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.


2 Answers 2


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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