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Suppose $\mathbf T$ is an algebraic torus defined over $\mathbb Q$, it splits as $\mathbf T_s\cdot\mathbf T_a$, respectively split and anisotropic over $\mathbb R$. Are the subtori $\mathbf T_s$ and $\mathbf T_a$ defined over $\mathbb Q$?

I cannot believe this is false but don't know how to prove it and didn't find a reference either.

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  • $\begingroup$ @nfdc23 I don't think he means $\mathbf T_s$ should be defined and split over $\mathbb Q$. An analogous statement that's actually true is that if $X$ is a scheme over $\mathbb Q$ that over $\mathbb C$ is isomorphic to a union of copies of points and $\mathbb A^1$, while none of the points and the $\mathbb A^1$s need be defined over $\mathbb Q$, the union of all the points and the union of all the $\mathbb A^1$s both are (because they are Galois-invariant). The subtle point is that this kind of argument works over $\mathbb C$ but not over $\mathbb R$. $\endgroup$ – Will Sawin May 7 '17 at 15:08
  • $\begingroup$ @WillSawin:Ah, ok. By saturation considerations with lattices inside $\mathbf{Q}$-vector spaces, it suffices to work in the isogeny category of tori and with rationalized character groups. So the question for $\mathbf{T}_s$ is if for a continuous representation of $\Gamma={\rm{Gal}}(\overline{\mathbf{Q}}/\mathbf{Q})$ on a finite-dimensional $\mathbf{Q}$-vector space $V$ and a chosen complex conjugation $\tau\in\Gamma$, the subspace $V^{\tau}$ is $\Gamma$-stable. That is easily contradicted using $S_3$-representations, such as in your answer. $\endgroup$ – nfdc23 May 8 '17 at 3:25
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The multiplicative group of the field $\mathbb Q( \sqrt[3]{2})$ provides a counterexample. Tensored with $\mathbb R$, it is $\mathbb C^\times \times \mathbb R^\times$, so it admits a one-dimensional compact subtorus. But the only one-dimensional subtorus defined over $\mathbb Q$ is $\mathbb Q^\times$ which is split. (To check this, use the fact that the group of cocharacters of the torus defined over $\overline{\mathbb Q}$ is $\mathbb Z^3$, with $S_3$ acting by a standard representation, and thus the only one-dimensional invariant subspace is the invariant element).

Let me give you some help believing this. Number theorists know that the information about some object defined over the rational numbers that you can see by tensoring with $\mathbb R$ is like the information of a geometric object that you can see locally. This is what the phrase "infinite place" is getting at. If something splits into two pieces locally, there might be no way to extend that into a global splitting. This can be hard to see, but $\mathbb R$ does not seem like a small local piece of $\mathbb Q$, but in fact it's roughly accurate - note that the completion $k[[t]]$ of the ring $k[t]$ of functions on a curve is very closely related to the local ring $k[t]_0$.

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    $\begingroup$ you probably mean ``$\mathbb{R}$ does not seem like a small local piece of $\mathbb{Q}$'', right? $\endgroup$ – Laurent Moret-Bailly May 7 '17 at 7:04

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