# When a group ring is a local ring [closed]

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.

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## closed as off-topic by Benjamin Steinberg, Karl Schwede, Steven Sam, abx, Stefan KohlAug 5 '14 at 8:56

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Hint: let $t$ generate your cyclic group $G$ of order $p$, and consider $(t-1)^p \in kG$. – user91132 Aug 27 '11 at 18:00
For the general problem, can you see why the augmentation ideal is nilpotent? – Greg Marks Aug 27 '11 at 18:29
Marco, if you read the FAQ you'll see that this site is not quite the best fit for your question. The FAQ suggests other places, like the sister (brother?) site math.stackexchange.com, where questions like yours will feel much more at home. – Mariano Suárez-Alvarez Aug 27 '11 at 19:21
I think this question does belong to MO. – Qfwfq Aug 27 '11 at 21:10
@unknowngoogle: this is a standard piece of knowledge, proved in pretty much any textbook which deals with the subject. It is quite not research math in any sense. – Mariano Suárez-Alvarez Aug 28 '11 at 4:05

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 $j.$

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I may be missing something...why are you going to the trouble to prove that the elements in the quotient (or is that the set difference?) are units? You mention that there is a unique module of $kG$ up to isomorphism...that seems to be equivalent to $kG$ being local by a simple argument. Do you agree? – Eric Auld Oct 25 '15 at 23:51
You mean unique simple module. As I said, I wrote my answer bearing in mind that the OP was writing an undergraduate thesis, so I wasn't sure how much background he had, and wanted to assume as little as possible. – Geoff Robinson Oct 25 '15 at 23:54

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).

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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.

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There's a way to go from quasi-Frobenius to local, though :) – Mariano Suárez-Alvarez Aug 27 '11 at 19:19
Indeed. I figured this question would be closed, and wasn't really sure the best way to give some useful information without giving it all away. I thought giving the reference where that answer is found--but not the answer itself--would be a nice middle ground for an undergrad who's trying to learn about the research process. – David White Aug 27 '11 at 20:30

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.

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Hi Julian: you need $k$ algebraically closed to say it that way, I think (doesn't really matter for the $p$-group case). – Geoff Robinson Aug 27 '11 at 21:07
(or, at least, a splitting field). – Geoff Robinson Aug 27 '11 at 21:08
what an overkill! – user91132 Aug 28 '11 at 7:36
@Konstantin: it isn't such a difficult theorem by modern standards. – Geoff Robinson Aug 28 '11 at 12:02
@Geoff: of course not, but I think I prefer the more direct proof: induct on $|G|$ and pick $z \in Z(G)$ of order $p$; then $\mathfrak{m}^n \subseteq (z-1)kG$ for some $n$ by induction so $\mathfrak{m}^{np} \subseteq (z-1)^pkG = 0$. – user91132 Aug 28 '11 at 17:25