TITLE: Hasse principle for rational times square
QUESTION [22 upvotes]: Does a Hasse principle hold for the property of being a rational times a square ?
Let $a \in \mathbb{K}$ be an element of a number field. Assume that at every place $\mathbb{K}_v$ of $\mathbb{K}$, $a$ can be written as $a=q k^2$, with $q \in \mathbb{Q}$ and $k \in \mathbb{K}_v$. Is it true that $a$ can always be written as $q k^2$ with $q \in \mathbb{Q}$ and $k\in \mathbb{K}$ ?
EDIT : there's a restriction at the real places of $\mathbb{K}$. One should assume that the sign of $a$ is the same at every real place.

REPLY [4 votes]: No, in general the Hasse principle for the property of being a rational number times a square does not hold. I consider the question in the form of GH from MO. I give a counter-example with a non-normal extension $K/\mathbb{Q}$ of degree 6 (the accepted answer of David Speyer gives a counter-example with a normal extension of degree 16). It suffices to take the concrete Galois extension $L/\mathbb{Q}$ of degree 12 with Galois group $G=A_4$ and with cyclic decomposition groups from the answer of Jeremy Rouse, and to take $K=L^H$, where $H$ is a subgroup of $G$ of order 2.
Consider the homomorphism of $\mathbb{Q}$-tori
$$ 
\varphi\colon\  \mathbb{G}_{m,\mathbb{Q}}\times_{\mathbb{Q}}R_{K/\mathbb{Q}}\mathbb{G}_{m,K} \ \to\  R_{K/\mathbb{Q}}\mathbb{G}_{m,K}\,,\quad (q,k)\mapsto qk^2.
$$
We wish to show that $Ш^1(\mathbb{Q},\ker \varphi)\ne 0$.
Set $T=R_{K/\mathbb{Q}}\mathbb{G}_{m,K}/\mathbb{G}_{m,\mathbb{Q}}$.
We write $T[2]$ for the subgroup of elements of order dividing 2 in $T$.
The homomorphism
$$ 
 \mathbb{G}_{m,\mathbb{Q}}\times_{\mathbb{Q}}R_{K/\mathbb{Q}}\mathbb{G}_{m,K} \ \to\ T,\quad (q,k)\mapsto \mathbb{G}_{m,\mathbb{Q}}\cdot k
$$
indices a homomorphism $\ker\varphi\to T[2]$ fitting into a
short exact sequence
$$ 1\to \mathbb{G}_{m,\mathbb{Q}}\to\ker\varphi\to T[2]\to 1, $$
from which we obtain a canonical isomorphism
$$ Ш^1(\mathbb{Q},\ker\varphi)\overset{\sim}{\to} Ш^1(\mathbb{Q}, T[2]).$$
Write $M=T[2]$. It suffices to show that $Ш^1(\mathbb{Q}, M)\ne 0$.
We have a reduction $Ш^1(\mathbb{Q},M)= Ш^1(L/\mathbb{Q},M)$,
see Sansuc, J.-J. Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres. J. Reine Angew. Math. 327 (1981), 12–80, Lemma 1.1(ii).
Since all the decomposition groups of $L/\mathbb{Q}$ are cyclic, we have  $Ш^1(L/\mathbb{Q},M)=Ш^1_\omega(G,M)$ (see my question for the definition of $Ш^1_\omega(G,M)$ ).  By the answer of Kasper Andersen $Ш^1_\omega(G,M)\ne 0$. Thus  $Ш^1(\mathbb{Q},\ker\varphi)\neq 0$, and our extension $K/Q$ is a counter-example to the Hasse principle.
It would be interesting to construct explicitly an element $a\in K$ for which the Hasse principle fails (i.e., $a=qk^2$ locally, but not globally.)