Hasse principle for rational times square

Question

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.

Answer 1

$\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}$I've been meaning for a while to come back and talk about the group theory of this situation. I'm going to use GH from MO's fixed formulation: For every place $u$ of $\QQ$, there should be a rational number $q_u$ such that $a/q_u$ is square in $\KK_v$ for all $v$ above $u$. Note that all of the examples of "No Hasse" principle in my other answer do not incorporate GH from MO's fix, and disappear once it is included.

I will assume that $\KK/\QQ$ is Galois, with Galois group $H$. Others are welcome to work out the non-Galois case. Assume that the local condition holds.

Observation: $\KK(\sqrt{a})$ is Galois over $\QQ$. Proof: It is equivalent to show, for any $\sigma \in H$, that $\sigma(a)/a$ is square in $\KK$. Let $u$ be a place of $\QQ$. Then there is a $q_u \in \QQ$, and elements $x_v \in \KK_v$ for each $v$ over $u$, such that $a = q_u x_v^2$ in $\KK_v$. Then $\sigma(a) = q_u \sigma(x_v)^2$ in $\KK_{\sigma(v)}$. So $\sigma(a)/a = \sigma(x_{\sigma^{-1}(v)})^2/x_v^2$ in $\KK_v$. So $\sigma(a)/a$ is locally a square everywhere and hence a square.

Let $G = \mathrm{Gal}(\KK(\sqrt{a})/\QQ)$. So we have a short exact sequence $$1 \to \mathbb{Z}/(2 \mathbb{Z}) \to G \to H \to 1 \ (\ast)$$ which, since $\mathrm{Aut}(\mathbb{Z}/(2 \mathbb{Z}))$ is trivial, must be a central extension.

We have $a = q x^2$, for $q \in \QQ$ and $x \in \KK$, if and only if $\KK(\sqrt{a}) = \KK(\sqrt{q})$. This happens if and only if the extension $(\ast)$ is split.

If $\# H$ is odd, then any central extension is split and we are done; the Hasse principle holds. This has been observed in other answers.

But the local hypothesis puts additional constraints on the sequence $(\ast)$. Suppose that $\tau \in H$ has even order, and let $\sigma$ be a lift of $\tau$ to $G$. Let $(w,v,u)$ be a tower of places in $(\KK(\sqrt{a}), \KK, \QQ)$ respectively, with Frobenius elements $\sigma$ and $\tau$ corresponding to $w$ and $v$, and assume that $w$ is unramified with odd characteristic. Then $\KK_v:\QQ_u$ is even degree, unramified with odd residue characteristic, which means that all units of $\QQ_u$ is square in $\KK_v$. So $a$ is square in $\KK_v$, and we deduce that $\KK_v(\sqrt{a}) = \KK_v$. So $\sigma$ and $\tau$ have the same order. In short, we have shown:

If $\tau \in H$ has even order, and $\sigma$ is a lift of $\tau$ to $G$, then $\sigma$ has the same order as $\tau$. $(\dagger)$

There are many groups $H$ for which one check that $(\dagger)$ implies $(\ast)$ is split -- for example, cyclic groups, or $(\mathbb{Z}/2 \mathbb{Z})^n$. So that is a number more cases in which the Hasse principle holds.

However, $(\dagger)$ does not always imply splitting of $(\ast)$. The smallest example I can find is that $G$ is the $32$-element group $\left( \begin{smallmatrix} 1 & \mathbb{Z}/4 & \mathbb{Z}/2 \\ 0 & 1 & \mathbb{Z}/4 \\ 0 & 0 & 1 \end{smallmatrix} \right)$ and $H$ is the quotient by $\left( \begin{smallmatrix} 1 & 0 & 1 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{smallmatrix} \right)$.

All nilpotent groups are realizable as Galois groups over $\mathbb{Q}$, so thiere is some tower $\KK(\sqrt{a})/\KK/\QQ$ which yields this group. And this $a$ will not be globally $q x^2$. Will it obey the local condition? At the unramified primes, yes. If $v$ is a place of $\KK$ which is not split over $\QQ$, then $v$ splits further in $\KK(\sqrt{a})$, so $a$ is square in $\KK_v$. If $v$ is a place of $\KK$ which is split over $\QQ$, then $\KK_v \cong \QQ_u$ and $\sigma(a)/a \in \QQ_u^2$ for every $\sigma$, so we can find some $q \in \QQ$ such that $\sigma(a) /q$ is in $\QQ_u^2$ for every $\sigma$.

I'm not sure about the ramified primes.

If someone understands enough about the constructive Galois problem to rig up an extension with this Galois group, it would be fun to see.

Answer 2

$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}\def\LL{\mathbb{L}}\def\Gal{\mathrm{Gal}}\def\FF{\mathbb{F}}$Here are some examples of $\mathbb{K}$ for which this Hasse principle holds, and for which it does not.

Hasse principle $\KK= \QQ(\sqrt{b})$ a quadratic extension of $\QQ$. Let $\LL$ be the normal closure of $\KK(\sqrt{a})$ over $\QQ$. We break into cases according to $\Gal(\LL/\QQ)$.

Case 1 $\Gal(\LL/\QQ)$ is either $\ZZ/4 \ZZ$ or $D_8$ (the dihedral group of order $8$).

In this case, there is a conjugacy class $\sigma$ in $\Gal(\LL/\QQ)$ of order $4$ whose image in $\Gal(\KK/\QQ)$ is the nontrivial class. By Cebatarov, there are infinitely many primes $p$ of $\QQ$ whose Frobenius class is this $\sigma$.

Let $p$ be such a prime, chosen relatively prime to $2$ and $a$. Then $p \mathcal{O}_K$ is prime, with residue field $\mathbb{F}_{p^2}$ and $a \equiv q k^2 \bmod p \mathcal{O}_K$ for some $q \in \mathbb{F}_p$ and some $k \in \mathbb{F}_{p^2}$. But every element of $\mathbb{F}_p$ is square in $\mathbb{F}_{p^2}$, so $a$ is square in $\mathcal{O}/p \mathcal{O}_K$. We deduce that $p$ splits in $K(\sqrt{a})$ and thus in $\LL$, contradicting our choice of Frobenius class.

Case 2 $\Gal(\LL/\QQ)$ is $(\ZZ/2 \ZZ)^2$.

Then $\LL \cong \QQ(\sqrt{b}, \sqrt{c})$ for some $c \in \QQ$, and we have $K(\sqrt{a}) = K(\sqrt{c})$. But then $a=c k^2$ for some $k \in \KK$, as desired. $\square$.

No Hasse principle If we only formulate the Hasse principle as "for all but finitely many primes", then we have a counterexample whenever $\KK/\QQ$ is Galois of odd degree. The reason is simple: For any $a$ whatever in $\KK$, the condition will be satisfied at any unramified prime of odd characteristic.

Proof: Let $\pi$ be such a prime of $\mathcal{O}_{\KK}$, with residue field $\mathbb{F}_{p^f}$. Note that $f$ divides $[\KK:\QQ]$, so it is odd.

Then $\KK_{\pi}^{\times}/\QQ_p^{\times} \cong \FF_{p^f}^{\times}/\FF_p$ (because the extension is unramified). The quotient has order $p^{f-1} + \cdots + p+1$ which is odd, since $f$ is odd. So every element of $\KK_{\pi}$ is of the form $q k^2$ for $q \in \QQ_p^{\times}$ and $k \in \KK_{\pi}$, and it is easy to use an approximation argument to make $q \in \QQ$.

It is easy to choose $a$ to also work at the archimedean places, the even places and the ramified places, and thus get a counterexample to the whole claim.

Answer 3

I do not know if this is helpful but the question is clearly related to the Shafarevich-Tate group of some algebraic torus. Namely, let's consider algebraic torus $T=Res_{K/\mathbb Q}\mathbb G_m/\mathbb G_m$, where $Res_{K/\mathbb Q}\mathbb G_m$ is the Weil restriction of $\mathbb G_m$ from $K$ to $\mathbb Q$. Then from Hilbert 90 $T(\mathbb Q)=K^\times/\mathbb Q^\times$. Consider the short exact sequence $0\rightarrow T[2] \rightarrow T \xrightarrow{\times 2} T \rightarrow 0$, and the corresponding exact sequence of cohomology groups gives an embedding of $$(K^\times/\mathbb Q^\times)/(K^\times/\mathbb Q^\times)^{\times 2}\simeq K^\times/(K^{\times 2}\cdot \mathbb Q^\times) \hookrightarrow H^1(\mathrm{Gal(\overline{\mathbb Q}/\mathbb Q)}, T[2](\overline{\mathbb Q})) $$ And for every $p$, the points $T(\mathbb Q_p)=(\prod_{\mathfrak p|p}K_{\mathfrak p}^\times)/\mathbb Q_p^\times$ and so $$ (\prod_{\mathfrak p|p}K_{\mathfrak p}^\times/K_{\mathfrak p}^{\times 2})/ \mathbb Q_p^\times\hookrightarrow H^1(\mathrm{Gal(\overline{\mathbb Q_p}/\mathbb Q_p)}, T[2](\overline{\mathbb Q_p})) $$ So if you want to prove that $\alpha \in K^\times/(K^{\times 2}\cdot \mathbb Q^\times)$ is zero if and only if it is zero in $(\prod_{\mathfrak p|p}K_{\mathfrak p}^\times/K_{\mathfrak p}^{\times 2})/ \mathbb Q_p^\times$ for each $p$ you need to prove that the image of $K^\times/(K^{\times 2}\cdot \mathbb Q^\times)$ in $$ \underline{Ш}^1(T[2])= \mathrm{Ker}(H^1(\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q,T[2](\overline{\mathbb Q}))\rightarrow \bigoplus _{p}H^1(\mathrm{Gal(\overline{\mathbb Q_p}/\mathbb Q_p)}, T[2](\overline{\mathbb Q_p}))) $$ is zero. In particular this is true if $\underline{Ш}^1(T[2])$ is 0.

Answer 4

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.)