# Cross product in arbitrary field

1. May 12, 2015

### 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 $\mathbb{R}^3$

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

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

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

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

$$[ 2] \left(\begin{array}{c} [ 1] \\ [ 4] \\ [ 2] \\ \end{array}\right) = \left(\begin{array}{c} [ 2] \\ [ 3] \\ [ 4] \\ \end{array}\right)$$

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 $\mathbb{C}$.

2. May 12, 2015

### 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.