When a group ring is a local ring Hi there, I'm stuck with my undergraduate thesis on the following proposition:
If $k$ is a field of characteristic $p > 0$ and $G$ is a finite $p$-group, then the group ring $kG$ is local.
In particular where $k = \mathbb{Z}_p$ and $G$ is cyclic of order $p$.
Thanks for any help.
 A: Dear Marco, it is well-known that if $F$ is a field of characteristic $p>0$ and $G$ is a group then the augmentation ideal of the group algebra $FG$ is nilpotent if and only if $G$ is a finite $p$-group (see the book "D. Passman: The algebraic structure of group rings",  Lemma 1.6 of Chapter 3, page 70). In that case the Jacobson radical $J$ of $FG$ clearly coincides with the augmentation ideal which has codimension $1$ in $FG$, and you are done. You can also find the description of the Jacobson radical of the group algebra of a finite $p$-group over a field of characteristic $p>0$ in the book "R. Pierce: Associative algebras" (Corollary in Section 4.7).
A: Since you say this is an undergraduate thesis, I will take a few steps back. The augmentation ideal $I$ of the group algebra $kG$ is $\{ \sum_{g \in G} \alpha_{g}g: \sum_{g \in G} \alpha_g = 0\}.$ It is easy to see (in several ways) that $I$ is a two-sided ideal of $kG.$ One way is to note that it is the annihilator of the trivial module, which is $1$-dimensional, with a $k$-basis $\{v \}$ such that $vg = v $ for all $g \in G.$ No element of $I$ is a unit, as $I$ is a proper ideal. It remains to prove that every element of $kG \backslash I$ is invertible. Again, there are several ways to do this: one is to note that if $M$ is a simple (sometimes called irreducible) $kG$-module, then $G$ fixes a non-zero vector of $M$, so $M$ must be the trivial module. I leave this to you to do, or to research. Then it follows that $I$ annihilates every simple $kG$-module. Then you can use the fact that every finite dimensional $kG$-module has a composition series to see that $I^{n}$ annihilates the regular module $kG$ for some integer $n.$ It follows in particular that every element of $I$ is nilpotent.
Every element of $kG \backslash I$ is of the form $\lambda 1_G + j$ for some $j \in I$ and 
nonzero $\lambda \in k.$ Then it is relatively easy to see that $1_{G} + \frac{j}{\lambda}$ is invertible, using the nilpotency of $j.$  
A: Maschke's Theorem says that if $p \not | |G|$ then $kG$ is semisimple. You've probably learned this fact. What you may not know is that even if $p$ does divide $|G|$ as in your situation, you still know $kG$ is quasi-Frobenius. You should look into a homological algebra textbook, e.g. "Lectures on Modules and Rings" by T.Y. Lam, to learn more about such rings.  There are many theorems and lots of counter-examples which may help you. As an example, one property of quasi-Frobenius rings $R$ is that the class of injective $R$-modules is the same as the class of projective $R$-modules.
A: Use the following theorem by Brauer, that you can find in many books about representation theory of finite groups:
The number of simple modules is equal to the number of conjugacy classes whose elements have order coprime to the characteristic of the field $p$, the so-called $p$-regular classes.
