5
$\begingroup$

The main reduction step in proving the Kronecker-Weber Theorem is to reduce the theorem to proving the following result:

Theorem. If $K/\mathbb{Q}$ is a cyclic extension of degree $p^n$ which is unramified outside $p$, then $K \subset \mathbb{Q}(\zeta_{p^{n+1}})$.

My question: Is it possible to construct a cyclic extension of degree $p^n$ (take $n=2$ for convenience) which is ramified both at $p$ and at another prime $q\neq p$? It is clear that such a prime must be tamely ramified, but how does one construct $K$?

$\endgroup$
1
$\begingroup$

Take a prime $q \equiv 1 \bmod p^n$; then the field of $q$-th roots of unity has a cyclic subfield of degree $p^n$. Compose it with the cyclic field ramified exactly at $p$ an look for a suitable cyclic subextension different from those ramified at only one prime.

$\endgroup$
1
$\begingroup$

Actually you can get hold of all the cyclic extensions of the type you are looking for. Things are clearer if we start taking a "higher" point of view. For a number field $K$ and an (odd, for simplification) prime $p$, fix a finite set $S$ of primes of $K$ containing all the primes above $p$, and denote by $G(S)$ the Galois group over $K$ of the maximal pro-$p$-extension of $K$ which is unramified outside $S$ (i.e. the compositum of all the finite $p$-extensions of $K$ with that property), and by $X(S)= G(S)^{ab}$ its maximal abelian quotient. By CFT, the $\mathbf Z_p$-module $X(S)$ is of finite type, of the form $\mathbf Z_p^{r} \times T(S)$, where $T(S)$ is the (finite) $\mathbf Z_p$-torsion. The free factor $\mathbf Z_p^{r}$ is the Galois group over $K$ of the compositum of all the $\mathbf Z_p$-extensions of $K$, i.e. of the Galois extensions of $K$ whose Galois groups are isomorphic to $(\mathbf Z_p , +)$. It is known that such extensions are unramified outside {$p$}, so that for your purpose, we must look at the torsion $T(S)$. Using Galois cohomology, it can be shown that a minimal system of generators of $T(S)$ is obtained by "abelianizing" a minimal system of relations of the pro-$p$-group $G(S)$; see e.g. H. Koch's "Galois theory of $p$-extensions", chapter 11.

The aforementioned relations are generally unknown, but they are explicit (just as the K-W.theorem is explicit) in the case $K=\mathbf Q$. In your situation where $S$={p, q}, $G(S)$ can be described minimally by 2 generators $t_p , t_q$ with 1 single relation of the form $t^{q-1}$ modulo commutators (op. cit., example 11.11). This means in particular that $G(S)$ is a free pro-$p$-group (i.e. $T(S)=0$) iff $q\neq1\mod p$, in which case you cannot get your desired cyclic extension. In the other cases, take $q\equiv1\mod p^n$ as in the answer of @franz lemmermeyer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.