Compute the Jacobson radical of $R=\mathbb{F}_3[S_3]$ and find all simple $R$-modules.

Question

Compute the Jacobson radical of $R=\mathbb{F}_3[S_3]$ and find all simple $R$-modules.

I need to find a surjective map and compute its kernel. But I am not sure which map. Any hints?

Answer 1

The starting idea is that when the characteristic of the field and the order of the group are not coprime, you are going to have a nonzero Jacobson radical. Because the ring is Artinian, the Jacobson radical must be a nilpotent ideal. If we succeed in finding a nilpotent ideal $I$ such that $R/I$ has zero radical, then we can be assured that $J(R)=I$. (You will need to understand and fully justify this strategy for yourself.)

Like the document you referenced, I like to use $\sigma$ for the generator of order $3$ and $\tau$ for the generator of order $2$.

By the Freshman's dream theorem, $(1-\sigma)^3=1-\sigma^3=0$, so $1-\sigma$ is nilpotent. We'd also like to prove that $(1-\sigma)r$ is nilpotent for any other $r\in R$, but since $1-\sigma$ isn't central, this isn't immediately obvious.

But $(1-\sigma)$ does satisfy a weaker property that does the trick, namely that $(1-\sigma)R=R(1-\sigma)$. The rule $(1-\sigma)\tau=\tau(1+\sigma)(1-\sigma)$ allows you to commute $(1-\sigma)$ from the left of the product $(1-\sigma)r$ to the right to get $r'(1-\sigma)$ for some $r'\in R$. Likewise $(1-\sigma)r'=r''(1-\sigma)$. Then $(1-\sigma)R=R(1-\sigma)$ is a two sided ideal and
$$ ((1-\sigma)r)^3=(s-\sigma)r(s-\sigma)r(s-\sigma)r\\= (s-\sigma)rr'(s-\sigma)(s-\sigma)r\\=r'r''(s-\sigma)(s-\sigma)(s-\sigma)r=0 $$

This establishes that $(1-\sigma)R$ is a nilpotent ideal, and hence is contained in the Jacobson radical.

Now let's focus on the kernel of the map $F_3[S_3]\to F_3[S_3/\langle\sigma\rangle]$. Of course, everything in $\langle\sigma\rangle$ is mapped to the identity of the quotient, so $1-\sigma$ is in the kernel.

If $\sum r_g g \mapsto 0$, we know that $\sum_{g\in\langle\sigma\rangle}r_g=0$ and $\sum_{g\in G\setminus\langle\sigma\rangle}r_g=0$

Given that, a little reverse engineering (or just plain long division) shows that $$(\sigma-1)((r_{\sigma^2}+r_\sigma)1+r_{\sigma^2}\sigma+(r_{\sigma\tau}+r_{\sigma^2\tau})\tau+r_{\sigma^2\tau}\sigma\tau=\sum r_gg$$

demonstrates that $\sum r_gg$ is in $(1-\sigma)R$. So the kernel is equal to this ideal.

By Maschke's theorem, the second group ring is semisimple; moreover, it is commutative. The only two possibilities are that it is a $F_9$ or that it is $F_3\times F_3$. It can't be $F_9$ though: $(1-\tau)(1+\tau)=0$ shows there are zero divisors in $F_3[S_3/\langle\sigma\rangle]$.

Thus $R/J(R)$ has two distinct maximal ideals corresponding to two nonisomorphic simple modules. As an exercise you can attempt to describe the maximal right ideals of the original ring that correspond to these.