Extension by adjoining roots of unity and ramification

Question

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!

Answer 1

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 $L(\zeta)/L$.

Answer 2

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.

Answer 3

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.