# Showing that an extension is Galois

Let $E=\mathbb{Q}(\zeta_{p})$ where $\zeta_{p}$ is the primitive $5^{\text{th}}$ root of unity.

(a) Show that $E/\mathbb{Q}$ is Galois,

(b) Compute $[E:\mathbb{Q}]$,

(b) Compute $\text{Gal}(E/\mathbb{Q})$.

The (?)s denote something that I think is correct, but not sure and don't know how to show.

What I have so far:

(a) Clearly $\mathbb{Q}(1,e^{\frac{\pi i}{5}},e^{\frac{2\pi i}{5}},e^{\frac{3\pi i}{5}},e^{\frac{4\pi i}{5}})=\mathbb{Q}(e^{\frac{\pi i}{2}})=\mathbb{Q}(\zeta_{p})=E$ is the splitting field of $f(t)=t^{5}-1$ since $\zeta_{p}$ generates all of $e^{\frac{\pi i}{5}}$, $e^{\frac{2\pi i}{5}}$, $e^{\frac{3\pi i}{5}}$, and $e^{\frac{4\pi i}{5}}$, and $1\in\mathbb{Q}$. Also, $f(t)$ is irreducible by the rational root test (?). Hence $E/\mathbb{Q}$ is finite and normal. To show $E/\mathbb{Q}$ is Galois I just have to show that it is separable (?) - how do I do that?

(b) Now, $[E:\mathbb{Q}]=\deg(m_{\zeta_{p}/\mathbb{Q}})=\deg(t^{5}-1)=5$ where $m_{\zeta_{p}/\mathbb{Q}}$ is the minimal monic irreducible polynomial of $\zeta_{p}$ over $\mathbb{Q}$ (?).

(c) Finally, the Galois group is a subgroup of $S_{5}$ which consists of all permutations which preserve the validity of all polynomials relations on the roots. My thinking is the $\text{Gal}(E/\mathbb{Q})$ should be the whole of $S_{5}$ (?) if so, how do I show that?

• Surely $f(t) = t^5 - 1$ is not irreducible, since $f(1) = 0$. Write $f = (t-1) \cdot g$ with a polynomial $g$ of degree 4 and check if $g$ is irreducible. This implies that $[E:\mathbb{Q}]$ is at most 4 and the Galois-group is contained in $S_4$. Mar 31 '17 at 13:58
• Of course! (I feel silly). $f(t)=t^{5}-1=(t-1)(t^{4}+t^{3}+t^{2}+t+1)$, and each of these are irreducible! Where do I go from there? Mar 31 '17 at 14:50
• Now $g$ is the minimal polynomial of $\zeta_5$ and it follows that $[E : \mathbb{Q}]=4$. It remains to compute the Galois-group. Where can you send $\zeta_5$? Well it has to be another primitive 5th root of unity, i.e. $\zeta_5^j$ for $j \in (\mathbb{Z}/5\mathbb{Z})^\times$. By the way, extensions like this are called "Cyclotomic Fields" and you may want to read the wikipedia article on those. Mar 31 '17 at 15:04
• Every extension over a field of characteristic zero is a separable extension.
– Xam
Mar 31 '17 at 16:10
• Ok, so $E$ is the splitting field of $g(t)$ which is irreducible. Hence $E/\mathbb{Q}$ is finite and normal. Since $\text{char}(\mathbb{Q})=0$, and any extension of a field of characteristic zero is separable, $E/\mathbb{Q}$ is Galois. Is that right? As in my question (but now with the correct statement that $g$ is irreducible, and not $f$) $[E:\mathbb{Q}]=\deg(t^{4}+t^{3}+t^{2}+t+1)=4$. I don't understand how to get the Galois group from there. Mar 31 '17 at 16:53