Proposition 5

The following properties are equivalent:

(i) The quaternion algebra F is a field.

(ii) There is no nonzero element $q\in F$ such that $T_F(q) = N_F(q) = 0$.

(iii) Every element of F of square zero is zero.

(iv) There does not exist any nonzero vector $(x, y, z, t)$ in $K^4$ such that

(v) There does not exist any nonzero vector $(x, y, z)$ in $K^3$ such that

(vi) The quadratic algebra E is a field, and $\gamma$ is not the norm of an element of E.

An element $q$ of F is invertible if and only if $N_F(q)$ is not 0. The equivalence of (i) and (iv) therefore follows from formula (31) of III, §2, No. 5, p. 445; it is clear that (i) implies (iii) and that (iv) implies (v).

The equivalence of (ii) and (iii) follows from Remark 3 of VIII, p. 362.

Suppose that F is not a field. If $\gamma (4\alpha +\beta^2)\not= 0$, then the algebra F is central simple of degree 4 over K; it is isomorphic to the algebra $\mathbf{M}_2(K)$ (VIII, p. 120, Theorem 1) and therefore contains a nonzero element of square zero. If $\gamma = 0$, then we have $j^2= 0$. If $4\alpha +\beta^2= 0$, then we have $(2i-\beta )^2= 0$ and $2i-\beta \not= 0$ if K has characteristic different from 2. Finally, if K has characteristic 2 and $\beta$ is zero, then we have $T_F(q) = 0$ for every $q\in F$ (III, §2, No. 5, p. 445, formula (31)). Since F is not a field, there exists a nonzero element $q$ of F such that $N_F(q) = 0$; we have $q^2= 0$. This proves the implication (iii)$\Rightarrow$(i).

Let $q=x+yi$ be an element of E. We have $N_{E/K}(q) =x^2+\beta xy-\alpha y^2$. Suppose that property (v) holds. We have $N_{E/K}(q)-\gamma \not= 0$ and $N_{E/K}(q)\not= 0$ if $q\not= 0$, and therefore (vi).

Finally, suppose that property (vi) holds. Let $q$ be a nonzero element of F; we write it as $u+vj$ with $u$ and $v$ in E. If $v$ is zero, then $q$ is invertible. If $v$ is not zero, then we have $N_F(q) = N_F(v)N_F(v^{-1}u+j) =$ $N_{E/K}(v)(N_{E/K}(v^{-1}u)-\gamma )$ by III, §2, No. 5, p. 443, formula (24). Since $\gamma$ is not a norm, $N_F(q)$ is not zero, and $q$ is invertible.