# Is this a Galois extension?

1. Nov 18, 2016

### PsychonautQQ

1. The problem statement, all variables and given/known data
Let Q < L < Q(c) where c is a primitive nth root of unity over Q. Is [L:Q] a Galois extension?

2. Relevant equations

3. The attempt at a solution
L must be equal to Q(d) where d is a non primitive nth root of unity. [Q(d):Q] is not a galois extension because the minimal polynomial of d over Q is x^d -1, and this polynomial has d roots, not all of which are in Q. Is this correct?

2. Nov 18, 2016

### Staff: Mentor

What does it mean for $\mathbb{Q}\subsetneq \mathbb{Q}(d)$ to be a Galois extension? Can the minimal polynomial of $d$ have other roots than powers of $d$? By the way $x^d-1$ is not the minimal polynomial, but only a factor of this: $x-1$ divides $x^d-1$ and the result is still in $\mathbb{Q}[x]$. So why should we call such a polynomial minimal though?

3. Nov 19, 2016

### PsychonautQQ

Q<Q(d) is a galois extension if Q(d) is a finite normal and separable extension. We know it is separable because Char(Q)=0. If Q(d) is a finite and normal extension then it is a splitting field for the minimal polynomial of d in Q. The minimal polynomial of d will be a factor of x^d-1, but I'm not sure if the roots of it's minimal polynomial are all powers of d; if they are, then I suspect this is a Galois extension.

4. Nov 19, 2016

### Staff: Mentor

You don't need the powers. Only the definition of normal. Firstly, how did you find $d$? And why has it to be of the claimed form, i.e. its minimal polynomial a divisor of $(x^d-1)$? Btw, it's a bit confusing to denote the element by $d$ as well as the degree of $x^d-1$. One is a complex and one a natural number. And then, does this situation differ at any point from the one you started at?

5. Nov 21, 2016

### PsychonautQQ

my bad, I'm more lost than I thought I suppose. I was thinking something like if the degree of d is n then the minimal polynomial of d will be a divisor of x^n-1. Am I way off track here?

6. Nov 21, 2016

### Staff: Mentor

You should proceed step by step.

What we have is $\mathbb{Q} \subseteq \mathbb{Q}(c)$ with a primitive $h-$th root of unity, i.e. $c^h=1$. (I choose $h$ because you burnt $n$ in your question.) Let us define the minimal polynomial of $c$ by $m_c(x)$. Then $m_c(x)\,\vert \,(x^h-1)$ and $m_c(x)$ is the product of all $(x-c^k)$ with $(k,h)=1$ being coprime. The Galois group of $\mathbb{Q}(c)$ over $\mathbb{Q}$ is isomorphic to the cyclic group $\mathbb{Z}_h^* = \mathbb{Z}_{\varphi(h)}$.

Now we assume a field $\mathbb{Q} \subseteq L \subseteq \mathbb{Q}(c)$ and we know, that its Galois group is a cyclic subgroup $\mathbb{Z}_n$ of $\mathbb{Z}_{\varphi(h)}$, i.e. $n\cdot m = \varphi(h)$ for some $m$. We may further assume that $\mathbb{Z}_n$ is generated by a (single) automorphism of order $n$, which is also an automorphism $\sigma$ of $\mathbb{Q}(c)$ and therefore $\sigma^{\varphi(h)}=(\sigma^m)^n = 1$. So $\mathbb{Z}_n=\{\sigma^m,\sigma^{2m},\ldots,\sigma^{(n-1)m},\sigma^{nm}=1\}$ is the Galois group of $L$. This means in return that $L$ consists of all elements $a$, that can be substituted by $\sigma^{m}(a)$.

This is what we know from the given situation. And it might be helpful to have examples like $h=12, 16$ or $24$ in mind. Now from here we can either list all elements $a \in L$ given a basis of $\mathbb{Q}(c)$ over $\mathbb{Q}$, i.e. powers of $c$, or construct a primitive element $d$ (and show that $L=\mathbb{Q}(d)$).