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$\zeta_{15}$ is $15$th primitive $n$th root of unity.

Question: Find the structure of the group $Gal(\mathbb{Q}(\zeta_{15})/\mathbb{Q})$

I know that if $p$ is prime then $G=Gal(\mathbb{Q}(\zeta_{p})/\mathbb{Q})=\mathbb{Z_{p}}^*$ but when $p$ is not prime, I am not show how to solve this if not prime.

Does my required group have any relation to some cyclic group $\mathbb{Z_n}$, i.e. abelian

Would really appreciate your guidance as I have not got my head around these concepts of cyclotomic fields and Galois structure. Thanks

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  • $\begingroup$ It's the same thing as for primes. The Galois group is $\Bbb{Z}_{15}^*$ $\endgroup$ – Crostul Apr 3 '16 at 11:11
  • $\begingroup$ Why is this the case? I would like to be able to construct some kind of proof $\endgroup$ – thinker Apr 3 '16 at 11:11
  • $\begingroup$ en.wikipedia.org/wiki/Cyclotomic_field $\endgroup$ – Crostul Apr 3 '16 at 11:12
  • $\begingroup$ I have seen it, and it does not give an explanation in sufficient detail $\endgroup$ – thinker Apr 3 '16 at 11:13
  • $\begingroup$ Google "cyclotomic fields". You will find a lot of proofs for this fact, and other things $\endgroup$ – Crostul Apr 3 '16 at 11:15
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Let $G$ be the Galois group of $\Bbb Q(\zeta_n)$ over $\Bbb Q$. An element $f \in G$ is entirely determined by its image on $\zeta_n$.

Since $f(\zeta_n)^k=f(\zeta_n^k)=1 \iff \zeta_n^k=1$ (recall that $f$ is a field automorphism), you know that $f(\zeta_n)$ is a primitive $n$-th root of unity: $$f(\zeta_n)=\zeta_n^{k_f}$$ for some integer $1≤k_f≤n$ coprime with $n$.


Therefore, you have a well-defined map $$\alpha : G \to (\Bbb Z/n\Bbb Z)^* \qquad \alpha(f)=[k_f]_n$$ You can check that this is a group isomorphism. It is clearly injective. Since $|G|=[\Bbb Q(\zeta_n):\Bbb Q]=\text{deg}(\Phi_n)=\phi(n) = |(\Bbb Z/n\Bbb Z)^*|$, $\alpha$ is bijective.

Finally, $(f \circ g)(\zeta_n)=f(g(\zeta_n)) = f(\zeta_n^{k_g})=\zeta_n^{k_g \, k_f}$ shows that $$\alpha(f \circ g) = [k_{f \circ g}]_n = [k_f]_n[k_g]_n=\alpha(f)\alpha(g).$$


More generally, let $K$ be a field of characteristic coprime with $n$ (e.g. if $\text{car}(K)=0$), and suppose that $R_n = \{x \in \overline K \mid x^n=1\}$ denotes the set of $n$-th roots of unity in an algebraic closure $\overline K$ of $K$ (notice that $R_n$ is always a cyclic group for the multiplication, since it is a finite subgroup of $(K^*,\cdot)$).

Then the extension $K(R_n)$ over $K$ is Galois (it is separable because $n$ is coprime with the characteristic of $K$) and you can show similarly that the Galois group of $K(R_n)$ over $K$ embeds in $(\Bbb Z/n\Bbb Z)^*$.

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  • $\begingroup$ so in this case $\zeta_{n}$ acts as a cyclic generator? $\endgroup$ – thinker Apr 3 '16 at 14:10
  • $\begingroup$ Yes, absolutely: the $n$-th roots of $1$ (i.e. $x$ such that $x^n=1$) are of the form $x=\zeta_n^j$, i.e. the set $R_n$ of the $n$-th roots of $1$ is actually the subgroup generated by $\zeta_n$. Indeed, $R_n$ is a finite subgroup (for the multiplication) of $\overline{ \Bbb Q}^*$ (where $\overline{\Bbb Q}$ is an algebraic closure of $\Bbb Q$. You may know that every finite subgroup of $(K^*,\cdot)$ (where $K$ is any field) is actually cyclic. $\endgroup$ – Watson Apr 3 '16 at 14:13
  • $\begingroup$ Then, $\Bbb Q(\zeta_n)$ over $\Bbb Q$ is Galois because it is separable (any extension of a field of characteristic $0$, as $\Bbb Q$, is separable) and normal (it is the splitting field of $X^n-1$ : any root of $X^n-1$ belongs to $R_n = \langle \zeta_n \rangle$, and therefore to $\Bbb Q(\zeta_n)$). $\endgroup$ – Watson Apr 3 '16 at 14:16

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