Ramification criteria for Kummer extensions

Question

Let $K$ be a number field containing $n$-th roots of unity. The usual Kummer theory provides a correspondence between between abelian subgroups $A \subset K^*/(K^*)^n$ and abelian extensions of K of exponent dividing $n$ explicitly by $A\mapsto K(A^{\frac{1}{n}})$.

Let $L=K(A^{\frac{1}{n}})$ be such Kummer extension, $\frak{P}$ $ \subset \mathcal{O}_L$ a prime ideal lying over a prime ideal $\frak{p}$ $ \subset \mathcal{O}_K$ of $K$.

The usual criterion to decide over ramification behavior of $\frak{P}$ over $\frak{p}$ is to take a uniformizer $p$ of $\frak{p}$ and evaluate it under the normalized valuation associated with $\frak{P}$. It's unramified iff it's valuation equals one.

Question: Does in setting of Kummer extension $L=K(A^{\frac{1}{n}})$ exist a criterion for ramification behavior of a prime $\frak{P}$ over $\frak{p}$ depending only on properties of the abelian group $A$ "generating" $L$.

That is, "how much" information about ramification behavior of $L$ over $K$ can be more less directly read up from $A$?

The question is motivated by a constrution in this cript on descent on elliptic curves, where on page $5$ is constructed certain Kummer extension $L(U^{\frac{1}{n}})$ for

$$ U := \{a \in L^× \ \vert n \vert v(a) \text{ for all valuations associated to places } v \in S_L \}/(L^* )^n $$

for $S_L$ certain finite set of places/ primes. And the question is why this extension is unramified outside place in $S_L$ and those divisible by $n$.

Does it come from certain ramification criterion specific for Kummer extensions?

Answer 1

$K( A^{1/n})$ is unramified if and only if the image of $A$ in $K_{\mathfrak p}^* / (K_{\mathfrak p}^*)^n$ lies in a certain cyclic subgroup of order $n$, which is the cyclic subgroup generated by a primitive $|\mathcal O_K/\mathfrak p|-1$th root of unity if $\mathfrak p\nmid n$ or is something harder to describe if $\mathfrak p\mid n$.

This is just because the ramification depends only on the induced extension of $K_{\mathfrak p}$, which is also generated by the $n$th roots of elements of $A$ and thus is the Kummer extension of $K_{\mathfrak p}$ defined by the image of $A$ in $K_{\mathfrak p}^* / (K_{\mathfrak p}^*)^n$. Every local field has a unique unramified extension of exponent n, whose Galois group is $\mathbb Z/n$, and thus corresponds to a cyclic subgroup of order $n$ inside $K_{\mathfrak p}^* / (K_{\mathfrak p}^*)^n$ under the Kummer duality between $K_{\mathfrak p}^* / (K_{\mathfrak p}^*)^n$ and the Galois group. It remains to identify this subgroup. If $\mathfrak p\nmid n$ then $|\mathcal O_K/\mathfrak p|-1$ is a multiple of $n$ because the field $\mathcal O_K/\mathfrak p$ contains the $n$th root of unity and this implies the generator of the $|\mathcal O_K/\mathfrak p|-1$th roots of unity generates a cyclic subgroup of order $n$. Adjoining the $n$th root of this generator gives an unramified extension by Hensel's lemma.

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If $A$ consists of $a$ such that $n\mid v(a)$ for all valuations associated to places not in $S_L$ then clearly for $\mathfrak p\notin S_L$ the image of $A$ consists of elements of $K_{\mathfrak p}^*$ with valuation a multiple of $n$. If also $\mathfrak p$ does not lie over $n$ then $K_{\mathfrak p}^* /(K_{\mathfrak p}^*)^n \cong (\mathbb Z/n \times \mathbb Z/n)$ with the projection to the first $\mathbb Z/n$ given by the valuation mod $n$ and the second $\mathbb Z/n$ is the cyclic subgroup generated by the roots of unity. So indeed the image of $A$ lies in this cyclic subgroup.