Describe the structure $\operatorname{Gal}(\mathbb{Q}(\zeta_4)/\mathbb{Q})$

Question

I know that if $n$ is prime then $G=\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \simeq \Bbb Z_{n-1}$

But I am unsure what $G$ is when $n$ is not prime. For example when $n=4$:

$\operatorname{Gal}(\mathbb{Q}(\zeta_4)/\mathbb{Q}) \simeq \Bbb Z_4^*$

What is the structure of the group $\Bbb Z_4^*$? I am guessing it is cyclic, but I do not know how to explicitly find its generators.

Also, what is the basis of $\mathbb{Q}(\zeta_4)$ over $\mathbb{Q}$? Is it $\{\zeta, ..., \zeta^4\}$?

Answer 1

For general $n$, the units of the ring $\Bbb Z/n\Bbb Z$, which I’ll write $(\Bbb Z/n\Bbb Z)^*$, do not form a cyclic group. For, when $n=p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$, then $(\Bbb Z/n\Bbb Z)^*\cong(\Bbb Z/p_1^{e_1}\Bbb Z)^*\oplus(\Bbb Z/p_2^{e_2}\Bbb Z)^*\oplus\cdots\oplus(\Bbb Z/p_m^{e_m}\Bbb Z)^*$, the $p_i$ all being prime.

For odd primes $p$, $(\Bbb Z/p^e\Bbb Z)^*$ is always cyclic, but the structure of $(\Bbb Z/2^e\Bbb Z)^*$ is $C_2\oplus C_{2^{e-2}}$, where by $C_m$ I mean an abstract cyclic group of order $m$. (The explanation of why these groups have this structure is a story for another night.)

Answer 2

I mean, $\zeta_4 = i$ so the extension has degree $2$, so the Galois group is a finite group of order $2$ and there's only one of those, $\Bbb Z/2\Bbb Z$. Like all quadratic extensions, the basis is $\{1, i\}= \{1,\sqrt{-1}\}$. If you want the exact structure, it's mercifully simple: $i\mapsto \overline{i}$ is the automorphism, i.e. complex conjugation is the only one.

Answer 3

If $\zeta_n$ is a primitive $n^{th}$ root of unity, then $$Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}_n)^*.$$

What the elements of the group look like:

If $\sigma \in Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, then there is an $a \in \{1,2,\dots,n\}$ with $a$ relatively prime to $n$ such that $$\sigma(\zeta_n) = (\zeta_n)^a$$

For the structure of $(\mathbb{Z}_n)^*$, see Lubin's answer.