Nilpotent elements of index $2$ in group algebra $FA_4$

Question

Let $A_4 = K_4 \rtimes C_3$ be alternating group on $4$ symbols and $F$ be finite field containing $4$ elements. By definitions of group algebra and augmentation ideal, there exist a natural map $$\phi : FA_4 \rightarrow F\left(C_3\right)$$ which fixes elements of $C_3$ and sends elements of $K_4$ to identity. Also kernel of $\phi$ is $\Gamma(K_4)$, where $\Gamma(K_4)$ is ideal having elements $\{k-1, k \in K_4\}$ as additive basis over $FC_3$. There is a standard result which says $\Gamma(K_4)$ will be nilpotent ideal in this case. How can I find those elements which are nilpotent of index $2$ in $\Gamma(K_4)$ ?

Answer 1

Ok, this isn't too difficult, but nobody else has answered...

For notation, it is easier to write the group algebra $FK_4$ as $F[s,t]/((s-1)^2,(t-1)^2)$, where $s$ is just $(1\; 2)(3\; 4)$ and $t$ is $(1\; 3)(2\; 4)$. Then any element of $FK_4$ can be written uniquely as $\alpha+\beta s+\gamma t+\delta st$. The trace of such an element is $\alpha+\beta+\gamma+\delta$. Let $\sigma=(1\; 2\; 3)\in A_4$. As you say, the kernel $\Gamma(K_4)$ is the set of elements of $FA_4$ of the form: $$u=u_1 + u_\sigma \sigma + u_{\sigma^2} \sigma^2$$ where $u_1, u_\sigma,u_{\sigma^2}\in FK_4$ all have trace zero. Note that $\sigma\cdot (\alpha+\beta s+\gamma t+\delta st) = (\alpha + \delta s + \beta t + \gamma st)\cdot \sigma$. Denote $\sigma^{-1} u \sigma$ by $u^{\sigma}$.

Now $$u^2= (u_1^2 + u_\sigma u_{\sigma^2}^{\sigma^2} + u_{\sigma^2} u_\sigma^\sigma) + (u_1 u_\sigma + u_\sigma u_1^\sigma + u_{\sigma^2} u_{\sigma^2}^{\sigma^2})\sigma + (u_1u_{\sigma^2} + u_{\sigma^2}u_1^{\sigma^2} + u_\sigma u_\sigma^\sigma)\sigma^2.$$ Write $$u_1 = \alpha +\beta s+ \gamma t+\delta st,\; u_\sigma = A+Bs+Ct+Dst, \; u_{\sigma^2}=a+bs+ct+dst$$ where $\alpha+\beta+\gamma+\delta=A+B+C+D=a+b+c+d=0$. Then it's not too difficult to check that $u^2=0$ if and only if both of the following hold: $$a^2+b^2+c^2+ab+ac+bc+(A+B)(\alpha+\gamma)+(B+C)(\beta+\gamma)=0$$ $$A^2+B^2+C^2+AB+AC+BC+(b+c)(\alpha+\gamma)+(a+b)(\alpha+\beta)=0.$$ Magma tells me that the ideal generated by these two polynomials is prime (and remains so over e.g. a field with 32 elements), so the subset you want is an irreducible subvariety. (This doesn't however tell you much about the set of $K$-rational points, which is what you asked for.)