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Let $n \in \mathbb N^*$, $p(x) = x^n-1$ and $F$ be the splitting field of $p(x)$ over the rationals. Show that the Galois group of $F$ over $\mathbb Q$ is abelian.

Let us set $\omega\triangleq\mathrm{cis}(2\pi/n)$. This proof is trivial for $n$ equal to $1$ and $2$, so let's start with odd $n\neq1$.

$$p(x) = (x-1)q(x)\,,$$

where $q$ is seen to be irreducible in $\mathbb Q[x]$ by plotting its roots on the trigonometric circle. $F$ is actually the splitting field of $q(x)$:

$$F = \mathbb Q(\omega,\dots,\omega^{n-1})\,.$$

The Galois group is the set of automorphisms in $F$, which are the morphisms from $F$ to $F$ that map $\omega$ to a root of $q$:

$$\mathrm{Gal}(F) = \{\omega\overset{\large\sim}\longmapsto\omega^k\,:\,k = 1,\dots,n-1\}\,.$$

Therefore, it has order $n-1$.

For even $n\neq2$,

$$p(x) = (x+1)(x-1)r(x)\,,$$

$$F = \mathbb Q(\omega,\dots,\omega^{\frac n2-1},\omega^{\frac n2+1},\dots,\omega^{n-1})$$

and

$$\begin{align} \mathrm{Gal}(F) &= \{\omega\overset{\large\sim}\longmapsto\omega^k\,:\,k = 1,\dots,n-1;\,k\neq n/2\}=\\ &= \{\omega\overset{\large\sim}\longmapsto\pm\omega^k\,:\,k = 1,\dots,n/2-1\}\,, \end{align}$$

meaning $|Gal(F)| = n-2$.

Now what? Why does $\mathrm{Gal}(F)$ have to be abelian in both cases?

One idea I had was to prove it's cyclic instead, which immediately implies commutativity. I think the automorphism that maps $\omega$ to $\omega^{-2}$ always has the order of the group. Should I try that or am I missing an easier path?

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1 Answer 1

up vote 4 down vote accepted

You don't need to compute the order of the Galois group to show that it's abelian. You also haven't computed the order correctly: for example, when $n = 9$ the map $\omega \mapsto \omega^3$ does not extend to an automorphism (it isn't invertible). In general you've made some assumptions about irreducibility which are false; for example you claim that

$$\frac{x^9 - 1}{x - 1}$$

is irreducible and it is actually equal to

$$(x^2 + x + 1)(x^6 + x^3 + 1).$$

In general the full factorization of $x^n - 1$ is somewhat complicated. See cyclotomic polynomial.

The Galois group is also not cyclic in general.

Instead of trying to compute the Galois group, try showing that the Galois group embeds into an abelian group. (There is a natural group to consider here because an automorphism is determined by where it sends $\omega$, but my point is you don't have to show that the Galois group is this entire group and you also don't want to because this is false in general, and even when it's true it's not easy.)

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I've done exercises with $x^n-a$, $n$ and $a\neq1$ known, and indeed, the Galois group does not turn out to be abelian. Based on your answer, I have another idea: when $a$ is $1$, $\omega\mapsto\omega^k$ is invertible if and only if $k$ is invertible modulo $n$; but the set of such numbers is an abelian group, so I could argue $\mathrm{Gal}(F)\simeq\mathbb Z_n^\times$. –  Luke Sep 21 '12 at 16:32
    
@Luke: again, that's not true in general (for example if the ground field is $\mathbb{C}$ then the extension is also $\mathbb{C}$ so the Galois group is trivial). What you can prove is that the Galois group embeds into $(\mathbb{Z}/n\mathbb{Z})^{\times}$ and you can't say anything stronger than this in general. –  Qiaochu Yuan Sep 21 '12 at 17:47

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