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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?

Jose Brox
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  • 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? – rschwieb Oct 19 '18 at 20:13
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    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 $n$? – Jyrki Lahtonen Oct 20 '18 at 05:08
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    For an example see this old answer of mine. A 9-dimensional division algebra with center $\Bbb{Q}$ inside $M_3(\Bbb{Q}(\cos(2\pi/7))$. – Jyrki Lahtonen Oct 20 '18 at 05:10
  • 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. – Jyrki Lahtonen Oct 21 '18 at 14:42
  • @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) – Jose Brox Oct 22 '18 at 09:53
  • @JyrkiLahtonen Sorry for the delay, had to leave for the weekend... I know that the division algebras subject is rich, but I'm no expert (I'm still educating myself on it). That's what I wanted to see in this problem what could be said that is as independent of the ground field as possible, informally, what instances come from the structure of matrices. The universal object here is the ring of generic matrices of order $n$ over the free (commutative) field, so you are right that the characteristic is important, sorry. We can restrict to char $0$ and work over... – Jose Brox Oct 22 '18 at 10:04
  • @JyrkiLahtonen ...$\mathbb{Q}(X)$ with $X$ uncountable and $n$ fixed. Perhaps I have erred the construction, you see that we want the class where M$_n(F)$ is a model for each $F$ (char $0$) (if this makes any sense); it would be then in this sense in which $F:=\mathbb{R}$ matters. I actually didn't put much thought before asking, I just wrote the ideas that I saw that seemed to restrict the problem. I need to think more about your interesting examples. If you have any ideas about the right way of looking at this, I'd love to read them (my purpose is to learn more) – Jose Brox Oct 22 '18 at 10:34
  • The case $F=\Bbb{R}$ probably matters, but I know nothing about models, so cannot comment. The one point I want to make is that the reasons why we have so few division algebras over $\Bbb{R}$ are mostly topological. Most notably the fact that odd degree polynomials have zeros in $\Bbb{R}$. So $\Bbb{R}$ is anything but "universal" among characteritsic zero fields in this sense, or? $\Bbb{C}$ or some other algebraic closure might be, but over $\Bbb{C}$ (or any other algebraically closed field) we don't have any non-trivial finite dimensional division algebras with the base field as its center. – Jyrki Lahtonen Oct 25 '18 at 05:07

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