Ramification of n-th root extension

Question

Let $K$ be a local field of characteristic zero and with normalized valuation $v$ and with residue characteristic $p$, and let $n$ be relatively prime to $p$. Assume $K$ contains all $n$th roots of unity. Let $a \in K$, and let $L=K[x]/(x^n-a)$. The discriminant of $L/K$ is $\pm n^na^{n-1}$. Silverman Arithmetic of Elliptic Curves claims on p. 213 that $L/K$ is unramified if and only if $$v(a) \equiv 0 \bmod n$$ I’m confused with why the $\bmod n$ arises. I’m forgetting the connection between the discriminant and a prime being unramified - I thought it was a prime is unramified if and only it does not divided the discriminant.

Answer 1

Let $\pi \in K$ with $v(\pi)=1$
If $n$ divides $v(a)$, then one can write $a=u\cdot \pi^{nk}$ for some $k \in \Bbb N$ and $u \in \mathcal O_K^\times$. $\mathcal O_K^\times$ is a direct product of the finite group of roots of unity in $K$ whose order is coprime to $p$ and a pro-p group (cf. Neukirch ANT II.5.7). For an abelian pro-p group $G$, the map $g \mapsto g^n$ is a bijection, thus we can write $u=\zeta \cdot w^n$ where $\zeta$ is a root of unity whose order is coprime to $p$. One gets $L=K(\sqrt[n]{a})=K(\pi^kw\sqrt[n]{\zeta})=K(\sqrt[n]{\zeta})$. Thus $L$ is obtained from $K$ by adjoining a root of unity whose order is coprime to $p$, such extensions are known to be unramified.

If $n$ does not divide $v(a)$, then if we consider the unique extension of $v$ to $L$ (let it be denoted by $v$, as well), we get $v(a)=n \cdot v(\sqrt[n]{a})$ which implies that $v(\sqrt[n]{a})$ is not an integer, thus the extension $L=K(\sqrt[n]{a}) /K$ is not unramified.

Answer 2

Also worth pointing out --- the discriminant of the polynomial is not the same as the discriminant of the field extension. For example, let $p$ be an odd prime and suppose that $-1$ is not a quadratic residue. Then

$$X^2 + p^2$$

is a polynomial with discriminant $-4p^2$ which has positive valuation. But you get the same field when you adjoin a root of $X^2 + 1$ which has discriminant $-4$. The point is that $\mathbb{Z}_p[\sqrt{-p^2}]$ is not the full ring of integers inside $\mathbb{Z}_p[\sqrt{-1}]$ but has index $p$, and so the discriminant picks up a factor of this index squared.