Let $k$ be a field and let $Q$ be a quaternion algebra over $k$.
It is well known that, if $\mathrm{char}\,k\neq 2$, one can define $Q$ as the $k$-algebra of dimension $4$ generated by elements $x,y$ and relations $x^2 =a$, $y^2 =b$ and $xy = - xy$ where $a,b\in k$ and $b\neq 0$. Call this $Q=(a,b)$.
One can also define a "quaternion" algebra if $\mathrm{char}\,k = 2$, taking symbols $x,y$ and imposing relation $x^2 +x =a , y^2 =b, xy = yx + y$. Call this $Q=[a,b)$.
It is clear from the definition that, in case $\mathrm{char}\,k\neq 2$, the isomorphism class of $(a,b)$ depends only on the classes of $a,b$ in $k^*/(k^*)^2$ or, more precisely, letting $a = u^2 \alpha,b = v^2 \beta$ we get that $(a,b) = (u^2\alpha ,v^2 \beta ) \simeq (\alpha, \beta) $ and the isomorphism is given by sending $x\mapsto u^2 x$ and $y\mapsto v^2 y$.
The question is: do we have a similar description for $[a,b)$, namely with $\mathrm{char}\,k=2$? It seemed natural to me to look at the classes of $a,b$ in $k/\wp(k)$ but I don't get the right relations.
Any help would be appreciated, especially welcome is a good reference for dyadic quaternion algebras.
Edit.
So far I have been able to see that, given $k$ of $\mathrm{char}\,k=2$ and $Q=[a,b)$ a quaternion algebra, then we can replace $a$ with any of its representatives in the class $[a]\in k/\wp{k}$, namely we can replace $a\mapsto \alpha + (u^2 + u)$ and we obtain an isomorphic quaternion algebra.
However, I fail to see what we can do with $b$ (if anything can be said). Looking at the book suggested in the comments, it looks like in general a quaternion algebra is not equivalent to give a pair of elements $a,b$ and some relations, unless the characteristic is not 2.
So in the end I would say that $[a,b)$ depends on the class of $a$ in $k/\wp{k}$ as well as on the choice of $b\in k$. Does that make sense?
