What does Lang mean here by "the usual criterion"?

Question

Let $K$ be a number field containing the $n$th roots of unity, $S$ a finite set of places containing all the archimedean ones and all those which divide $n$, $K_S$ the group of $S$-units of $K$, and $L = K(\sqrt[n]{x} : x \in K_S)$. According to Serge Lang, Algebraic Number Theory (pg 216):

"$L/K$ is unramified outside of $S$ by the usual criterion, namely $L$ is obtained by adjoining roots of equations $X^n - \alpha $, where $\alpha \in K^{\ast}$ is a unit outside $S$."

I don't know what he means by "the usual criterion," so I don't understand why $L/K$ should be unramified outside of $S$.

One idea I had was to suppose the existence of an $\alpha \in K_S \cap \mathcal O_K$ which generated $L$ over $K$. In that case, if we let $f(X) = X^n - \alpha$, and $\mu \in \mathcal O_K[X]$ the minimal polynomial of $\sqrt[n]{\alpha}$ over $K$, then $$\mathscr D(L/K) \supseteq \mu'(\sqrt[n]{\alpha}) \mathcal O_L \supseteq n \sqrt[n]{\alpha}^{n-1} \mathcal O_L$$ where $\mathscr D$ is the different. So if $v$ is a place of $K$ which is not in $S$, and $w$ is a place of $L$ lying over $v$, then $ord_w(n \sqrt[n]{\alpha}^{n-1} \mathcal O_L)$, and hence $ord_w\mathscr D(L/K)$, is $0$.

I don't think such an $\alpha$ exists in general.

Answer 1

The next sentence on that page is part of the reasoning: If $A$ is a root, then $f'(A)$ is divisible only by the primes of $S$. The criterion used is the second part of Proposition 7 in Chapter II, $\S$4 (p. 48).

In your situation $\alpha\in K$ may not be in $\cal{O}_K$, but it is in $\cal{O}_{K_\frak{p}}$, since $\cal{O}_{K_{\frak{p}}}$ contains all the units of $K_\frak{p}$, and $\alpha$ is such a unit. This is enough, because ramification can be detected locally, so we can replace the number field extension $L/K$ by $L_{\frak{P}}/K_{\frak{p}}$. Once we do this, the criterion is:

If $L=K(\beta)$ for some $\beta$ satisfying a monic polynomial $g(X)\in \cal{O}_K[\rm{X}]$ such that $\overline{g}\in k_{\frak{P}}[X]$ has no multiple root, then $\frak{P}$ is unramified over $\frak{p}$ and $k_{\frak{P}}=k_{\frak{p}}(\overline{\beta})$.

Suppose $\beta$ is a root of of $g(X)=X^n - \alpha$. Then $\overline{\beta}\in k_\frak{P}$ is a multiple root of $\overline{g(X)}$ if and only if $g'(\overline{\beta})=n\overline{\beta}^{n-1}=0$. Now $\overline{\beta}$ can't be zero, because $\overline{\alpha} = \overline{\beta}^n$ is a unit, so $\overline{\beta}$ is a multiple root if and only if $\frak{p}$ divides $n$. By the definition of $S$, this does not happen.

Answer 2

Thanks to Prometheus for answering. Here is another answer similar to the approach I originally had. We have $$L = K(\sqrt[n]{K_S}) = K(\sqrt[n]{K^{\ast n} K_S})$$ and $K^{\ast n}K_S$ is a subgroup of $K^{\ast}$ containing $K^{\ast n}$, with the quotient $K^{\ast n}K_S/K^{\ast n} \cong K_S/K_S^n$ finite. Thus if we pick coset representatives $u_1^na_1, ... , u_t^na_t$ for $K^{\ast n}$ in $K^{\ast n}K_S$ (say $u_i \in K^{\ast}, a_i \in K_S$), we have $$L = K(\sqrt[n]{u_i^na_i} : i = 1, ... , n) = K(\sqrt[n]{a_1}, ... , \sqrt[n]{a_t})$$ so for $v \not\in S$, it suffices to show that $v$ is unramified in each $K(\sqrt[n]{a})$, since a compositum of unramified extensions is unramified. Now apply the argument in Prometheus's answer, or my argument using the local different above (here $a$ is a unit in $K_v$, so $\sqrt[n]{a}$ is an integral generator of $K_v(\sqrt[n]{a})$ ).