Determinant function $D$ when $\text{char}(K)=2$

Question

Let $K$ be a field and $V=K^n$. The determinant function $D:V^n\rightarrow K$ is characterized by the three property:

1. $D$ is linear in each row
2. Switching two rows changes the sign.
3. $D(e_1,\dots,e_n)=1$

From 2,

$$D(a_1,\dots, b,\dots, b\dots, a_n)=-D(a_1,\dots, b,\dots, b\dots, a_n)$$

So, $2D(a_1,\dots, b,\dots, b\dots, a_n)=0$.

I think $D(a_1,\dots, b,\dots, b\dots, a_n)=0$ but what if $\text{char}(K)=2$?

Answer 1

The correct general definition is the one with $D(v_1,\cdots,v_n)=0$ for all $v_1,\cdots, v_n$ such that $v_i=v_j$ for some $i\ne j$. This implies the switcheroo one: in fact, \begin{align}&D(a_1,a_2,\cdots,a_n)+D(a_2,a_1,\cdots, a_n)=\\=&D(a_1,a_2,\cdots,a_n)+D(a_1,a_1,\cdots,a_n)+D(a_2,a_2,\cdots,a_n)+D(a_2,a_1,\cdots,a_n)=\\=&D(a_1,a_2+a_1,\cdots,a_n)+D(a_2,a_2+a_1,\cdots,a_n)=D(a_1+a_2,a_1+a_2,\cdots,a_n)\end{align}

Hence $D(a_1,a_2,\cdots,a_n)=-D(a_2,a_1,\cdots, a_n)$. The other way around just doesn't work in characteristic $2$.

Answer 2

Usually one defines the determinant via $$ 2. \;\; D(a_1,\ldots, b,\ldots, b,\ldots a_n)=0, $$ that is the determinant is zero whenever two rows are the same. This definition works perfectly fine if $\text{char}K=2$ and is also equivalent to your definition if $\text{char}K\neq 2$.