Show that $\mathbb{Q}(\zeta_n)$ is Galois over $\mathbb{Q}$ and $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}))\cong \mathbb{Z_n}^*$ [duplicate]

Question 1: For $n \in \mathbb{N}$ explain why $\mathbb{Q}(\zeta_n)$ is Galois over $\mathbb{Q}$

We must show that it is normal and separable.

$\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is normal since it is a splitting field for $X^n-1 \in\mathbb{Q}[X]$ (I think... Is this correct?).

It is separable because it is of characteristic $0$

Question 2: $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}))\cong \mathbb{Z_n}^*$

Is this because the primitive nth root of unity $\zeta_n$ acts as a generator for a cyclic group? I am unable to fill out a decent proof on this.

• For your first question, the only non-trivial (though easy) thing to check is that it is indeed a splitting field. A priori, you only added one root of $X^n-1$, but it's easy to see that adding one primitive root is enough to add all the roots. – Captain Lama Apr 3 '16 at 14:01
For your second question, if you take $\sigma$ in $G=Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, $\sigma(\zeta_n)$ is an $n$-th primitive root of unity, so it can be written $\sigma(\zeta_n)=\zeta_n^k$ for some $k$ coprime to $n$. So you can introduce $\chi :\sigma \mapsto k$ from $G$ to $(\mathbb{Z}/n\mathbb{Z})^{\times}$, which is a well defined and injective group homomorphism. It's an isomorphism since the cardinal of $G$ is the degree of the extension $\mathbb{Q}(\zeta_n)/\mathbb{Q}$, i.e the degree of the $n$-th cyclotomic polynomial $\Phi_n$, that is $\varphi (n)$. ($\Phi_n$ is irreducible over $\mathbb{Q}$ -which is not trivial to prove- so it is the minimal polynomial of $\zeta_n$ over $\mathbb{Q}$).
• But how do you know that the degree is $\varphi(n)$? Doesn’t this depend on the irreducibility of the cyclotomic polynomial $\Phi_n$? When $n$ is divisible by at least two odd primes, or by $4$ and an odd prime, I think the irreducibility is not an elementary fact. (As always, I can easily be wrong.) – Lubin Apr 3 '16 at 16:33
• Here is the defintion I know of $\Phi_n$: $\Phi_n(X)=\Pi_{\zeta \in I} (X-\zeta)$ where $I$ is the set of primitive $n$-th root of unity. So, from this I deduce the that the degree is $\varphi (n)$, $\varphi$ being the Euler charasteristic function. As far as I know, the degree doesn't depend on the irreducibilty. And as I pointed out, the irreducibility, unless $n$ is a prime, is not trivial. – A.B. Apr 3 '16 at 17:28
• But if $\Phi_n$ is reducible (as it is over some extensions of $\Bbb Q$), then your chosen root of unity $\zeta_n$ will not be conjugate to every other primitive $n$-th root of unity, and your Galois group will be isomorphic to a proper subgroup of $(\Bbb Z/(n))^*$. – Lubin Apr 3 '16 at 17:36
• @Lubin I don't really understand your concern (which may be due to the lack of mastery on my side) since the chosen $n$-root of unity $\zeta_n$ in question is a primitive one. – A.B. Apr 3 '16 at 18:02
• It’s primitive in the sense that it generates the (multiplicative) group of all $n$-th roots of unity. It may or may not (over an unspecified base field $k$) be Galois conjugate to all the other $n$-th roots of unity. Here’s an example with $n=5$: Consider the fifth roots of unity over $\Bbb F_{19}$. There, $X^4+X^3+X^2+X+1=(X^2+5X+1)(X^2+15X+1)$. – Lubin Apr 4 '16 at 1:55