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This question is somehow linked to this previous question I asked earlier.


Let's consider the following family $F$ of subspaces of dimension $2$ of $\mathbb R^4$. A subspace $A$ is in $F$ if it admits a basis of the form $(Y_1,Y_2)$ where

$$\begin{cases} Y_1=(0,1,\xi,\xi^2), \\ Y_2=(1,0,\xi^2,\xi).\end{cases},$$

with $\xi\notin\mathbb Q$.

I am interested in the following problem:

For every $A\in F$, does there exist a rational subspace $B$ of $\mathbb R^4$ of dimension $2$, such that $\dim(A\cap B)\geqslant 1$?

Some context.

  • By rational subspace, I mean a subspace of $\mathbb R^4$ which admits a rational basis, in other words a basis $(a,b)$ with $a,b\in\mathbb Q^4$.

  • For instance, the vector $v=(0,\xi,3\xi,\xi-\xi^2)$ is in a rational subspace of dimension $2$ since:

$$v=\xi\begin{pmatrix} 0 \\ 1 \\ 3 \\ 1\end{pmatrix}+\xi^2\begin{pmatrix} 0 \\ 0 \\ 0 \\ -1\end{pmatrix}.$$

  • I conjecture the answer to be no, but I can't prove it so far.

  • We can reformulate the problem the following way: does there exist a vector $v\in A\setminus\{0\}$ such that

$$\exists \xi_1,\xi_2\in\mathbb R,\quad \exists r_1,r_2\in\mathbb Q,\quad v=r_1\xi_1+r_2\xi_2\quad ?$$

  • We can write the problem in coordinates, which leads us to find $\alpha,\beta,\xi_1,\xi_2\in\mathbb R$ and $a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2\in\mathbb Q$, such that

$$\alpha Y_1+\beta Y_2=a \xi_1+b\xi_2,$$

which gives

$$\begin{cases} \beta=a_1\xi_1+b_1\xi_2 \\ \alpha=a_2\xi_1+b_2\xi_2 \\ \alpha\xi^2+\beta\xi=a_3\xi_1+b_3\xi_2 \\ \alpha\xi+\beta\xi^2 = a_4\xi_2+b_4\xi_2,\end{cases}$$

but it does not seem to help at all.


Since I am quite stuck with this problem, any help, hints, or new ways to look at the problem would be much appreciated.

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    $\begingroup$ In your reformulation you write $\exists r_1,r_2\in\Bbb{Q}$?, but do you mean $\exists r_1,r_2\in\Bbb{Q}^4$ in stead? I also believe the answer is indeed no, I will try to write up an answer shortly. $\endgroup$ – Servaes Oct 25 '18 at 9:49
  • $\begingroup$ @Servaes No, I meant $r_1,r_2\in\mathbb Q$. Thank you for the answer, I'll read it! $\endgroup$ – E. Joseph Oct 25 '18 at 13:22
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    $\begingroup$ But then the statement makes no sense; if $\xi_1,\xi_2\in\Bbb{R}$ and $r_1,r_2\in\Bbb{Q}$ then $v=r_1\xi_1+r_2\xi_2\notin A$. $\endgroup$ – Servaes Oct 27 '18 at 15:18
  • $\begingroup$ I guess in the system should be $a_3\xi_1$ instead of $a_2\xi_1$. $\endgroup$ – Alex Ravsky Jan 12 at 23:21
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Not only does such a rational subspace $B$ exist for all $A\in F$, but in fact there is a single rational subspace $B$ that intersects all $A\in F$ nontrivially: Note that for any $\xi\in\Bbb{R}$ we have $$(0,1,\xi,\xi^2)+(1,0,\xi^2,\xi)=(1,1,\xi+\xi^2,\xi+\xi^2)=1\cdot(1,1,0,0)+(\xi+\xi^2)\cdot(0,0,1,1),$$ so every $A\in F$ intersects the rational subspace $B$ spanned by $(1,1,0,0)$ and $(0,0,1,1)$ nontrivially.

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