# A family of subspaces of $\mathbb R^4$ which does not intersect non-trivially a rational subspace

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.

• 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. – Servaes Oct 25 '18 at 9:49
• @Servaes No, I meant $r_1,r_2\in\mathbb Q$. Thank you for the answer, I'll read it! – E. Joseph Oct 25 '18 at 13:22
• But then the statement makes no sense; if $\xi_1,\xi_2\in\Bbb{R}$ and $r_1,r_2\in\Bbb{Q}$ then $v=r_1\xi_1+r_2\xi_2\notin A$. – Servaes Oct 27 '18 at 15:18
• I guess in the system should be $a_3\xi_1$ instead of $a_2\xi_1$. – Alex Ravsky Jan 12 at 23:21

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.