I apologize in advance if this is easy, but I've tried Googling, and had no luck. 


I'm currently working on a proof, and I realized in the course of writing that this proof will break if out there in the world there exists a division algebra $D$ with the following properties:


1. The characteristic of $D$ (as a normal ring) is $p$.
2. The degree of $D$ (the square root of its dimension over its center) is divisible by $p$.


I would really appreciate if anyone knows an example, or a proof that there are none. I could easily add hypotheses to rule out this case, but that would make things a bit messier, so I don't want to do that unless it's really necessary. Unfortunately, I know zippo about division algebras in characteristic p, other than that every finite one is actually just a field. 


**EDIT**: Thanks for the references. Doing a little reading based on Mikhail's suggestion, I found a result which is good enough for my purposes: no such division algebra exists whose center is a perfect field, which was proven in 1934 by the remarkably named [Abraham Adrian Albert](http://www.ams.org/journals/tran/1934-036-02/S0002-9947-1934-1501750-X/home.html). Those wanting more details on his life can read this [detailed and impressive obituary](http://www.ams.org/journals/bull/1974-80-06/S0002-9904-1974-13622-0/S0002-9904-1974-13622-0.pdf) by Jacobson. The best detail is that his Ukrainian father decided to ditch his last name (which is now unknown!) in Victorian England, and adopt the prince consort's name instead.



Can a division algebra have degree divisible by its characteristic?
There are counterexamples for each $p$. The easiest maybe is the following: Let $F$ be the field of order $p^p$, and $\sigma$ be an automorphism of $F$ of order $p$. Let $D=F((t))$ be the set of Laurent series of the form $\sum a\_it^i$ with the usual addition. Define multiplication by $ta=a^\sigma t$. Then $D$ is a division algebra with center $Z=\mathbb F\_p((t^p))$, so $[F:Z]=p^2$.


