On $\mathrm{Gal}(\mathbb{Q}(e^{\frac{2i\pi}{n}}/Q)$

Question

I want to show that $\mathrm{Gal}(\mathbb{Q}(e^{\frac{2i\pi}{n}}/Q)$ is abelian but without using or proving the fact that $\mathrm{Gal}(\mathbb{Q}(e^{\frac{2i\pi}{n}}/Q) \cong (\mathbb{Z}/n\mathbb{Z})^{\times} $.

I know that $\mathrm{Gal}(\mathbb{Q}(e^{\frac{2i\pi}{n}}/Q)$ is the splitting field of $\Phi_n $ and is therefore isomorphic to a subgroup of $S_n$ whose cardinality is $\phi(n)$ but I can't manage to prove that $\mathrm{Gal}(\mathbb{Q}(e^{\frac{2i\pi}{n}}/Q)$ is abelian.

Thanks for your help!

Well you'll have to use some arithmetic about the roots of this polynomial. E.g. can you map one primitive $n$-th unit root to another? Would that extend to an automorphism of the field? Uniquely? — Torsten Schoeneberg, Dec 09 '24 at 21:21

score 3 · Answer 1 · answered Dec 09 '24 at 21:44

There is no need to know the size of that Galois group to see it is abelian, and it is much simpler to think of it as the splitting field over $\mathbf Q$ of $x^n - 1$.

Let $\sigma$ be an automorphism of the field $F = \mathbf Q(r)$ where $r$ is a root of unity of order $n$. Since $r$ generates $F$ as a field extension of $\mathbf Q$, $\sigma$ is completely determined by $\sigma(r)$, which has to be an $n$-th root of unity, so $\sigma(r) = r^a$ where $a$ is some integer.

Let $\tau$ be a second automorphism of $F$ and write $\tau(r) = r^b$. Then $$ \sigma(\tau(r)) = \sigma(r^b) = (\sigma(r))^b = (r^a)^b = r^{ab}. $$ and similarly $\tau(\sigma(r)) = r^{ba} = r^{ab}$. Thus $\sigma\tau$ and $\tau\sigma$ are equal at $r$ and thus they are equal on the field $\mathbf Q(r) = F$. So $\sigma\tau = \tau\sigma$ as automorphisms of $F$, which shows the field automorphisms of $F$ are commutative under composition.

Very simple and direct argument. +1 – Paramanand Singh Dec 18 '24 at 05:16 — Paramanand Singh, Dec 18 '24 at 05:16

Dietrich Burde · Accepted Answer · 2024-12-10T09:29:10.410

For $\sigma$ in $G=Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ we have that $\sigma(\zeta_n)$ is an $n$-th primitive root of unity, so it can be written $\sigma(\zeta_n)=\zeta_n^k$ for some $k$ coprime to $n$. This yields an injective group homomorphism $G \hookrightarrow (\mathbb{Z}/n\mathbb{Z})^{\times}$. Hence $G$ is abelian. Of course you can show that the homomorphism is also surjective, but this is not necessary for your question. So we did not use or prove the equality of these groups, only that $G$ is a subgroup of the unit group of $\Bbb Z/n\Bbb Z$.

On $\mathrm{Gal}(\mathbb{Q}(e^{\frac{2i\pi}{n}}/Q)$

2 Answers2