TITLE: A normal extension over $\mathbb{Q}$
QUESTION [6 upvotes]: Let $f(x)$ be an irreducible polynomial of degree $5$ in $\mathbb{Q}[x]$. Suppose $a$ and $b$ are distinct roots of $f$ and that $\mathbb{Q}(a)=\mathbb{Q}(b)$. Show that $\mathbb{Q}(a)$ is a normal extension of $\mathbb{Q}$.
Now both $\mathbb{Q}(a)$ and $\mathbb{Q}(b)$ have degree $5$ and $f$ has at least one real root and I also think that the given roots can not be conjugate of each other, otherwise $f$ wouldn't be an irreducible polynomial of degree $5$. But I couldn't get the statement. Any help would be great.

REPLY [6 votes]: Let $K$ denote the splitting field of $f$ over $\mathbb{Q}$, and let $G := \text{Gal}(K/\mathbb{Q})$ and $H := \text{Gal}(K/\mathbb{Q}(a))$. If we enumerate the roots of $f$ as $\{a,b,c,d,e\}$, then $G < S_5$. Furthermore, since $f$ is irreducible,
$$
[G:H] = [\mathbb{Q}(a):\mathbb{Q}] = 5
$$
Now, $H = \{\sigma \in G : \sigma(a) = a\}$. However, since $\mathbb{Q}(a) = \mathbb{Q}(b)$,
$$
H = \{\sigma \in G : \sigma(b) =b\}
$$
Hence, $H < S_3$, so $|H| \in \{1,2,3,6\}$.
If $|H| = 6$, then $|G| = 30$, but $S_5$ does not have a subgroup of order 30. Similarly, $|H| \neq 3$. If $|H| = 2$, then $H$ (and therefore $G$) contains a transposition. However, $G$ already contains a 5-cycle, and so $G$ must be equal to $S_5$. This contradicts the fact that $|G| = |H|[G:H] \leq 30$. Hence,
$$
|H| = 1
$$
and so $K = \mathbb{Q}(a)$, whence $\mathbb{Q}(a)/\mathbb{Q}$ is normal.