Is every antisymmetric k-linear transformation alternate?

Question

Definition 1. Let $V$ and $W$ be linear spaces over field $\mathbb{F}$. A function $f:V^k\to W$ is $k$-linear if it is linear over each entry separately.

Definition 2: A $k$-linear function $f$ is said to be alternate if $$ f(v_1,\dots,v_i,\dots,v_j,\dots,v_k) = 0 $$ if $v_i = v_j$ for some $1\le i<j\le k$.

Definition 3: A $k$-linear function $f$ is said to be antisymmetric if $$ f(v_1,\dots,v_i,\dots,v_j,\dots,v_k) = -f(v_1,\dots,v_j,\dots,v_i,\dots,v_k) $$ for every $1\le i<j\le k$.

I have proved that if a $f$ satisfies (2) then it satisfies (3). I want to prove or disprove the reverse. I tried to find one myself and also searched the web and found no counterexamples so I wonder if it is true, but I tried to show it and I'm stuck. Any ideas?

Answer 1

They are equivalent if the field does not have characteristic $2$.

Over a field of characteristic $2$, antisymmetry is equivalent to symmetry: alternation is a strictly stronger property. As a trivial example, let $V=k$ where $k$ has characteristic $2$. Then there is no nonzero alternating form on $V\times V$. Yet $(a,b)\mapsto ab$ is (anti)symmetric on $V\times V$.

Answer 2

Here's my attempt, can someone check my arguments please?

Assume $f$ satisfing (3). Let $1\le i<j\le k$ for some $i,j$. Then $$ f(v_1,\dots,v_i,\dots,v_i,\dots,v_k) = -f(v_1,\dots,v_i,\dots,v_i,\dots,v_k) $$ then $$ f(v_1,\dots,v_i,\dots,v_i,\dots,v_k)+f(v_1,\dots,v_i,\dots,v_i,\dots,v_k) = 0 $$ and if $char(\mathbb{F})\ne 2$ then this implies that $f(v_1,\dots,v_i,\dots,v_i,\dots,v_k)=0$ so $f$ is alternate. If $char(\mathbb{F})=2$ then we may have $f(v_1,\dots,v_i,\dots,v_i,\dots,v_k)+f(v_1,\dots,v_i,\dots,v_i,\dots,v_k) = 0$ with $f(v_1,\dots,v_i,\dots,v_i,\dots,v_k)\ne 0$.

Is this correct?