Example of a Non-Commutative Division Ring With Finite Characteristics 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?
 A: 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.
A: 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.
