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Let $F$ be a field and M$_n(F)$ the ring of $n\times n$ matrices. By a domain we mean a not necessarily commutative ring without zero divisors. We consider subdomains $R$ of the ring M$_n(F)$. Examples are the diagonal embbeding of $\mathbb{Z}$ inside M$_2(\mathbb{Q})$ and the integer quaternions inside a matrix representation of the quaternions. Some natural questions arise:

  1. Knowing $F$, can we classify all the subdomains $R$, for all $n$?
  2. Which are the subdomains which are universal for all $F$, in the sense that they are subdomains of generic matrices?

Regarding 1, it would be interesting to see some references for specific kinds of fields.

Regarding 2, I have the following ideas: $R$ being a domain ring implies $F\cdot R$ being a domain algebra, so we can restrict to algebras. Since M$_n(F)$ is a PI-ring, the subdomains $R$ are PI, so they are right Ore domains and have right algebras of quotients $Q(R)$. On the other hand, M$_n(F)$ already has inverses for all the elements of $R$. Suppose we could prove that $Q(R)$ is inside M$_n(F)$ (or M$_m(F)$ with $m>n$) without losing the information on the "relative position" of $R$, and consider the case $F:=\mathbb{R}$. Since $Q(R)$ is a division $\mathbb{R}$-algebra, it must be $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$. Since the problem is universal, these are the only possible cases for all fields. So the problem is reduced to linear representations of $F$, $F(i)$ and $F(i,j,k)$, the subdomains of those division algebras, and the possible positions of $R$ relative to $Q(R)$ inside the matrix algebras (not trivial!).

Is really $Q(R)$ inside some matrix ring? What more can be said? Is there a better approach?

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  • $\begingroup$ R being a domain ring implies F⋅R being a domain algebra What do you mean? Would you say $\mathbb Q\cdot \mathbb Z$ is a "domain algebra" in your first example? $\endgroup$ – rschwieb Oct 19 '18 at 20:13
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    $\begingroup$ I don't understand the reasoning leading up to quaternions. I guess it's because I don't understand your definition of universal. Are you assuming that the characteristic is zero? Are you aware of the fact that there exists infinitely many pairwise non-isomorphic division algebras with center $\Bbb{Q}$ and dimension $n^2$ for any natural number $n$? $\endgroup$ – Jyrki Lahtonen Oct 20 '18 at 5:08
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    $\begingroup$ For an example see this old answer of mine. A 9-dimensional division algebra with center $\Bbb{Q}$ inside $M_3(\Bbb{Q}(\cos(2\pi/7))$. $\endgroup$ – Jyrki Lahtonen Oct 20 '18 at 5:10
  • $\begingroup$ I want to apologize. I'm sure you are aware of the richness of Brauer groups of number fields (among other things). But I'm a bit at loss as to what universality means here. $\endgroup$ – Jyrki Lahtonen Oct 21 '18 at 14:42
  • $\begingroup$ @rschwieb I mean a domain subalgebra of the matrix algebra, so yes, we would have $\mathbb{Q}=\mathbb{Q}\mathbb{Z}$ diagonally in the first example. Since the problem seems "hard", I was using a broad-brush approach in order to be able to say something... we would have domain subrings as "orders" of domain subalgebras, so we would classify the subalgebras as first step, and then descend to the classification of subrings inside each subalgebra (I forgot to add this step at the end) $\endgroup$ – Jose Brox Oct 22 '18 at 9:53

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