Question in proof of lemma before theorem 7 of section 5.6 of hoffman/kunze linear algebra I have some question in proof of a lemma in section 'Multilinear Function'.

Lemma before Theorem 7
The proof says that $\pi_r L$ is alternating form because $\pi_r L(\alpha_{\tau 1}, \dots, \alpha_{\tau r}) = (\text{sgn} \tau) \pi_r L(\alpha_1, \dots, \alpha_r)$ for any permutation $\tau$. But that is not the definition of alternating form.

Definition of alternating form
I tried to prove that by using fact that if  $\pi_r L(\alpha_{\tau 1}, \dots, \alpha_{\tau r}) = (\text{sgn} \tau) \pi_r L(\alpha_1, \dots, \alpha_r)$ for any permutation $\tau$, then $\pi_r L (\dots, \alpha, \dots, \alpha, \dots) = - \pi_r L (\dots, \alpha, \dots, \alpha, \dots)$. But if $K = Z_2$, then value of $L(\dots, \alpha, \dots, \alpha, \dots)$ may be $1$ because $1 = -1$ in $Z_2$.
Why do that $\pi_r(L\alpha_{\tau 1}, \dots, \alpha_{\tau_r}) = (\text{sgn} \tau) \pi_r L(\alpha_1, \dots, \alpha_r)$ for every permutation $\tau$ implies that $\pi_r L$ is alternating?
i'm sorry for my poor english.
 A: Indeed, the proof presented for the Lemma is incorrect.  The criterion for a multilinear form to be alternating which it relies on only works if $2$ is not a zero divisor in the base ring $K$ (in which case the argument which you attempted to fill in the gap does work).
To give a correct proof of the Lemma for arbitrary $K$, you can instead directly investigate the value of $\pi_rL(\alpha_1,\dots,\alpha_r)$ if $\alpha_i=\alpha_j$.  To sketch the proof, let $\tau$ be the permutation that swaps $i$ and $j$.  Then we can partition the symmetric group $S_r$ into cosets of the subgroup $\{1,\tau\}$.  This partitions all the permutations of $\{1,\dots,r\}$ into pairs of the form $\{\sigma,\sigma\tau\}$.  Now note that since $\tau$ leaves $(\alpha_1,\dots,\alpha_r)$ unchanged, $L_{\sigma}$ and $L_{\sigma\tau}$ have the same value on $(\alpha_1,\dots,\alpha_r)$.  But since $\sigma$ and $\sigma\tau$ have opposite signs, this means that their terms in the sum that defines $\pi_rL(\alpha_1,\dots,\alpha_r)$ will cancel out, and so $\pi_rL(\alpha_1,\dots,\alpha_r)$ will be $0$.
