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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.

Thanks for your help

Bart Michels
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thinker
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    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 03 '16 at 14:01
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    Related : http://math.stackexchange.com/questions/1725821. You could have commented my answer… – Watson Apr 03 '16 at 14:07

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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 $\mathbb{Q}$).

A.B.
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  • 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 03 '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 03 '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 03 '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 03 '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 04 '16 at 01:55
  • In my entire first comment, the degree I am talking about is that of $\Phi_n$ (not that of the extension). Of course the degree of the extension depends on the irreducibility of $\Phi_n$, which is granted over $\mathbb{Q}$ though not elementary. What I understand from your last comment is that $\Phi_5$ is not irreducible over $\mathbb{F}_{19}$, on which I agree. But I don't see how this last fact alters my argument over $\mathbb{Q}$ ($\Phi_5$ is still irreducible over $\mathbb{Q}$). – A.B. Apr 04 '16 at 06:19
  • Sorry, if you are taking for granted that $\Phi_n$ is irreducible, then my comment should simply have been to point out that this is a deep(ish) fact that is essential to the proof that the Galois group is $(\Bbb Z/(n))^*$, and I guess your edit takes care of that. I gave you the plus-one, no matter what. – Lubin Apr 04 '16 at 17:42
  • @Lubin Thank you for the +1. My edits were mainly typos. I took the irreducibility over $\mathbb{Q}$ for granted is that, while it is not easy to prove, it seems to me less advanced than anything involving Galois groups, which now I see is based only on the artificial fact that in class we did as an exercise long before getting into the galois stuff. Anyway, your insistence was positive since it forced me to clear my own confusion on these matters. – A.B. Apr 04 '16 at 19:07