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It is known that, for any group $G$, there exists a group $H$ containing $G$ such that the group ring $F[H]$ for some field $F$ is primitive, see Formanek, Edward; Snider, Robert L., Primitive group rings, Proc. Am. Math. Soc. 36 (1972), 357-360 (1973). ZBL0223.16007.

The group and faithful, simple module is constructed as follow:

$G_1 = G$, $M_1 = F[G_1]$

$G_2 = \textrm{Aut}_F(M_1)$, $M_2 = F[G_2] \oplus M_1$

and so on with $H = \bigcup G_i$ and $M = \bigcup M_i$.

We can consider $M$ both a left- and a right-module. Is it simple and faithful on both sides?

If so, then consider $\textrm{End}_{F[H]} M$ (where $M$ is a left-module) as it is left-simple we know that this is a division ring by Schur's lemma. However, there is also a ring homomorphsim from $F[G]$ into $\textrm{End}_{F[H]} M$ by the action of right scalar multiplication. And as $M$ is right-faithful, it has no non-zero annihilators and thus the homomorphsim is injective.

It seems, therefore, we have embedded $F[G]$ into a division ring?

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  • $\begingroup$ Cool paper, by the way. I wonder how far one can go replacing $F$ with something more general. $\endgroup$
    – rschwieb
    Sep 29, 2022 at 15:50

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It seems, therefore, we have embedded 𝐹[𝐺] into a division ring?

Well, you are probably asking because that is an absurdity, right? $F[G]$ can be chosen to have zero divisors.

It seems the proof claims only that $_{F[H]}M$ and $M_{F[H]}$ are simple and faithful. But the problem lies here:

However, there is also a ring homomorphsim from 𝐹[𝐺] into $End_{𝐹[𝐻]}𝑀$ by the action of right scalar multiplication.

This would be true if $M$ were an $F[H], F[H]$ bimodule (or even just an $F[H], F[G]$ bimodule) because then we're guaranteed that $(am)b=a(mb)$ , which says that the action of $F[G]$ on the right is left $F[H]$ linear. The bimodule condition is not asserted and in light of the absurdity we could derive from it, is apparently false.

So I think that is the issue: you simply have an abelian group that is a simple module on both sides using two natural actions by the same ring: but not a fully fledged bimodule that would permit the embedding into a division ring.

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  • $\begingroup$ Yes I agree it is absurd. Does the fact its a bimodule not follow from it being a left and right module? $\endgroup$
    – user491484
    Sep 29, 2022 at 14:59
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    $\begingroup$ Being a bimodule is a stronger condition than being both a left and a right module. The condition $(am)b=a(mb)$ is not guaranteed. As an example, $FG$ is a left $FG$-module via left multiplication. But $G$ acts on the right of $FG$ by $a\odot g = g^{-1}a$ (which extends to a right $FG$-module structure). But $h(a\odot g) = hg^{-1}a$ and $(ha)\odot g =g^{-1}ha$ and so unless $G$ is abelian, these actions don't give a bimodule. $\endgroup$ Sep 29, 2022 at 15:09
  • $\begingroup$ Ahh I was imagining a different module structure where $G$ acts on $FG$ just by multiplication on the right $\endgroup$
    – user491484
    Sep 29, 2022 at 15:15
  • $\begingroup$ Right. Being a left and right module at the same time doesn't "link" the two actions with the condition you need to make your embedding work. $\endgroup$
    – rschwieb
    Sep 29, 2022 at 15:23
  • $\begingroup$ If it is just group multiplication then they are linked however by associativity of the underlying group $\endgroup$
    – user491484
    Sep 29, 2022 at 15:47

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