# Finding a subspace of dimension $3$ which does not intersect a rational subspace of dimension $2$

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?

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