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Some context.

  • By rational subspace, I mean a subspace of $\mathbb R^5$ which admits a rational basis. In other words, a basis formed with vectors of $\mathbb Q^5$.

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

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

The question.

Let's consider the set $\mathcal I$ of subspaces $A$ of $\mathbb R^5$ of dimension $3$, which admits a basis of the form $(Y_1,Y_2,Y_3)$ where the coordinates of the $Y_i$ are of the form

$$r_1+r_2\sqrt 6+r_3\sqrt {10}+r_4\sqrt{15},\qquad r_1,r_2,r_3,r_4\in\mathbb Q.$$

For instance, the subspace

$$A=\mathrm{Span}(Y_1,Y_2,Y_3)$$

where

$$Y_1=\begin{pmatrix} 1 \\ \sqrt{10}-\sqrt{15} \\ \sqrt{6} \\ 1+\sqrt{10}+2\sqrt{15} \\ 0 \end{pmatrix}, Y_2=\begin{pmatrix} 1+\sqrt 6 \\ 0 \\ \sqrt{6} \\ 1-\sqrt 6-\sqrt{10}+3\sqrt{15} \\ 0 \end{pmatrix},Y_3=\begin{pmatrix} \sqrt {15} \\ \sqrt{6}\\ \sqrt{10}\\ 1 \\ 1\end{pmatrix}$$

is in $\mathcal I$.

Can we find a subspace $A\in\mathcal I$ such that for all rational subspace $B$ of dimension $2$,

$$A\cap B=\{0\}\quad ?$$

What I tried.

We can notice that the answer to the question is positive if, and only if, we can find $A\in\mathcal I$ such that for all $\alpha,\beta,\gamma\in\mathbb R$

$$\dim_{\mathbb Q}\mathrm{Span}_{\mathbb Q}(Z_1,\ldots,Z_5)\leqslant 2$$

where $(Z_1,\ldots,Z_5):=\alpha Y_1+\beta Y_2+\gamma Y_3$.

We can also notice that an admissible $A\in\mathcal I$ would verify that for all linearly independent vectors $X_1,X_2\in\mathbb Q^5$,

$$\det(Y_1,Y_2,Y_3,X_1,X_2)\ne 0.$$

If we develop the last determinant, since

$$\dim_{\mathbb Q}\mathrm{Span}_{\mathbb Q}(1,\sqrt 6,\sqrt{10},\sqrt{15})$$

we get a system of $4$ equations and $10$ rational unknowns (the coordinates of $X_1$ and $X_2$).

We can reformulate the problem in this way:

Can the resulting system of $4$ equations has rational solutions?

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  • $\begingroup$ If you explicitly write down the 4 equations in 10 unknowns, it might be interesting to check if it has rational solutions. $\endgroup$ – Tito Piezas III Jun 2 '19 at 16:22

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