If $p(x)=x^n-1$, prove that the Galois group of $p(x)$ over the field of rational numbers is abelian.

Here's what I have so far.

Denote the Galois group $G(K,\mathbb{Q})$, where $K$ is the splitting field for $p(x)$ over $\mathbb{Q}$.

By setting $x^n-1=0$, we find the $n$th roots of unity $\omega, \omega^2,\cdots,\omega^n=1$, where $\omega=e^{2\pi i/n}$. Then, the splitting field $K=\mathbb{Q}(\omega)$.

By a theorem, $K$ is a normal extension of $\mathbb{Q}$. We now wish to examine $G(K,\mathbb{Q})=G(\mathbb{Q}(\omega),\mathbb{Q})$. By defintion, this is the group of automorphisms of $\mathbb{Q}(\omega)$ that keep every element of $\mathbb{Q}$ fixed. In other words, if $a\in \mathbb{Q}$, $\sigma(a)=a$ for all $\sigma\in G(\mathbb{Q}(\omega),\mathbb{Q})$.

Suppose $\sigma,\tau \in G(\mathbb{Q}(\omega), \mathbb{Q})$. We know the group structure is given by composing automorphisms. To show that this group is abelian, we need to show that $(\sigma \circ \tau)(b)=(\tau \circ \sigma)(b)$ for all $b\in \mathbb{Q}(\omega)$.

We know that all elements of $\mathbb{Q}$ are fixed. That is, if $a\in\mathbb{Q}$,



Now, consider $\sigma(\omega)=\sigma(e^{2\pi i/n})$. We have $(\sigma(e^{2\pi i/n}))^n=\sigma(e^{2\pi i})=\sigma(1)=1$. This implies that $\sigma(e^{2\pi i/n})=$ an $n$th root of unity. Thus, $\sigma$ just permutes roots of unity.

This is where I am confused. If the automorphism permutes roots of unity, it doesn't seem to necessarily be the case that $(\sigma\circ\tau)(\omega)=(\tau\circ\sigma)(\omega)$.

Please let me know where to go from here (or where I've gone wrong in my argument). Thanks.


Your argument up to the point where you're stuck seems completely correct to me.

To finish it, suppose that $\sigma(\omega) = \omega^j$ and that $\tau(\omega) = \omega^k$ for some $j,k \in \mathbb{N}$. Then just observe that $$ (\sigma \circ \tau)(\omega) = \sigma(\tau(\omega)) = \sigma(\omega^k) = \sigma(\omega)^k = (\omega^j)^k = \omega^{jk} $$ and similarly $$ (\tau \circ \sigma)(\omega) = \tau(\sigma(\omega)) = \sigma(\omega^j) = \tau(\omega)^j = (\omega^k)^j = \omega^{jk} $$


You are correct that the group of all permutations of roots of unity is not abelian. But not all permutations actually give a field automorphism.

The important fact is this: once you know $\sigma(\omega)$, you also know $\sigma(\omega^2)$, $\sigma(\omega^3)$, etc.


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