The main goal is to find whether or not a subspace of $\mathbb R^5$ of dimension $3$ intersects a rational subspace of dimension $2$. By rational subspace, we mean a subspace of $\mathbb R^5$ which admits a rational basis (i.e. a basis formed with vectors with rational entries). This is what motivates this question.

The question.

Let $Y_1,Y_2,Y_3$ be three vectors of $\mathbb R^5$ such that all the coordinates of the $Y_i$ are in

$$\mathbb Q(\sqrt 2,\sqrt 3,\sqrt 6),$$

i.e. the coordinates of the $Y_i$ are of the form

$$a+b\sqrt 2+c\sqrt 3+d\sqrt 6,\qquad a,b,c,d\in\mathbb Q.$$

Let $X_1,X_2\in\mathbb Q^5$ be two vectors with rational entries such that $(X_1,X_2)$ is free over $\mathbb R$.

Does there exist such $(Y_1,Y_2,Y_3)$ such that for all such $(X_1,X_2)$, the matrix $M\in\mathrm M_5(\mathbb R)$ with columns $Y_1,Y_2,Y_3,X_1,X_2$, i.e.

$$M:=(Y_1\vert Y_2\vert Y_3\vert X_1\vert X_2),$$ satisfies

$$\det M\ne 0\quad ?$$


I have tried many choices of vectors $Y_1,Y_2,Y_3$, but it always result in a system of four rational equations that I can not solve. The goal would be to show that the system has no rational solution.

Any ideas or references which would be related to this matter would be of great help.

  • $\begingroup$ Can you recall what means $(X_1,X_2)$ is free over $\mathbb R$? It means linearly independent over $\Bbb R$? $\endgroup$ – Alex Ravsky Jan 12 '19 at 17:47
  • 1
    $\begingroup$ @AlexRavsky Yes, by free over $\mathbb R$ I mean *linearly independent over $\mathbb R$. $\endgroup$ – E. Joseph Jan 13 '19 at 9:21
  • $\begingroup$ What's the point of writing $\mathbb Q(\sqrt 2,\sqrt 3,\sqrt 6)$ over $\mathbb Q(\sqrt 2,\sqrt 3)$? The second includes elements which come from products of $\sqrt 2$ and $\sqrt 3$; i.e. $\sqrt 6$. $\endgroup$ – YiFan Jan 22 '19 at 1:25
  • $\begingroup$ Also, I'm probably missing something here, but wouldn't the choice $X_1=X_2$ guarantee that the determinant vanishes? $\endgroup$ – YiFan Jan 22 '19 at 1:28

Okay, so my previous answer was clearly wrong, but now I know why: it is because it is impossible.

The main point is the following: Let $a,b,c,d$ be bilinear skew-symmetric forms from $\mathbb{Q}^5$ to $\mathbb{Q}$. Then, there exists some rational vectors $x,y$ such that $(x,y)$ is free and $a(x,y)=b(x,y)=c(x,y)=d(x,y)=0$.

Proof: A standard reasoning shows that $a$ cannot be nondegenerate, so there exists some nonzero $x$ such that for all $y$ $a(x,y)=0$.

Let $H_b=\{y,b(x,y)=0\}$, $H_c$ and $H_d$ to be the same for $c,d$. These are three hyperplanes of $\mathbb{Q}^5$, so their intersection has dimension at least $2$ so it contains a vector $y$ such that $(x,y)$ is free.

Therefore, $a(x,y)=0$ because $x$ is degenerate for $a$, and $b(x,y)=c(x,y)=d(x,y)=0$ because $y$ is in all hyperplanes.

So how is this related to the OP’s question?

Let $Y_1,Y_2,Y_3$ be vectors in $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{6})^5$. So $\varphi : (u,v) \in (\mathbb{Q}^5)^2 \longmapsto \det (Y_1,Y_2, Y_3, x,y) \in E=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{6})$ is bilinear and skew-symmetric.

$E$ is a $\mathbb{Q}$-vector space of dimension $4$, so let $a,b,c,d$ be the coordinates of $\varphi$ in any basis. They are skew-symmetric $\mathbb{Q}$-bilinear forms, so there exists a free pair $(x,y)$ of rational vectors vanishing $a,b,c,d$, thus $\varphi(x,y)=0$.

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Questions seems easy/managable, but any attempts I made end inconclusive.

+1 I also tried to find a required set $(Y_1,Y_2,Y_3)$, but failed. I have a suspicion that it does not exist.

I obtained the following reduction of the matrix $M$, which can be useful to others in order to answer the question. We can reduce $M$ applying to it the following operations, which do not change its (non)singularity.

If the set $(Y_1,Y_2,Y_3)$ is linearly dependent over $\Bbb R$, then there is nothing to solve because $\det M=0$. So we assume that $(Y_1,Y_2,Y_3)$ is linearly independent over $\Bbb R$. Then swapping the rows of the matrix $M$, we can assure that the $3\times 3$ matrix in the left upper corner of $M$ is non-singular. Now it is easy to check that multiplying any of the first three columns by non-zero elements of a field $\Bbb Q(\sqrt{2},\sqrt{3},\sqrt{6})$ and adding any of the first three columns to an other, we can reduce the matrix $M$ to the following form.

$$\begin{pmatrix} 1 & 0 & 0 & x_{11} & x_{21} \\ 0 & 1 & 0 & x_{12} & x_{22} \\ 0 & 0 & 1 & x_{13} & x_{23} \\ y_{11} & y_{21} & y_{31} & x_{14} & x_{24} \\ y_{12} & y_{22} & y_{32} & x_{15} & x_{25} \end{pmatrix} $$

Next, subtracting from the last two rows the first three, multiplied by elements of $\Bbb Q$ we can achieve that all $y_{ij}$ have a form $b_{ij}\sqrt{2}+c_{ij}\sqrt{3}+d_{ij}\sqrt{6}$.

Next we can reduce the matrix $M$ multiplying any of the last two columns by non-zero elements of a field $\Bbb Q$ and adding any of the last two columns to an other. There are several reduced form of $M$, with the most complicated calculations, probably, when the form is

$$\begin{pmatrix} 1 & 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & x_{13} & x_{23} \\ y_{11} & y_{21} & y_{31} & x_{14} & x_{24} \\ y_{12} & y_{22} & y_{32} & x_{15} & x_{25} \end{pmatrix} $$

The determinant $\Delta$ of this matrix equals

$$x_{14}x_{25}- x_{24}x_{15}+x_{15}y_{21}+x_{24}y_{12}-x_{14}y_{22}-x_{25}y_{11}+y_{11}y_{22}-y_{12}y_{21}+$$ $$ x_{13}x_{24}y_{32}+ x_{15}x_{23}y_{31}-x_{13}x_{25}y_{31}-x_{13}y_{21}y_{32}- x_{14}x_{23}y_{32}+x_{13}y_{22}y_{31}+x_{23}y_{11}y_{32}-x_{23}y_{12}y_{31}.$$

Since $\Bbb Q(\sqrt{2},\sqrt{3},\sqrt{6})$ is a vector space over $\Bbb Q$ of dimension $4$ and we have a freedom to choose $6$ arbitrary rational numbers $x_{ij}$, there is a hope to find such numbers which annul $\Delta$. But for me is not clear how to do this, because an equation $\Delta=0$ is non-linear and I didn’t find a promising start to solve it by excluding one of $x_{ij}$.

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