let $f = \prod f_i^{e_i} \in F[x]$ be an irreducible factorization. Let the set of roots of each $f_i$ (assumed to be distinct) be $Z_i$. Let $G$ be the Galois group of $f$ and $Z = \bigcup Z_i$ the roots of $f$. Then certainly $G$ acts on $Z$, inducing an embedding $G \subseteq \Sigma(Z)$. However, this won't be a transitive subgroup of $\Sigma(Z)$ unless there is only one $f_i$ (so $f$ is a prime power). Hence, $G < \Sigma(Z)$. $G$ will, however, act transitively on each $Z_i$, so indeed these will end up being the orbits of this action.