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Suppose $R$ is a noncommutative ring. What is the sufficient and necessary condition for $GL_n(R)$ embedding into $GL_m(F)$ for some field $F$?

In particular, what if $R$ is the group ring $\mathbb{Z}D_8$ or $\mathbb{Z}_2[D_8,t^{\pm1}]$, where $D_8$ is the dihedral group of order 8.

Thanks!.

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    $\begingroup$ Then please edit the question to reflect what you are actually asking. $\endgroup$
    – Tobias Kildetoft
    Commented Mar 15, 2016 at 9:45
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    $\begingroup$ Well if $R$ was the group ring of nonlinear group then the answer would certainly be no. $\endgroup$
    – Derek Holt
    Commented Mar 15, 2016 at 9:54
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    $\begingroup$ @ToddLeason You're right, I don't know what I was thinking. However, the following should hold: the answer is positive if $R$ embeds in $M_m(F)$ for some $m$ and a field $F$. $\endgroup$
    – Emil Jeřábek
    Commented Mar 15, 2016 at 11:30
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    $\begingroup$ @EmilJeřábek: I guess your comment would imply a positive solution for $\mathbb{Z}[D_8]$ by embedding in $\mathbb{C}[D_8]$ and using Wedderburn's theorem. Actually, this would imply a positive solution for group rings of finite groups. $\endgroup$
    – Matthias Wendt
    Commented Mar 15, 2016 at 14:06
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    $\begingroup$ In particular, $\mathbb Z[C_4]$ should embed as a ring in $M_4(\mathbb Q(i))$ so that the generator of $C_4$ maps to the diagonal matrix with $1,i,-1,-i$ on the diagonal, and $GL_2(\mathbb Z[C_4])$ then embeds in $GL_8(\mathbb Q(i))$ by replacing each entry with the corresponding $4\times4$ block. Where does this go wrong? $\endgroup$
    – Emil Jeřábek
    Commented Mar 15, 2016 at 19:56

2 Answers 2

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This question is studied and partially answered in https://arxiv.org/abs/0904.3153. The answer depends not only on $R$ but also on $n$.

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I'm really just expanding on other people's comments, so I've made this answer community wiki.

If $R$ is a subring of a finite dimensional algebra over a field $K$ ($d$-dimensional, say), then $R$ embeds in $M_d(K)$ via the regular representation. So $M_n(R)$ embeds in $M_{nd}(K)$, and so $\text{GL}_n(R)$ embeds in $\text{GL}_{nd}(K)$.

This isn't a necessary and sufficient condition (as the much deeper results referred to in Mark Sapir's answer show), but is enough to deal with both examples in the question, since $\mathbb{Z}[D_8]$ is a subring of the finite-dimensional $\mathbb{Q}$-algebra $\mathbb{Q}[D_8]$, and $\mathbb{Z}_2[D_8,t^{\pm1}]$ is a subring of the finite dimensional $\mathbb{Z}_2(t)$-algebra $\mathbb{Z}_2(t)[D_8]$ (I think the OP uses $\mathbb{Z}_2$ to mean the integers mod $2$, but if it means the $2$-adic integers, then it's a subring of the finite dimensional $\mathbb{Q}_2(t)$-algebra $\mathbb{Q}_2(t)[D_8]$).

In particular, even Maschke's theorem isn't needed.

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