Proposition 2

a) Suppose that the field K has characteristic different from 2. If $\gamma$ or $4\alpha +\beta^2$ is nonzero, then the center of F is equal to K; otherwise, it has dimension 2 and is generated by 1 and $ij-ji$.

b) Suppose that the field K has characteristic 2. If $\beta \not= 0$, then the center of F is equal to K; if $\beta = 0$, then the algebra F is commutative.

By formula (30) of III, §2, No. 5, p. 444, we have

An element $q=x+yi+zj+tk$ of F is central if and only if it commutes with $i$ and $j$, that is, we have

(5) $2z+\beta t=-\beta z+ 2\alpha t= 0$ and $2\gamma t=\beta \gamma t= 2y=\beta y= 0$.

First suppose that the characteristic of K is different from 2. If $\gamma$ is nonzero, then the equalities in (5) imply $y=t= 0$ and then $z= 0$; consequently, we have $q\in K$. If $\gamma = 0$ and $4\alpha +\beta^2\not= 0$, then they imply $y=z=t= 0$, so we have $q\in K$. If $\gamma = 4\alpha +\beta^2= 0$, then the system (5) reduces to $y= 2z+\beta t= 0$, so we have $q=x+t/2(ij-ji)$, which completes the proof of a).

Now suppose that the field K has characteristic 2. The system (5) can then be written as $\beta t=\beta z=\beta y= 0$; assertion b) follows.