# Quaternion algebra of characteristic 2?

I've been reading up on quaternion algebras recently and noticed the vast majority of theorems are contingent on setting the characteristic $p \neq 2$. In particular, this is true for the central theorem that all quaternions over the reals are either isomorphic to the 2x2 real matrix algebra or form division rings.

I'm wondering why this is so: what's so special about the characteristic two that it causes quaternions to "break down" algebraically, so to speak?

Dually, are there any concrete examples that can help illustrate why characteristic two is such an anomaly in so far as quaternion algebras are concerned?

Thanks!

• Well in characteristic 2, $ij=-ij=ji$, so the standard quaternion algebra $\Bbb F_2[Q_8]$ is actually commutative. Mar 21, 2015 at 1:09
• @pval the first half of the comment is just fine, but in the last half, the group algebra over the quaternion group is quite different from what I think the OP is referring to Mar 21, 2015 at 3:01
• @rschwieb I'm somewhat rusty on this stuff. I meant the usual quaternions and I guess labeling them as a group algebra gives a different meaning to $-1$ Mar 21, 2015 at 3:25
• @pval yes, right. At any rate, the group algebra is 8 dimensional while the quaternion algebras are 4 dimensional. And group algebras almost always have a nontrivial ideal :) Mar 21, 2015 at 3:28
• If you describe quaternion algebras by concrete formulas, then you're bound to avoid characteristic $2$ just as quadratic field extensions are different in characteristic $2$: outside characteristic $2$ we can always complete the square and write a quadratic (Galois) field extension by adjoining a pure square root. This isn't true in characteristic $2$. For quaternion algebras from concrete formulas, you often avoid characteristic $2$ since there $-1 = 1$. A good definition over all fields is: central simple algebra of dimension $4$. Theorem: if it's not $M_2(F)$ then it's a division ring!
– KCd
Jul 16, 2022 at 23:57

GROVE attributes these to Huppert in lecture notes 1968/69. This is on page 120. With a field $F$ of characteristic 2, choose any two elements $a,b \in F,$ create a four dimensional vector space with basis $1,i,j,k,$ and create an associative algebra with this multiplication table:
$$\begin{array}{c|ccc|} & i & j & k \\ \hline i & a & k & aj \\ j & 1+k & b & bi + j \\ k & i + aj & bi & ab + k \\ \hline \end{array}$$