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Let $p$ be a prime, $n\geq 1$, $\zeta=\zeta_{p^n}$ a primitive $p^n$th root of unity, $L$ a number field, and $\wp$ a prime ideal of the ring of integers of $L$ lying above $p$.

Suppose that $L(\zeta)$ is a non-trivial extension of $L$. Is $L(\zeta)/L$ necessarily ramified at $\wp$? I think so, but how do you prove this?

Thanks!

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No. Enough ramification may have already happened when extending from $\mathbb{Q}$ to $L$.

For a counterexample let $p=3, n=1$, and let $L=\mathbb{Q}[\sqrt3]$. Then ${\frak p}=(\sqrt3)$ is the only prime ideal of $L$ above the rational prime $(3)$, and $e({\frak p}:3)=2$. This time $L(\zeta)=\mathbb{Q}[\sqrt3,i]$. Because $(3)$ is inert in $\mathbb{Q}[i]/\mathbb{Q}$, it follows that ${\frak p}$ must also be inert in the extension $L(\zeta)/L$.

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  • $\begingroup$ Thanks, but I'm actually hard pressed to find another example (with $p\neq3$). Anyone? $\endgroup$ – Jyrki Lahtonen Apr 1 '13 at 19:39
  • $\begingroup$ @JyrkiLahtonen, I extended your idea to work with any $p>2$ and any $n\geq 1$. See my answer below. $\endgroup$ – Álvaro Lozano-Robledo Apr 2 '13 at 2:43
  • $\begingroup$ @unity: If you can follow Álvaro's solution, do consider accepting that in place of mine. After all, it is more general. $\endgroup$ – Jyrki Lahtonen Apr 2 '13 at 19:37
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Let $p>2$ be any odd prime, and $n\geq 1$ also arbitrary. Let $\zeta=\zeta_{p^n}$ be a primitive $p^n$th root of unity. Let $q\neq p$ be another odd prime, and consider $K=\mathbb{Q}(\zeta,\sqrt{q})$. The extension $K/\mathbb{Q}$ is the compositum of $\mathbb{Q}(\zeta)$ and $\mathbb{Q}(\sqrt{q})$, which are disjoint (one ramifies only at $p$, the other one ramifies at $q$ and perhaps at $2$), so it is Galois, with Galois group $$G\cong (\mathbb{Z}/p^n\mathbb{Z})^\times \times \mathbb{Z}/2\mathbb{Z}\cong \mathbb{Z}/\varphi(p^n)\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}.$$ Clearly, the ramification index of $p$ in $K/\mathbb{Q}$ is $\varphi(p^n)$.

Let $H$ be the subgroup of order $2$ generated by $(\varphi(p^n)/2 \bmod \varphi(p^n),1 \bmod 2)$, and let $L=K^H\subset K$ be the fixed field of $H$. Notice that the inertia subgroup $I_p$ at $p$ is generated by $(1 \bmod \varphi(p^n),0\bmod 2)$. Since $I_p\cap H$ is trivial, it follows that $K/L$ is quadratic, unramified at $p$. Moreover, $\zeta\not\in L$, because if $\zeta\in L$, then $\mathbb{Q}(\zeta)\subseteq L$, and therefore $\mathbb{Q}(\zeta)=L$ (because $K/\mathbb{Q}(\zeta)$ is quadratic). But $\mathbb{Q}(\zeta)$ is not the fixed field of $H$, but the fixed field of the group generated by $(0\bmod \varphi(p^n),1\bmod 2)$, so $\mathbb{Q}(\zeta)\neq L$ and we have reached a contradiction. Hence $\zeta\not\in L$.

Now consider $L(\zeta)/L$. Clearly, since $\zeta\not\in L$, we have $L\subsetneq L(\zeta)\subseteq K$. But since $K/L$ is quadratic, we must have $L(\zeta)=K$. We have shown above that $L(\zeta)=K/L$ is unramified at $p$, and quadratic, so we are done.

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    $\begingroup$ Well done! I had to think this thru, and the following special case of your construction helped. Consider the case $p=5$, $q=3$, $n=1$. Here the generator of $H$ is the automorphism determined by $\zeta\mapsto \zeta^4=\overline{\zeta}, \sqrt3\mapsto -\sqrt3$. Thus $L=\mathbb{Q}[\sqrt3(\zeta-\zeta^4)]$. Here the number $$\sqrt3(\zeta-\zeta^4)=2i\sqrt3\sin(2\pi/5)=i\sqrt3\sqrt{\frac{5+\sqrt5}2}.$$ Anyway, the facts $L(\sqrt3)=K=L(\zeta)$ also make it clear $5$ is totally ramified in $L/\mathbb{Q}$ and ${\frak p}$ is unramified in $K/L$ as it is so in $\mathbb{Q}(\sqrt3)/\mathbb{Q}$. $\endgroup$ – Jyrki Lahtonen Apr 2 '13 at 7:58
  • $\begingroup$ Äh, I meant that $(5)$ is unramified in $\mathbb{Q}(\sqrt3)/\mathbb{Q}$. $\endgroup$ – Jyrki Lahtonen Apr 2 '13 at 9:50
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The only primes that ramify in the extension $L(\zeta)/L$ are precisely the ones that divide the discriminant of $L(\zeta)$. By class field theory and the finiteness of the class group, there always exists an everywhere unramified extension. The maximal abelian unramified extension is called the Hilbert class field. In fact, an everywhere unramified extension over $\mathbb{Q}$ would necessarily be a cyclotomic extension.

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  • $\begingroup$ Yes... So? Can $L(\zeta)/L$ be everywhere unramified? $\endgroup$ – unity Apr 1 '13 at 2:33
  • $\begingroup$ Give me a second to update my answer $\endgroup$ – Brent J Apr 1 '13 at 2:40
  • $\begingroup$ Why is an everywhere unramified extension of a number field $L$ necessarily cyclotomic? $\endgroup$ – unity Apr 1 '13 at 3:02
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    $\begingroup$ I think that an everywhere unramified extension of $\mathbb Q$ is just $\mathbb Q$ itself, right?Then I cannot understand your statements... Thanks for any explanations in advance. $\endgroup$ – awllower Apr 1 '13 at 12:45
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    $\begingroup$ ...there is no everywhere unramified extensions of $Q$, except for $Q$ itself. Again thanks for any explanation in advance. $\endgroup$ – awllower Apr 23 '13 at 4:24

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