# Is $\mathbb{Q}(\zeta_5)/\mathbb{Q}(\sqrt5)$ the maximal finite abelian extension of $\mathbb{Q}(\sqrt5)$ unramified away from $5\infty$?

The following problem appears in a homework question posed by B. Conrad (2(i) here: http://math.stanford.edu/~conrad/249BPage/homework/hmwk9.pdf):

Using class field theory, prove that $\mathbb{Q}(\zeta_5)/\mathbb{Q}(\sqrt5)$ is the maximal finite abelian extension of $\mathbb{Q}(\sqrt5)$ that is unramified away from $5\infty$ and has degree prime to $5$.

I think there's an easy counterexample to this claim. If $K:= \mathbb{Q}(\sqrt5)$, consider the extension $K(\zeta_{25})/K$. The Galois group $G$ injects into $(\mathbb{Z}/25\mathbb{Z})^\times$, so $G$ is abelian and $5 \nmid |G|$. Any finite prime of $K$ ramifying must contain $25$, so the only option is $(\sqrt5)$. Hence the extension is unramified away from $5\infty$. Finally, $K(\zeta_{25}) \nsubseteq K(\zeta_{5})$.

What is wrong?

• $|G| = 10$ and $5|10$. – TokenToucan Dec 22 '15 at 23:09
• @CuddlyCuttlefish you should consider posting this as an answer – Stella Biderman Jan 2 '16 at 19:58
• @StellaBiderman done :) – TokenToucan Jan 3 '16 at 0:14

$$[\mathbb Q(\zeta_{25}):\mathbb Q] = \phi(25) = 20$$ $$[\mathbb Q(\sqrt 5) : \mathbb Q] = 2$$
And that $\mathbb Q (\sqrt 5) \subseteq \mathbb Q(\zeta_{25})$, by working out that $\mathbb Q(\sqrt 5) = \mathbb Q(\zeta_5 + \zeta_5^{-1})$.
Then we have that $\mathbb Q(\zeta_{25})$ is a normal extension of $\mathbb Q(\sqrt 5)$ of degree $10$ (since degrees multiply in towers). It is abelian by the theorem about the galois group obtained by adjoining roots of unity. Then the order of the galois group is $10$, which contradicts your statement that $5 \nmid |G|$.
It looks to me like you probably just calculated $\phi(25)$ incorrectly (maybe as $16$ instead of $20$?)
• Yes, miscalculating $\phi(25)$ was exactly my mistake. – rj7k8 Jan 3 '16 at 0:23