Is $\Bbb Z_p^2$ a Galois group over $\Bbb Q$?

Question

$\newcommand{\Q}{\Bbb Q} \newcommand{\N}{\Bbb N} \newcommand{\R}{\Bbb R} \newcommand{\Z}{\Bbb Z} \newcommand{\C}{\Bbb C} \newcommand{\A}{\Bbb A} \newcommand{\ab}{\mathrm{ab}} \newcommand{\Gal}{\mathrm{Gal}} \newcommand{\prolim}{\varprojlim} $ Fix a prime number $p$, and denote by $\Z_p$ the additive group of $p$-adic integers.

Why is there no Galois extension $K/\Q$ such that $\Gal(K/\Q) \cong \Z_p^2$ as topological groups?

The answer is no, according to a remark in Topics in Galois Theory (second edition, p. 16, just before §2.2), by J.-P. Serre. By Kronecker–Weber theorem, we have $\Gal(\Q^{\ab} / \Q) \cong \widehat{\Z}^{\times}$ (since $\{\Q(\zeta_n) \mid n \geq 2\}$ is cofinal in the direct system of abelian extensions of $\Q$), so I think my question boils to prove that there is no continuous surjection $\widehat{\Z}^{\times} \to \Z_p^2$.

Notice that there does exist a continuous surjection $\widehat{\Z}^{\times} \to \Z_p$, via the projections $\widehat{\Z}^{\times} \to \Z_p^{\times} \cong (\Z / q \Z)^{\times} \times \Z_p \to \Z_p$ (where $q=p$ if $p>2$ and $q=4$ if $p=2$).

Remarks : this would show that the "infinite" inverse Galois problem is wrong. It is true that any profinite group is a Galois group over some field (Waterhouse). It is expected (Hilbert–Noether conjecture) that any finite group is a group over $\Q$ (easy for finite abelian groups, Shafarevitch theorem for solvable groups, wrong for profinite abelian groups). Fried and Kollar showed that any finite group is the automorphism group of some number field (not necessarily Galois over $\Q$).

Thank you!

Answer 1

There's indeed no continuous surjective homomorphism $\hat{\mathbf{Z}}^\times\to\mathbf{Z}_p^2$ for any prime $p$.

Indeed, we have $\hat{\mathbf{Z}}^\times\simeq\prod_p(\mathbf{Z}_p\times\mathbf{Z}/q_p\mathbf{Z})$. Any homomorphism into $\mathbf{Z}_p$ has to be trivial on $\mathbf{Z}_\ell$ for any $\ell\neq p$ and on $\mathbf{Z}/q_p\mathbf{Z}$ for all $p$. By continuity, it therefore factors through $\mathbf{Z}_p$. Since there's no surjective continuous homomorphism $\mathbf{Z}_p\to\mathbf{Z}_p^2$, we're done.

Answer 2

This is actually a CFT problem about the $\mathbf Z_p$-extensions of an arbitrary number field $K$. A convenient reference throughout will be chapter 13 of Washinton's book "Introduction to cyclotomic fields".

Since a $\mathbf Z_p$-extension is unramified outside $p$, let us introduce $K^{(c)}$ = the compositum of all the $\mathbf Z_p$-extensions of $K$, and $K^{(p)}$ = the maximal abelian pro-$p$-extension of $K$ which is unramified outside $p$. It is known (see below) that $Y(K)=Gal(K^{(p)}/K)$ is a noetherian $\mathbf Z_p$-module, say of rank $\rho$, whose maximal torsion-free quotient is nothing but Gal$(K^{(c)}/K) \cong \mathbf Z_p^{\rho}$. The problem now is to compute $\rho$.

Notations :

• $E(K)$ = group of units of $K$;

• $U^1(K_v)$ = group of principal units of the local field $K_v$, of $\mathbf Z_p$-rank = $[K_v:\mathbf Q_p] $;

• $U^1(K)$ = direct product of the $U^1(K_v)$'s for all $p$-places $v$ of $K$;

• $A(K)$ = the $p$-class group of $K$.

The relationship between semi-local and global CFT of $K$ is condensed in the following exact sequence of $\mathbf Z_p$-modules : $$E(K)\otimes\mathbf Z_p \to U^1(K) \to Y(K) \to A(K) \to 0.$$

The leftmost map is induced by the diagonal embedding; the middle is induced by the Artin map, its image is the inertia subgroup of $Y(K)$ common to all the $p$-places of $K$; in the rightmost map, $A(K)$ is isomorphic to the Galois group of the $p$-Hilbert class field of $K$ over $K$.

Taking the alternate sum of the $\mathbf Z_p$-ranks in this exact sequence immediately gives that $$\rho =\mathrm{rank}_{\mathbf Z_p}(Y(K)) =1+r_2+\delta(K),$$ where $r_2$ is the number of pairs of complex places of $K$, and the "Leopoldt defect" $\delta(K)$ is the $\mathbf Z_p$-rank of the kernel of the diagonal map above. The famous Leopoldt conjecture (proved for abelian fields by A. Brumer) asserts the nullity of $\delta(K)$. This shows in particular that $\rho = 1$ for $K = \mathbf Q$.

NB. A direct proof without Brumer's theorem is possible for $K=\mathbf Q$: when $K$ is totally real, another (perhaps better known) version of Leopoldt's conjecture is the non vanishing of a certain $p$-adic regulator, and this holds obviously for $\mathbf Q$.