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Let $R$ be a commutative Artinian ring and $J(R)$ its radical. Assume that the quotient $R/J(R)$ is a GI-ring.

(definitions that i use: I call a ring $S$ a GI-ring if its unit group, $\mathcal{U}(S)$, satisfy a group identity. And by a group identity a mean a (reduced) word $w(x_1, \ldots x_n)$ such that $w(u_1, \ldots, u_n)=1$ for all $u_i \in \mathcal{U}(S)$).

I was wondering if this group identity lift to $R$, so is $R$ necessarily also a GI-ring?

Since $R$ is artinian we know that $J(R)$ is nilpotent and thus units lift. But is not clear to me if some kinds of group identities of $\mathcal{U}(R/J(R))$ could lift to $R$. For example, assume that $R/J(R)$ is finite and thus satisfy the word $x^{n}=1$ with $n$ the cardinality of $R/J(R)$. Do this imply that $R$ is finite (and thus also GI)?

Is there something known? Or a 'easy' answer?

Thanks!

Edit:

as pointed out, $R$ is trivially GI since i supposed commutative. So i have to be more precise in what i had in mind: Let $G$ be a finite group, $R$ commutative Artinian with $1$. Then we know that $RG$ is still artinian and $J(RG)$ is nilpotent. Moreover $RG/J(RG) = R/J(R)G$. Assume $RG/J(RG)$ is GI. Do this identity lift to $RG$?. For example, if $R/J(R)$ is finite and thus $RG/J(RG)$ also, do $RG$ is GI? (even finite?)

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    $\begingroup$ Cross-posted at MSE. $\endgroup$
    – Michael Albanese
    Commented Dec 30, 2013 at 10:42
  • $\begingroup$ Yes indeed. I did that, because i had the feeling that people asking and answering questions on MO and MSE are not the same... $\endgroup$
    – Geoffrey Janssens
    Commented Dec 30, 2013 at 11:13
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    $\begingroup$ You should say that you have done this, otherwise someone on one site could spend a lot of time crafting an answer when you already have a perfectly good answer on the other site. $\endgroup$
    – Michael Albanese
    Commented Dec 30, 2013 at 11:19
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    $\begingroup$ Additionally it is considered bad manners to post it to both without waiting an appropriate amount of time in between. The number of people is no argument at all. There are way more people on MSE than on MO, the overlap is considerably and consists of some very good people. So you do not lose that much by positing only on MSE. The real point is something completely different: Is this a research-level question or not? MO is for research level questions, MSE is for everything else. (Although I admit that this particular question seems to fit the bill) $\endgroup$
    – Johannes Hahn
    Commented Dec 30, 2013 at 11:48
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    $\begingroup$ In any case, $R$ is trivially a GI-ring as it is assumed to be commutative... $\endgroup$
    – Salvatore Siciliano
    Commented Dec 30, 2013 at 17:05

1 Answer 1

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In general, it is not true that a GI-identity satisfied by $RG/J(RG)$ is also satisfied by the whole group algebra RG. For instance, if $R$ is a field of characteristic $p>0$ and $G$ is a finite $p$-group then $RG$ is a basic local algebra. In particular, if $G$ is non-abelian then $\mathcal{U}(RG/J(RG))$ is abelian whereas $\mathcal{U}(RG)$ is not.

However, at least when $R$ is a field, if $RG/J(RG)$ is a GI-ring then so is $RG$. This is clear when $R$ is finite, and it follows from Theorem 4.1 in "Y. Billig, D. Riley, V. Tasic: Nonmatrix varieties and nil-generated algebras whose units satisfy a group identity, J. Algebra 190 (1997), 241–252" when $R$ is infinite.

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