Cross product in arbitrary field

[jostpuur]

Let \mathbb{F} be an arbitrary field, and let a,b\in\mathbb{F}^3 be vectors of the three dimensional vector space. How do you prove that if a\times b=0, then a and b are linearly dependent?

Consider the following attempt at a counter example: In \mathbb{R}^3

<br /> \left(\begin{array}{c}<br /> 1 \\ 4 \\ 2 \\<br /> \end{array}\right)<br /> \times\left(\begin{array}{c}<br /> 2 \\ 3 \\ 4 \\<br /> \end{array}\right)<br /> = \left(\begin{array}{c}<br /> 10 \\ 0 \\ -5 \\<br /> \end{array}\right)<br />

holds. Since 5 is a prime number, \mathbb{Z}_5 is a field. In (\mathbb{Z}_5)^3

<br /> \left(\begin{array}{c}<br /> [ 1] \\ [ 4] \\ [ 2] \\<br /> \end{array}\right)<br /> \times \left(\begin{array}{c}<br /> [2] \\ [ 3] \\ [ 4] \\<br /> \end{array}\right)<br /> = \left(\begin{array}{c}<br /> [ 0] \\ [ 0] \\ [ 0] \\<br /> \end{array}\right)<br />

This might look like a counter example to the claim. One might consider the possibility that perhaps the claim is true for example when \mathbb{F} is a subfield of \mathbb{C}, but not in general?

A closer look reveals that the counter example attempt does not work, because

<br /> [ 2] \left(\begin{array}{c}<br /> [ 1] \\ [ 4] \\ [ 2] \\<br /> \end{array}\right)<br /> = \left(\begin{array}{c}<br /> [ 2] \\ [ 3] \\ [ 4] \\<br /> \end{array}\right)<br />

holds in (\mathbb{Z}_5)^3. Finding a counter example is difficult, and it seems that the claim is true after all. I only know how to prove the claim using determinants when \mathbb{F} is a subfield of \mathbb{C}.

 

[micromass]

Assume that $a$ and $b$ are linearly independent in the field $\mathbb{F}^3$, then you can expand them to a basis $(c,a,b)$. Because of linear algebra, this means that the matrix formed by $a$, $b$ and $c$ is invertible and thus has nonzero determinant. But this determinant is $c\cdot (a\times b)$. Since this is nonzero, it implies that $a\times b$ must be nonzero.