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?

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

Notice two things:

$$[\mathbb Q(\zeta_{25}):\mathbb Q] = \phi(25) = 20$$ $$[\mathbb Q(\sqrt 5) : \mathbb Q] = 2$$

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 $20$?)

  • $\begingroup$ Yes, miscalculating $\phi(25)$ was exactly my mistake. $\endgroup$ – rj7k8 Jan 3 '16 at 0:23

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