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.

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    $\begingroup$ Concerning the edit, the necessary condition needs to be strengthened as a whole (not just at the real places of $\mathbb{K}$): for any place $u$ of $\mathbb{Q}$, there is $q\in\mathbb{Q}$ such that for any place $v$ of $\mathbb{K}$ lying above $u$, we have $a=qk^2$ for some $k\in\mathbb{K}_v$. In fact the assumption $q\in\mathbb{Q}$ could be relaxed to the more natural condition $q\in\mathbb{Q}_u$. $\endgroup$
    – GH from MO
    Jun 4, 2016 at 16:01
  • $\begingroup$ Consider the question in the form of GH from MO. The answer seem to depend on $\mathbb{K}$. $\endgroup$ Jun 4, 2016 at 18:02
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    $\begingroup$ (cont.). Let $\mathbb{L}$ denote the normal closure of $\mathbb{K}$ in $\overline{\mathbb{Q}}$, and let $G=\mathrm{Gal}(\mathbb{L}/\mathbb{Q})$. If all the Sylow subgroups of $G$ are cyclic, then the answer seem to be YES. In general I would expect the answer NO. $\endgroup$ Jun 4, 2016 at 18:10

4 Answers 4


$\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}$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.

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    $\begingroup$ It is known that for every solvable group $G$, there exists a Galois extension $\mathbb{K}/\mathbb{Q}$ whose Galois group is $G$ and such that the decomposition group at all primes $p$ is cyclic. Does this help? $\endgroup$ Jun 14, 2016 at 7:12
  • $\begingroup$ Thanks! That does do it! The point is that the only cyclic subgroup of $G$ containing $z$ is $\langle z \rangle$, so this means that every place of $\KK$ either splits in $\KK(\sqrt{a)}$, or else is split over $\QQ$. I'll rewrite to spell this out when I get the chance. So any extension with this group and with all decomposition groups cyclic works. $\endgroup$ Jun 14, 2016 at 10:32
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    $\begingroup$ You can find a reference for this fact in the proof of Theorem 2.2. of arxiv.org/PS_cache/math/pdf/0612/0612528v4.pdf. $\endgroup$ Jun 14, 2016 at 11:08

$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\KK{\mathbb{K}}\def\LL{\mathbb{L}}\def\Gal{\mathrm{Gal}}\def\FF{\mathbb{F}}$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.

  • $\begingroup$ I now think that my "No Hasse principle" example, although it literally responds to what the original question asked, does not respond to its intent. See my comments on user42024's question for what I suspect was intended. $\endgroup$ Jun 5, 2016 at 12:32
  • $\begingroup$ The vanishing of a specific Tate-Shafarevich group as in user42024's answer is equivalent to the sufficiency of the finer necessary conditions in the comment of "GH from MO" (due to the vanishing of $Ш^1(K,T)$ that follows from Hilbert 90 and the local-to-global sequence for Brauer groups). That vanishing always (!) holds when $[K:\mathbf{Q}]$ is odd, so your "No Hasse principle" example shows that the necessary conditions as stated by "GH from MO" (and used by user42024) are genuinely stronger than the OP's necessary conditions; the stronger ones should be used. $\endgroup$
    – nfdc23
    Jun 5, 2016 at 16:41
  • $\begingroup$ @nfdc23: It seems that you claim that $Ш^1(\mathbb{Q},T[2])=0$ when $[K:\mathbb{Q}]$ is odd. If this is what you mean, could you sketch a proof? I can prove this assertion assuming that the degree of a normal closure $N$ of $K$ over $\mathbb{Q}$ is odd. $\endgroup$ Jun 5, 2016 at 19:10
  • $\begingroup$ I suppose nfdc23 assumed $K$ to be Galois as he considers $Ш^1(K,T)$ $\endgroup$
    – user42024
    Jun 5, 2016 at 20:17
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    $\begingroup$ @MikhailBorovoi: I meant to write the Tate-Shafarevich group over $\mathbf{Q}$, not $K$; a typo. The reason for the vanishing with odd degree (no Galois hypothesis needed!) is that in such cases the quotient map ${\rm{R}}_{K/\mathbf{Q}}(\mu_2) \rightarrow T$ modulo $\mu$ splits off as a direct summand (as a $\mathbf{Q}$-group), so the Tate-Shafarevich group for $T[2]$ over $\mathbf{Q}$ is thereby a direct summand of that of ${\rm{R}}_{K/\mathbf{Q}}(\mu_2)$ over $\mathbf{Q}$, or equivalently of $\mu_2$ over $K$. That in turn is trivial by Grunwald--Wang (as for $\mu_n$ whenever $8\nmid n$). $\endgroup$
    – nfdc23
    Jun 6, 2016 at 14:40

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.

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    $\begingroup$ One additional comment: Shafarevich-Tate group of a torus is known to be finite, so if you replace degree 2 by some prime p in your question, this kind of argument shows that for a fixed field K the statement is true for p big enough (bigger than the order of Sh(T)) $\endgroup$
    – user42024
    Jun 4, 2016 at 19:55
  • $\begingroup$ It is known that in general $Ш^1(T[4])\neq 0$, see Section 2 of Sansuc's paper, however it is not immediately clear how to construct a counter-example for $T[2]$. $\endgroup$ Jun 4, 2016 at 21:10
  • $\begingroup$ Could you please explain, why $Ш^1(T)=0$ implies $Ш^1(T[n])=0$? $\endgroup$ Jun 4, 2016 at 21:19
  • $\begingroup$ Ah, I had a simple argument in my head, with writing down long exact sequences of cohomology groups for $0\rightarrow T[n] \rightarrow T \rightarrow T\rightarrow 0$ for absolute Galois groups of $\mathbb Q$ and sum over $\mathbb Q_p$ correspondingly and looking on the natural map from one to another, but it appears to be wrong $\endgroup$
    – user42024
    Jun 4, 2016 at 21:50
  • $\begingroup$ I deleted the wrong part from the answer But I suppose it still should be true that $Sh(T[n])$ is finite $\endgroup$
    – user42024
    Jun 4, 2016 at 21:54

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.)


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