Exercise 4.5.20

Answers

Let $δ$ be the $2$ -linear function in Exercise 4.5.15 with $a = b = c = d = 1$ . Thus we have that

δ (\begin{matrix} a & b \\ c & d \end{matrix}) = ac + ad + bc + bd = δ (\begin{matrix} c & d \\ a & b \end{matrix}) = - δ (\begin{matrix} c & d \\ a & b \end{matrix}) .

The final equality holds since $𝔽$ has characteristic two. But now we have

δ (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}) = 1 \neq 0 .

jephianlin

2011-06-27 00:00