Is antisymmetry equivalent to alternativity?

Question

Reading my class's lecture notes, the definition of the determinant is given as follows:

A function $det: (F^n)^n \rightarrow F$ such that it satisfies the following:

1. Multilinearity: $det(\dots, v_j+v_j',\dots)=det(\dots, v_j,\dots) + det(\dots, v_j',\dots)$ and $det(\dots, c*v_j,\dots)=c*det(\dots, v_j,\dots)$ for $c \in F$
2. Antisymmetry: $det(\dots, v_j, \dots, v_i, \dots)=-det(\dots, v_i, \dots, v_j, \dots)$
3. Normalization: $det(e_1, \dots, e_n)=1$

Then there is a claim that alternatively, we can take Alternativity instead of Antisymmetry because they are equivalent.

Alternativity: $det(\dots, v_j, v_j, \dots)=0$

I can see that Antisymmetry implies Alternativity: $det(\dots, v_j, v_j, \dots)=-det(\dots, v_j, v_j, \dots)$, implying that $det(\dots, v_j, v_j, \dots)=0$.

And going from Alternativity to Antisymmetry, my guess was to use Multilinearity: $det(\dots, v_i, v_j, \dots)+det(\dots, v_j, v_i, \dots)=det(\dots, v_i+v_j, v_j+v_i, \dots) = 0$.

My question is, is there a way to go from Alternativity to Antisymmetry without relying on Multilinearity?

Answer 1

The properties of alternativity and antisymetry are defined only for multilinear applications (or at least, this is what we are dealing with right now).

Be careful that, in general (and for multilinear applications), the alternativity is a stronger property than antisymetry, and these two properties are equivalent if the base field has characteristic different than 2. As you can see with your proof, if $f$ is a multilinear antisymmetric application, you end up with an equality that looks like : \begin{equation} f(x,x,x_3,..., x_n) = -f(x,x,x_n,...,x_n) \end{equation} So what you actually get is \begin{equation} 2f(x,x,x_3,...,x_n) = 0 \end{equation}

If you want to deduce that $f(x,x,x_3,..,x_n) = 0$, you need to know that the base field has characteristic $\neq 2$.

Answer 2

$f$ defined by $f(x,y) = 0$ if $x = y$ and $f(x,y) = 1$ if $x \neq y$ respects alternativity. Not Antisymmetry. I think there is no way to not use multilinearity.