8
$\begingroup$

$\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!

$\endgroup$
2
  • $\begingroup$ (Notice also that every countably infinite group, as $(\mathbb{Z},+)$, cannot be profinite, so cannot be a Galois group) $\endgroup$
    – Watson
    Jul 13, 2018 at 10:05
  • $\begingroup$ Related: math.stackexchange.com/questions/2126379 $\endgroup$
    – Watson
    Aug 25, 2018 at 18:50

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ $\to\mathbf{Z}_p^2$ on the first line? $\endgroup$ Jul 13, 2018 at 14:16
  • $\begingroup$ Dear Y. Cor., thank you very much. Just to be sure, what is the easier way to show that $\mathbb{Z}_p^2$ is not a quotient group of $\mathbb{Z}_p$ ? (I easily see that it is not a quotient ring). $\endgroup$
    – Watson
    Jul 13, 2018 at 14:26
  • $\begingroup$ (Note for myself : every morphism $f : \mathbb{Z}_p \to \mathbb{Z}_l$ is trivial since $f(a)$ has to be $l^n$-divisible for every $a \in \mathbb{Z}_p$ and $n \geq 1$). $\endgroup$
    – Watson
    Jul 13, 2018 at 14:28
  • 1
    $\begingroup$ @Watson It's not quotient as topological group because $\mathbf{Z}_p$ has a dense cyclic subgroup, not its square (every cyclic subgroup being contained in a 1-dimensional subspace of $\mathbf{Q}_p$). $\endgroup$
    – YCor
    Jul 13, 2018 at 14:49
  • $\begingroup$ This is truly a wonderful answer. $\endgroup$
    – rae306
    Apr 24, 2021 at 11:19
2
$\begingroup$

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 $\mathbf Q$.

$\endgroup$
4
  • $\begingroup$ Dear @Nguyen Quang Do, thank you for your answer. If you don't mind I edited a bit your post. $\endgroup$
    – Watson
    Jul 14, 2018 at 13:32
  • $\begingroup$ Moreover, I would have a question on the "crucial" exact sequence $E(K)\otimes\mathbf Z_p \to U^1(K) \to Y(K) \to A(K) \to 0.$ I don't see how you define the map $E(K)\otimes\mathbf Z_p \to U^1(K)$ using the diagonal embeddings $O_K^{\times} \to O_{K_v}^{\times}$. And I'm not sure how to define the middle map using the Artin map $K_v^{\times} \to C_K \to \mathrm{Gal}_K$, wher $C_K$ is the idele class group of $K$. Or do you have a precise reference for this exact sequence? $\endgroup$
    – Watson
    Jul 14, 2018 at 13:35
  • 1
    $\begingroup$ Detailed calculations in the idelic language are in Wash. thm. 13.4 and Neukirch-Schmidt-Wingberg thm. 10.3.6. The "crucial" exact sequence is just a "functorial translation". With a little bit more details (dropping the notational reference to K): let $E^1$ be the subgroup of units of K which are congruent to 1 mod every prime above p. Inject $E^1$ into $U^1$ by the diagonal embedding, and denote by $\bar E^1$ the closure of the image. Since $E \otimes Z_p$ is the pro-p-completion of $E$ (because $E$ is of finite type), the natural map $E \otimes Z_p \to \bar E^1$ is our leftmost map $\endgroup$ Jul 14, 2018 at 16:25
  • 1
    $\begingroup$ As for the middle map, "Artin" is just an all purpose word to describe the following construction: by local CFT, the subgroup $U^1(K_v)$ of the pro-p-completion of $(K_v)^*$ is interpreted as the inertia subgroup of the local group $(G(K_v)^{ab})$, which in turn is identified with a decomposition subgroup of $G(K)^{ab}$, and we take its image under the projection $G(K)^{ab} \to Y(K)$ $\endgroup$ Jul 14, 2018 at 16:41

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.