Wedderburn's Little Theorem says that every finite Division Ring is commutative. What is about an infinite Division Ring with prime characteristics? Is this also a Field?


Let $F$ be any field with a non identity automorphism $\sigma$ and finite characteristic. (You could, for example, use the Frobenius endomorphism of a nonperfect field.)

The twisted polynomial ring $F[x;\sigma]$ is the set of polynomials written with coefficients on the left of powers of $x$, and the multiplication dictated by $xa=\sigma(a)x$.

Since $\sigma$ isn't the identity map, this yields a noncommutative Noetherian domain. This being the case, it has a "division ring of quotients" which must share the same finite characteristic with $F$ and $F[x;\sigma]$. Clearly it is also infinite.

  • $\begingroup$ Nice +1. Is there a quick way to see that the division ring of quotients exists for a Noetherian domain? I suppose it's a consequence of Goldie's theorem, but that seems like it must be overkill. $\endgroup$ – Stephen Mar 29 '17 at 13:40
  • $\begingroup$ @Stephen The path you mention is possible, but a perhaps more direct connection is that if a domain $D$ is right noetherian, then $xD\cap yD\neq 0$ for any nonzero $x,y$, which is what you need to prove it is right Ore. I don't remember how to do this off the top of my head, but if you had an $x_0D$ that wasn't essential, you can extend with $x_1D$ with trivial intersection with $x_0D$, and keep bootstrapping up this way. I think if you stop at finitely many factors and get an essential submodule, you have a contradiction. Therefore there's an infinite ascending chain fo the form $\oplus x_iD$ $\endgroup$ – rschwieb Mar 29 '17 at 13:52
  • $\begingroup$ @Stephen The result I know is true is that a domain is right Ore iff it has uniform dimension $1$ iff it has finite uniform dimension. The domains of infinite uniform dimension do not permit the ordinary construction of the right ring of quotients (although they may in fact embed in a division ring.) $\endgroup$ – rschwieb Mar 29 '17 at 13:54

Consider a generalization of the quaternions $\mathbb{H}$. It can be constructed in a similar way to the quaternions, and is called a quaternion algebra. You can form an algebra with basis $1,i,j, ij$ where $i^2=a,j^2=b, ij=-ji$, and $a,b\in\mathbb{F}$. If you choose an appropriate infinite non-algebraically closed field in positive characteristic not equal to 2, and the appropriate $a$ and $b$, your quaternion algebra will be a division ring, but will not be a field (since it is not commutative). You can read more about quaternion algebras on Wikipedia.

Edit: Thanks to Kevin Carlson, we have a link to the construction of an explicit example by Keith Conrad on Problem Set 4 in the following link: http://www.math.uconn.edu/~kconrad/ross2004/ . As Kevin mentions in the comments below, constructing explicit examples will likely take a bit of work.

  • $\begingroup$ You have to be a bit more careful: over some fields (algebraic closure is sufficent,) all choices of $a$ and $b$ will give a ring of matrices, not a division ring. I, for one, would be interested in more explicit examples. (The Wikipedia article claims every local field gives an example, but without justification.) $\endgroup$ – Kevin Carlson Jan 27 '15 at 5:37
  • $\begingroup$ You are right, thank you. I will change my wording to reflect this, and try to think of a more concrete example tomorrow. $\endgroup$ – Sergio Da Silva Jan 27 '15 at 5:40
  • $\begingroup$ I don't think your edit is correct. If your field is, for instance, separably but not algebraically closed, all the quaternion algebras will still be matrices. I think the justification for the non-obviousness of explicit examples is that it's probably not that easy. Keith Conrad develops a pretty general characteristic-2 example in problem set 4 here, but it takes a bit of work: math.uconn.edu/~kconrad/ross2004 $\endgroup$ – Kevin Carlson Jan 27 '15 at 5:45
  • $\begingroup$ Please give a particular example $\endgroup$ – Anirban Jan 27 '15 at 6:00
  • 1
    $\begingroup$ Sure, it's natural to imagine-but it already doesn't work over $\mathbb{C}$, so this isn't a positive characteristic issue. $\endgroup$ – Kevin Carlson Jan 27 '15 at 6:06

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