Jacobson radical of group ring is $\{\sum a_gg =0:\sum a_g =0\}$

Question

Let $G$ be a finite group and $F(G)$ the group ring over the finite field $F$.
Prove $Jac(F(G))=\{\sum a_gg =0:\sum a_g =0\}.$

I know that $Jac(F)=0$. Is there a possibility to show $Jac(F(G))=Jac(F)G$? Or how can I start to prove this? Any help is gretaly appreciated!

Answer 1

Think of the augmentation homomorphism $\epsilon :\mathbb{F}(G)\to \mathbb{F}, \epsilon (\sum a_{g}g)=\sum a_{g}g$, it is a surjective homomorphism, so its kernel (i.e the augmentation ideal) is a maximal ideal, this proves the inclusion $Jac(\mathbb{F}(G)) \subset Ker(\epsilon)$ . Remark that the augmentation ideal is generated by $g-e$,where $e$ is the identity element of $G$. I think you meant to write $$Jac(\mathbb{F}(G))=\left\{\sum a_{g}g \,\middle| \sum a_{g}=0\right\}$$ Remark: I proved only the inclusion $$Jac(\mathbb{F}(G)) \subset Ker(\epsilon)$$

Answer 2

The proposition is false. (Or, at least I cannot see what mistake you made that altered it from a true proposition.)

Take the field $F_3$ of $3$ elements and the cyclic group $C_2=\{e, c\}$. Then $F_3[C_2]\cong F_3\times F_3$ is a semisimple ring, clearly having Jacobson radical $\{0\}$.

But the augmentation ideal is nonzero: for example, $c-e$ is in the augmentation ideal (since that sum is really $(1)c+(-1)e$).

It is certainly true though that $J(F[G])$ is a subset of the augmentation ideal for the reason belkacem abderrahmane gave: the augmentation ideal has to be maximal when the coefficients of the group ring come from a field (or a simple ring for that matter.)

Is there a possibility to show $(())=()?$

No, there isn't. The next section contains a counterexample.

I know that $()=0$

Well, this is not necessarily true either. If we used $F_2$ instead and looked at $F_2[C_2]$, then in fact the Jacobson radical is equal to the augmentation ideal (and it isn't zero.)

The closest thing I can think of to a true version of your proposition is that if $G$ is a $p$-group and $F$ has characteristic $p$, then the Jacobson radical and augmentation ideal coincide.