I am trying to understand the proof of Proposition 1.6 in Silverman's Arithmetic of Elliptic Curves. Specifically, I am confused about how he uses the main theorem of Kummer theory. The theorem states:
If a field of characteristic zero contains the $m^{th}$ roots of unity, $\mu_m$, then its maximal abelian extension of exponent $m$, $M/K$, is obtained by adjoining the $m^{th}$ roots of all its elements. i.e. $M = K(a^{1/m} : a \in K)$.
Silverman is using this result to prove if $K$ is a field containing $\mu_m$, and $L/K$ is the maximal abelian sub-extension of $M/K$ that is unramified outside of a set of primes $S$, then $L/K$ is finite. He says that $L$ must be the maximal subfield of $M$ unramified outside of $S$, which makes sense to me. But then I am stuck. He uses a fact (that I will assume for now) that an extension of the form $K(a^{1/m})$ is unramified at prime/place $v$ if and only if $ord_v(a) \equiv 0 \text{ mod } m$. Given this, he says $L$ must be of the form $$ L = K(a^{1/m} : ord_v(a) \equiv 0 \text{ mod } m),$$ and I don't understand why that's true. Certainly, $K(a^{1/m} : ord_v(a) \equiv 0 \text{ mod } m) \subseteq L \subset M$. But how do we know a subset of the $a^{1/m}$ generate $L$? Is this true of all subfields of $M$?
Ah, I read the wikipedia page on Kummer theory and I think it answers my question. According to Wiki/Kummer theory, if $K$ contains the $m^{th}$ roots of unity, any abelian, exponent m extension of K will be generated by $m^{th}$ roots of elements of $K$. Thus $L$ is generated by $m^{th}$ roots of elements of $K$. Then you can use the above fact (that $K(a^{1/m})$ is unramified at prime/place $\nu$ if and only if $ord_\nu(a) \equiv 0$ mod $m$) to get the desired result.
Edit: Also while I'm at it, this SE question covers why $K(a^{1/m})$ is unramified $\leftrightarrow$ $ord_\nu(a) \equiv 0$ mod $m$.