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

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

The question.

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

$$Y_1=\begin{pmatrix} 1 \\ 0 \\ 0 \\ \sqrt 6 \\ \sqrt {15} \end{pmatrix},\quad Y_2=\begin{pmatrix} 0 \\ \sqrt{6} \\ 0 \\ -\sqrt {15} \\ \sqrt {10} \end{pmatrix}\quad\text { and }\quad Y_3=\begin{pmatrix} 0 \\ 0 \\ \sqrt{15} \\ \sqrt {10} \\ \sqrt {6} \end{pmatrix}.$$

Does there exist a rational subspace $B$ of $\mathbb R^5$, of dimension $2$, such that $A\cap B\ne\{0\}$?

I believe the answer to be negative.

What I tried.

For $Y\in A$, let's denote by $Y_1,\ldots,Y_5\in\mathbb R$ its coordinates. We can prove the following lemma.

Lemma. The answer to the question is negative, if, and only if,

$$\forall Y \in A,\quad \dim_{\mathbb Q}(\mathrm{Span}_{\mathbb Q}(Y_1,\ldots,Y_5)\ge 3.$$

I tried to investigate what such a rational subspace $B$ would look like. Let's take $B$, a rational subspace of $\mathbb R^5$ of dimension $2$, such that $A\cap B\ne\{0\}$.

Let $Y:=\alpha Y_1+\beta Y_2+\gamma Y_3$ be a vector in $A\setminus\{0\}$.

We have

$$\alpha Y_1+\beta Y_2+\gamma Y_3=\begin{pmatrix} \alpha \\ \beta\sqrt 6 \\ \gamma \sqrt {15} \\ \alpha\sqrt 6-\beta\sqrt {15} +\gamma\sqrt{10} \\ \alpha\sqrt{15}+\beta\sqrt{10}+\gamma\sqrt{6}\end{pmatrix}.$$

If we assume (to try to get somewhere) that

$$\dim_{\mathbb Q}(\mathrm{Span}_{\mathbb Q}(\alpha,\sqrt{6}\beta)=2,$$

then the last three coordinates of $Y$ must be a rational linear combination of the first two, i.e. there exists $x_1,\ldots,x_6\in\mathbb Q$ such that

$$\begin{cases}\gamma \sqrt {15}=x_1\alpha+x_2\beta\sqrt 6 \\ \alpha\sqrt 6-\beta\sqrt {15} +\gamma\sqrt{10}= x_3\alpha+x_4\beta\sqrt 6 \\ \alpha\sqrt{15}+\beta\sqrt{10}+\gamma\sqrt{6}=x_5\alpha+x_6\beta\sqrt 6\end{cases}$$

which can be rewrite

$$MX=0\quad\text{ with } \quad M=\begin{pmatrix} -x_1 & -x_2\sqrt 6 & \sqrt{15} \\ \sqrt{6}-x_3 & -\sqrt{15}-x_4\sqrt 6 & \sqrt{10} \\ \sqrt{15}-x_5 & \sqrt{10}- x_6\sqrt 6 & \sqrt 6 \end{pmatrix}\quad \text{ and }\quad X=\begin{pmatrix} \alpha \\ \beta\\ \gamma\end{pmatrix}.$$

So if $\det(M)\ne 0$, we have won, since $(\alpha,\beta,\gamma)=(0,0,0)$ is the only solution, thus $Y=0$ which is absurd.

Let's compute $\det(M)$ then:

$$\det(M)=A\sqrt{15}+B\sqrt{10}+C\sqrt 6+D,$$

with $A,B,C,D\in\mathbb Q$ depending on the $x_i$. Since

$$\dim_{\mathbb Q}(\mathrm{Span}_{\mathbb Q}(\sqrt 6,\sqrt{10},\sqrt{15}))=3,$$

and $A,B,C,D\in\mathbb Q$, if we assume $\det(M)=0$, we must have


The computations give

$$(\mathscr S)\quad \begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \\ -3x_4x_5+3x_3x_6+3x_1=0 \\ 6x_2-5x_3+15x_4=0 \\ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.\end{cases}$$

The question can now be reformulated as follow:

Does the system $(\mathscr S)$ has rational solutions?

If we try to solve the system, we can end up with this expression:

$\frac{20 \, x_{1} x_{3}^{2} x_{5} + 12 \, x_{1}^{3} - 30 \, x_{1}^{2} x_{3} + 20 \, x_{1}^{2} x_{5} + 30 \, x_{3}^{2} x_{5} - 30 \, x_{1} x_{5}^{2} - 75 \, x_{3} x_{5}^{2} + 186 \, x_{1}^{2} - 225 \, x_{3}^{2} + 120 \, x_{1} x_{5} - 90 \, x_{5}^{2} + 450 \, x_{1} + 1125 \, x_{3} + 180 \, x_{5}}{2 \, x_{1} x_{5} + 5 \, x_{3} x_{5} + 6 \, x_{5}}=0.$

The question is then to understand if this surface in $\mathbb R^3$ has rational points. If you are curious about it, this is what the numerator looks like:

enter image description here

Final remarks.

This question really interests me, but I feel quite stuck about it. May be my whole approach isn't going to help. Any hints, references or piece of solutions would be much appreciated.

  • 1
    $\begingroup$ @AlexRavsky The lemma was indeed maybe confusing, so I made it a little more clear I hope. $\endgroup$ – E. Joseph Jan 13 '19 at 9:25

If your question boils down to the system,

$$(\mathscr S)\quad \begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \\ -3x_4x_5+3x_3x_6+3x_1=0 \\ 6x_2-5x_3+15x_4=0 \\ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.\end{cases}$$

then, YES, this system has infinitely many rational solutions. If you let,

$$x_1 = x_4 x_5-x_3 x_6$$ $$x_2 = (5/6)(x_3-3x_4)$$ $$x_5 =\frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$

This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,

$$\text{Poly}_1 x_4^2+\text{Poly}_2 x_4+\text{Poly}_3=0$$

You simply solve the linear equation,

$$\text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$

for $x_6$, thus,

$$x_4 =-\frac{\text{Poly}_3}{\text{Poly}_2}$$

with free parameter $x_3$.


Above simultaneous equations shown below has numerical solutions:

$(\mathscr S)\quad \begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \\ -3x_4x_5+3x_3x_6+3x_1=0 \\ 6x_2-5x_3+15x_4=0 \\ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.\end{cases}$







The solution provided by Tito Piezas in this context

actually requires solving a cubic equation instead of

a quadratic equation as mentioned by him.

(Note by Tito Piezas): My solution is a cubic in the variable $x_3$, but only a quadratic in $\color{red}{x_4}$. This is easily verified using Mathematica.

  • $\begingroup$ Thanks a lot, I really appreciate the explicit solution. $\endgroup$ – E. Joseph Dec 2 '18 at 19:22
  • $\begingroup$ You are welcome $\endgroup$ – Sam Dec 2 '18 at 20:40
  • $\begingroup$ In my answer, I was careful to say it is a quadratic in the variable $\color{red}{x_4}$. Your comment applies to the variable $x_3$. $\endgroup$ – Tito Piezas III Dec 30 '18 at 2:54

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