This is a question from the proof of Proposition VIII.1.6 in Silverman's Arithmetic of Elliptic Curves. We have a number field $K$ and a place $v$; $\operatorname{ord}_v$ is the normalized valuation associated to $v$, and $K_v$ is the completion of $K$. By assumption, we know that $v(m) = 0$, and the author then states:
[S]ince the discriminant of [$X^m - a$] is $\pm m^m a^{m-1}$, we see that $K_v(\sqrt[m]{a}) / K_v$ is unramified if and only if $$\operatorname{ord}_v(a) \equiv 0 \pmod{m}.$$
I'm struggling to see why this is true. I assume the idea is to use the fact that an extension is ramified if and only if the corresponding prime divides the discriminant. In this case, we would then have that the extension is ramified if and only if $\operatorname{ord}_v(\pm m^m a^{m-1}) \geq 0$.
Some calculation shows that this is equivalent to $m\operatorname{ord}_v(a) - \operatorname{ord}_v(a) \geq 0$, but this is the same as just $\operatorname{ord}_v(a) \geq 0$, so something has gone wrong. How is Silverman getting to the condition modulo $m$?
The point is that $a$ is not uniquely determined: for any $b\in K$, we have $K(\sqrt[m]a) = K(\sqrt[m]{b^ma})$.
In particular, replacing $a$ with $a' = N(v)^{km}a$ for any $k$ will give the same extension, with the same ramification, even though the valuation of $a'$ is much bigger than the valuation of $a$.
The standard way of unpacking this is to assume that $a$ is $m$-power free, so that $0 \le \mathrm{ord}_v(a)<m$.
Perhaps a better way of unpacking this is to think of $a$ as an element of $K^{\times}/K^{\times m}$, the group of non-zero elements of $K$ modulo $m$-th powers. The valuation $v$ is still well-defined on this group, but only modulo $m$, and the correct condition for ramification is then $\mathrm{ord}_v(a) \equiv 0\pmod m$.