Showing Galois Group is Abelian

Question

I'm having trouble showing that $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is Abelian. First I want to be able to show that $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is Galois, but I'm also not sure how to do this. Any help is appreciated!

Answer 1

Find the conjugates of $\zeta_n$. If they are all in $\mathbb{Q}(\zeta_n)$, then it's Galois. It may be easier (but in the same spirit) to show that $\mathbb{Q}(\zeta_n)$ is the splitting field of a polynomial.

Answer 2

It's the splitting field for the polynomial $x^n-1$, hence it is Galois by the characterization of normal extensions as splitting fields for a set of polynomials.

To see it is abelian, it is most direct to show that

$$\text{Gal}\left(\Bbb Q(\zeta_n)/\Bbb Q\right)\cong (\Bbb Z/n\Bbb Z)^\times$$

by showing there is a map of the latter into the former and using the fact that the groups have the same cardinality, since $\varphi(n)=\left|(\Bbb Z/n\Bbb Z)^\times\right|$ by definition and is equal to $[\Bbb Q(\zeta_n):\Bbb Q]$ by direct computation, noting that $\varphi(n)=\text{deg}(\Phi_n(x))$, the $n^{th}$ cyclotomic polynomial, or by showing it is so for $n=p^\alpha$ and using

$$\Bbb Q(\zeta_{p^\alpha})\cap\Bbb Q(\zeta_{q^\beta})=\Bbb Q$$

for distinct primes, $p,q$ and using $[LK:k]=[L:k][K:k]$ when $L\cap K=k$.

The map in the isomorphism is, of course

$$\begin{cases}a\mapsto \sigma_a\\ \sigma_a(\zeta_n)=\zeta_n^a\end{cases}$$

which is easily seen to be a group homomorphism, since

$$\left(\zeta_n^{a}\right)^b=\zeta_n^{ab}$$

and injective since

$$\zeta_n^a=\zeta_n^b\iff n|(b-a)$$

but since $a,b\in(\Bbb Z/n\Bbb Z)^\times$, are represented by integers $1\le a,b\le n-1$ we see that $|b-a|\le n$ with equality only if $b=a$. But the only integer between $-(n-1)$ and $n-1$ which $n$ divides is $0$, since all multiples of $n$ are $nk$ and if $|k|\ne 0$, we have $|nk|> n-1$. So in order for $\zeta_n^a=\zeta_n^b$ it must be that $a=b$, hence injectivity and so we have an isomorphism because the two sets are finite of the same cardinality, so all injections are surjections as well.

Answer 3

First we'll show that $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is a Galois extension. For this, it suffices to show that $\#(\text{Aut}(\mathbb{Q}(\zeta_n)/\mathbb{Q}))=[\mathbb{Q}(\zeta_n):\mathbb{Q}]$. We have $[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\deg\Phi_n(x)=\varphi(n)$. Every $\sigma_k\in\text{Aut}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is of the form $\sigma_k(\zeta_n)=\zeta_n^k$ where $\gcd(n,k)=1$, so $$\#(\text{Aut}(\mathbb{Q}(\zeta_n)/\mathbb{Q}))=[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\varphi(n),$$ so we are done. Note that this also follows from the fact that (if $E$ is the splitting field over $F$ of $f(x)\in F[x]$) $\#(\text{Aut}(E/F))\le [E:F]$, with equality when $f$ is seperable over $F$, since $\mathbb{Q}(\zeta_n)$ is a splitting field for $x^n-1$ (clearly separable) over $\mathbb{Q}$, so $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is Galois.

Now, $\#(\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})=\varphi(n)$, and in fact $$\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong (\mathbb{Z}/n\mathbb{Z})^{\times}\cong\bigoplus_{i=1}^{\ell}(\mathbb{Z}/p_i^{e_i}\mathbb{Z})^{\times}$$ where $n=p_1^{e_1}...p_{\ell}^{e_{\ell}}$. It follows that $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is Abelian (and is called an Abelian extension).

The first isomorphism is given by the map $f:(\mathbb{Z}/n\mathbb{Z})^{\times}\to\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ defined as $a\pmod{n}\mapsto\sigma_a$, where $\sigma_a$ is the automorphism defined previously (it is easy to verify this is a homomorphism since $(\sigma_a\sigma_b)(\zeta_n)=\sigma_a(\zeta_n^b)=(\zeta_n^b)^a=\zeta_n^{ab}=\sigma_{ab}(\zeta_n)$, and since every Galois automorphism is of this form $f$ is also bijective). The second is a consequence of CRT.