Is every finite abelian group the Galois group of some finite extension of a non-archimedean local field?

Question

Some notation: if $F$ is a non-archimedean local field, we denote by $q_F$ the size of its residue field, whose characteristic is $p = p_F$.

This question is analogous to Every finite abelian group is the Galois group of some finite extension of the rationals, but the same approach does not apply directly. The reason is that for a local field $F$, the Galois group $\operatorname{Gal}(F(\zeta_n)/F)$ is generally not the entire multiplicative group $(\mathbf{Z}/n\mathbf{Z})^\times$, even when $F=\mathbf{Q}_p$.

It is straightforward to construct a local field extension whose Galois group is cyclic by considering the unramified extension: $$ \operatorname{Gal}(F(\zeta_{q_F^n - 1})/F) \cong \mathbf{Z}/n\mathbf{Z}. $$ However, it is unclear how to construct an extension $L/F$ such that $\operatorname{Gal}(L/F) \cong (\mathbf{Z}/n\mathbf{Z})^2$.

An alternative approach is to use local class field theory. This reduces the problem to determining which finite abelian groups arise as quotients of $F^\times$ by an open subgroup of finite index. However, the structure of $F^\times$ is intricate. If $F$ has characteristic $0$, we have $$ F^\times \cong \mathbf{Z} \oplus \mathbf{Z}/(q_F - 1)\mathbf{Z} \oplus \mathbf{Z}/p^a\mathbf{Z} \oplus \mathbf{Z}_p^d, $$ where $a$ is a non-negative integer and $d = [F:\mathbf{Q}_p]$. Since we are quotienting by an open subgroup of $F^\times$, and since $\mathbf{Z}_p$ is profinite with compact open neighborhood bases $p^n\mathbf{Z}_p$, we need to determine the possible finite quotients of $$ \mathbf{Z} \oplus \mathbf{Z}/(q_F - 1)\mathbf{Z} \oplus \mathbf{Z}/p^a\mathbf{Z} \oplus (\mathbf{Z}/p^N\mathbf{Z}_p)^d $$ where $N$ is a non-negative integer. Understanding this group's possible finite quotients remains a challenge.

See also this post In general, not every finite solvable group arises as Galois group over $\Bbb Q_p$. — Dietrich Burde, Mar 09 '25 at 11:34
See sections 2 and 3 of The Inverse Galois Problem for p-adic fields by David Roe: https://arxiv.org/abs/1809.10195 . He calls such groups $p$-realizable. — Viktor Vaughn, Mar 09 '25 at 15:59

score 5 · Accepted Answer · answered Mar 09 '25 at 15:58

If the group $$ G = \mathbf{Z} \oplus \mathbf{Z}/(q_F - 1)\mathbf{Z} \oplus \mathbf{Z}/p^a\mathbf{Z} \oplus (\mathbf{Z}/p^N\mathbf{Z}_p)^d. $$ has a subgroup $H$ such that $G/H$ has order $n$, then $nG \subset H \subset G$. When $n$ is not divisible by $p$, $$ nG = n\mathbf{Z} \oplus (n,q_F - 1)\mathbf{Z}/(q_F - 1)\mathbf{Z} \oplus \mathbf{Z}/p^a\mathbf{Z} \oplus (\mathbf{Z}/p^N\mathbf{Z}_p)^d, $$ so $$ G/nG \cong \mathbf{Z}/n\mathbf{Z} \oplus (n,q_F - 1)\mathbf{Z}/(q_F - 1)\mathbf{Z}, $$ which is a finite abelian group with at most two generators. Thus $G/H$ has at most two generators. Hence a finite abelian extension of $F$ with order not divisible by $p$ (its residue field characteristic) has a Galois group with at most two generators.

When $\ell$ is prime, a generating set of $(\mathbf{Z}/\ell\mathbf{Z})^3$ has size at least $3$. Thus no finite Galois extension of a local field has Galois group isomorphic to $(\mathbf{Z}/2\mathbf{Z})^3 \oplus (\mathbf{Z}/3\mathbf{Z})^3$.

Is every finite abelian group the Galois group of some finite extension of a non-archimedean local field?

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