- #1

jostpuur

- 2,116

- 19

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

[tex]

\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)

[/tex]

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

[tex]

\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)

[/tex]

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

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

[tex]

[ 2] \left(\begin{array}{c}

[ 1] \\ [ 4] \\ [ 2] \\

\end{array}\right)

= \left(\begin{array}{c}

[ 2] \\ [ 3] \\ [ 4] \\

\end{array}\right)

[/tex]

holds in [itex](\mathbb{Z}_5)^3[/itex]. 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 [itex]\mathbb{F}[/itex] is a subfield of [itex]\mathbb{C}[/itex].