5

Show that the Group Ring $F_p[G]$ where $F_p$ is finite field of order $p$ and $G$ is a $p$-Group (not necessarily abelian) has a unique maximal ideal, i.e. it is a local ring.

Attempt:

Consider the augmentation map, which is a ring homomorphism from $F_p[G]$ to $F_p$, taking $\sum a_g g$ to $\sum a_g$. This map is surjective and the kernel is the augmentation ideal. Because the image is a field, then the kernel must be maximal [right?], so the augmentation ideal contains the Jacobson Radical, which is the intersection of all maximal ideals. Also, this ideal is not the whole ring, so it does not contain any units.

This is where I'm stuck. Here are some paths that I'm trying to go down (they may be equivalent)

If we can show that the Jacobson radical is equal to a maximal ideal, which the augmentation ideal is, then the maximal ideal is unique, because if there were another maximal ideal then Jacobson has to be inside of it. Thus the augmentation ideal which is maximal would live in another maximal ideal, which is a contradiction. However, I don't know how to show that the augmentation ideal is equal to the Jacobson radical. I've seen something along the lines of trying to prove the augmentation ideal consists of nilpotent elements, but I didn't understand it very well: Prove that the augmentation ideal in the group ring $\mathbb{Z}/p\mathbb{Z}G$ is a nilpotent ideal ($p$ is a prime, $G$ is a $p$-group) Also if this were true, how do we conclude? This group ring isn't commutative, so nilpotent elements aren't necessarily in the Jacobson radical, right? Are somehow using

Also, I read: https://mathoverflow.net/questions/73856/when-a-group-ring-is-a-local-ring and I understand that if we can show that all elements outside of the augmentation ideal are invertible, i.e. units, then we are done, because one of the defining properties of being a local ring is the non-units form an ideal. However, I couldn't follow this either.

I think what's throwing me off most is how to deal with non-commutativity, and I don't have the best of grasp of which my definitions/properties stop working.

Community
  • 1
user324283
  • 133

1 Answers1

5

This group ring isn't commutative, so nilpotent elements aren't necessarily in the Jacobson radical, right?

While it does not necessarily contain all nilpotent elements, the Jacobson radical does contain each nil ideal. So, the nilpotent augmentation ideal is contained in the Jacobson radical.

Conversely, the augmentation ideal of the group ring over a field is maximal right ideal, so the Jacobson radical is contained in it. Thus the Jacobson radical is a maximal right ideal, and the ring has exactly one maximal ideal.

rschwieb
  • 160,592
  • You want a "is contained in" instead of "contains" in the first paragraph – Tobias Kildetoft Mar 19 '16 at 20:05
  • Thanks! Could you explain "True, but that would only be a problem if the ideal was nil and not nilpotent. The Jacobson radical does contain each nilpotent ideal. So, the augmentation ideal contains the Jacobson radical." A little bit more. Or give me a reference or link. – user324283 Mar 19 '16 at 21:07
  • @TobiasKildetoft Yes thanks, that was intended – rschwieb Mar 19 '16 at 21:38
  • @user324283 You can for example use Corollary 1.4 from the book by Assem-Simson-Skowronski (which states that the Jacobson radical contains any nilpotent ideal). – Tobias Kildetoft Mar 19 '16 at 21:43
  • @user324283 if $x$ is in a nil ideal, then $xR$ is a nilpotent ideal. This means $xr$ is nilpotent for all $r\in R$, and $1-xr$ is a unit for all $r$. By the quasi regularity characterization of the Jacobson radical, $x$ is in the radical. – rschwieb Mar 19 '16 at 21:50
  • @user324283 sorry, I think the argument actually applies to nil ideals now, and I changed the wording of what you asked about. You are right that it does not necessarily contain all nilpotent elements, but an entire ideal that us nil will be contained there. – rschwieb Mar 19 '16 at 21:54
  • Ok, from what I've gathered, if we have that the augmentation ideal is a nilpotent ideal (which is true, right?) then consider an $x$ in the ideal, for all $r$ in the group ring, we have $rx$ is in the ideal and that $(rx)^n=0$ since the ideal is nilpotent. Then we have $1=1-(rx)^n=(1-rx)(1+...+(rx)^{n-1})$ and so $(1-rx)$ is invertible. This proves $x$ is in the Jacobson radical, and so the augmentation ideal is contained in it--assuming it's nilpotent. – user324283 Mar 19 '16 at 23:30
  • @user324283 Yes... that's right. – rschwieb Mar 20 '16 at 02:50
  • @rschwieb Since the elements not in the augmentation ideal map to nonzero elements, and since any nonzero element of a field could not be nilpotent, so could we deduce that the nilpotent augmentation ideal is just the set of nilpotent elements of the group ring? – karparvar Mar 28 '18 at 21:49
  • @rschwieb please help me to solve my this problem https://math.stackexchange.com/questions/3025927/exponent-of-the-non-abelian-group-1jfg-for-finite-non-abelian-p-group-g ... thanks – neelkanth Dec 04 '18 at 18:27