There is a classical description of the behavior of prime ideals (ramified, inert, or split) in a Kummer extension of prime degree. I have seen this, for instance, in Gras Class Field Theory: From Theory to Practice, Lemmermeyer Reciprocity Laws, and a special case in Hilbert's Zahlbericht.

None of these authors gives a similar description for Kummer extensions of composite degree. Gras says that it is difficult to give the decomposition law for general degree, while Lemmermeyer says that "a similarly simple criterion in the case of Kummer extensions of prime power degree is not known, nor even for cyclic quartic extensions!"

Now if you just want to distinguish when a prime is unramified in these extensions, and not the splitting/inertia behavior, then I wonder if there is a reasonably simple criterion that people just don't bother to write down because it feels incomplete. The following is a partial criterion that seems to follow from the degree $p$ case by a simple inductive argument. (Note: this has been edited. The original post had par 2 of the statement below as a biconditional, but user463330's post shows how to easily produce an example showing the stronger statement is false.) My question now is, can part (2) of the following statement be modified into a biconditional giving true necessary and sufficient conditions for whether a prime ramifies in a cyclic, degree $p^r$ extension?

Let $K$ be a number field containing the $p^r$th roots of unity, and let $L = K(\sqrt[p^r]{\alpha})$ for some $\alpha \in K^{\times}$, not a $p$th power. Let $\mathfrak{p}$ be a place of $K$, and let $v_{\mathfrak{p}}$ be the corresponding valuation. Then

1. If $\mathfrak{p}$ does not divide $p$, then $\mathfrak{p}$ is unramified in $L/K$ if and only if $v_{\mathfrak{p}} (\alpha) \equiv 0 \pmod{p^r}$ 2. If $\mathfrak{p}$ divides $p$, let $e$ be the ramification index of $\mathfrak{p}$ in $K/\mathbb{Q}(\zeta_p)$ ($\zeta_p$ is a primitive $p$th root of unity). Then $\mathfrak{p}$ is unramified in $L/K$ if the congruence $$\alpha/x^{p^r} \equiv 1 \pmod{\mathfrak{p}^{(r(p-1)+1)e}}$$ has a solution $x$ in $K^{\times}$.

When $r=1$ and the extension has prime degree, then this is the classical criterion (and the converse of the second statement holds as well).


1 Answer 1


Consider only a cyclic extension of degree $4$ of $\mathbb{Q}(i)$.

Then your modulus is $(8)$ (if I am not wrong: $r=2$, $p-1=1$, $e=2$). But any local unit $x$ in $\mathbb{Q}_2(i)$ is the product of a power of $i$ by a unit congruent to $1$ modulo $2\mathbb{Z}_2[i]$ (try all residues).

So (if correct) $x^4$ is congruent to $1$ modulo $(8)$ and does not enter. Your criteria becomes $\alpha$ congruent to $1$ modulo $(8)$. This may be exact, but as you say, does not give the exact splitting of $2$.

But I see that if $\alpha$ is congruent to $1$ modulo $(8)$, it seems that $\pi$ must split in the quadratic extension of $K$ (is it correct?), which does not cover totally inert case?

Perhaps, to be certain, try numerical experiments with PARI/GP which computes easily relative discriminants.


The case of $p=2$ is perhaps particular? But for $p\neq2$, we still have the relation$$(1-\zeta)^{p-1} =p*\text{unit},$$which makes trouble.

I think that some papers of Massy have some computations on the subject, check out his webpage.


There are also papers with $p^r$-cyclic Kummer extensions, but I do not remember the authors$\ldots$ try with hunting on the Internet yourself.

  • $\begingroup$ You are right. If I adjoin a fourth root of $9+4i$ to $\mathbb{Q}(i)$, I get an extension in which $2$ is totally inert. I'll need to think about this more. (I still believe that both directions of 1 hold and that one direction of 2 holds.) $\endgroup$ Commented Jul 19, 2017 at 21:49
  • $\begingroup$ I found the error in my original thinking. Admittedly, the result keeps getting less interesting as I keep weakening it! I am starting to understand why no one writes down anything other than the prime degree case! I have searched some for $p^r$-cyclic Kummer extensions, but the one thing I've learned in 20 years of looking through math literature is that you can never search too much. $\endgroup$ Commented Jul 19, 2017 at 22:10

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