By the above, the algebra F is a free right E-module with basis $(1, j)$. Consequently, F[X] is a free right E[X]-module with basis $(1, j)$. We denote by $u$ the endomorphism of the right E[X]-module F[X] defined by $u(P) =$ $(X-q)P$ for $P\in F[X]$. The characteristic polynomial of $q$ is the determinant of $u$ viewed as an endomorphism of the K[X]-module F[X]. By Proposition 6 of III, §9, No. 4, p. 546, it is equal to N(det $u$), where N denotes the norm from E[X] to K[X]. Let us write $q$ as $x+jy$ with $x, y\in E$. The matrix of $u$ with respect to the basis $(1, j)$ is $\begin{pmatrix} X-x & -\gamma y \\ -y & X-\overline{x} \end{pmatrix}$ ; its determinant is equal to $D = (X-x)(X-\overline{x})-\gamma yy= X^2-T_F(q)X + N_F(q)$ (cf. formula (3)). Since D belongs to K[X], we have $N(D) = D^2$; Proposition 1 follows.

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.

(i) The algebra F is central simple.

(ii) For every nonzero element $x$ of F, there exists an element $y$ in F such that $T_F(xy)\not= 0$.

(iii) We have $(4\alpha +\beta^2)\gamma \not= 0$.

Suppose that these properties hold. Then for every $x$ in F, the reduced characteristic polynomial of $x$ is $X^2-T_F(x)X + N_F(x)$. In particular, $T_F(x)$ is the reduced trace of $x$, and $N_F(x)$ is its reduced norm.

(i)⇒(ii): If the algebra F is central simple, then it follows from Proposition 1 of VIII, p. 361 and the definition of the reduced trace (VIII, p. 340, Definition 2) that $T_F$ is its reduced trace; assertion (ii) follows (VIII, p. 343, Proposition 5).

(ii)⇔(iii): Let $(e_i)_{1 \leq i \leq 4}$ be a basis of F of type $(\alpha, \beta, \gamma)$ (III, §2, No. 5, p. 445). The matrix $(T_F(e_i e_j))$ is equal to

Its determinant is $-\gamma^2 (4\alpha + \beta^2)^2$. The equivalence of properties (ii) and (iii) follows from V, §8, No. 2, p. 49, Lemma 1.

(iii)⇒(i): Suppose $(4\alpha + \beta^2)\gamma \neq 0$. We then have $\gamma \neq 0$, and we have $\beta \neq 0$ if K has characteristic 2. By Proposition 2, the algebra F is central. Let x be an element of the Jacobson radical of F. For every $y \in F$, $xy$ is nilpotent, so $T_F(xy) = 0$ (Remark 3 of VIII, p. 362). Since (ii) is equivalent to (iii), we have $x = 0$. This proves that F is a semisimple K-algebra. Since its center is K, it is simple.

The last assertion follows from Proposition 1 of VIII, p. 361 and the definition of the reduced characteristic polynomial (VIII, p. 340, Definition 1).

Denote the characteristic of K by p. By Proposition 3, if $p \neq 2$, then every quaternion algebra over K of type $(\alpha, 0, \gamma)$ with $\alpha$ and $\gamma$ in $K^*$ is central simple. If $p = 2$, then every quaternion algebra of type $(\alpha, 1, \gamma)$ with $\alpha \in K$ and $\gamma \in K^*$ is central simple. Conversely, we have the following.

a) If $p \neq 2$, then there exist nonzero elements $\alpha$ and $\gamma$ of K such that the algebra A is isomorphic to the quaternion algebra of type $(\alpha, 0, \gamma)$.

b) If $p = 2$, then there exist an element $\alpha$ of K and an element $\gamma$ of $K^*$ such that the algebra A is isomorphic to the quaternion algebra of type $(\alpha, 1, \gamma)$.

By Wedderburn's theorem (VIII, p. 120, Theorem 1), there exist an integer $r \geq 1$ and a field D with center K such that A is isomorphic to $\mathbf{M}_r(D)$. We then have $r^2[D : K] = [A : K] = 4$. If $r = 2$, then A is isomorphic to $\mathbf{M}_2(K)$, and Proposition 4 follows from the example of VIII, p. 362. Otherwise, we have $r = 1$, and A is a field with center K. It then has a maximal commutative subfield E that is a separable extension of K; since A has degree 4 over K, the extension E has degree 2 over K (VIII, p. 265, Corollary 2). It is therefore quadratic (III, §2, No. 3, p. 439). Let $s$ be the conjugation of E (III, §2, No. 3, p. 440). By the Skolem–Noether theorem (VIII, p. 256, Corollary 1), there exists an invertible element $j$ of A such that we have $s(x) =jxj^{-1}$ for every $x$ in E. The field E is separable over K, so we have $s\not=$ Id$_E$, so that $j \notin E$. Since A is a vector space of dimension 4 over K, it is a left vector space of dimension 2 over E, so we have $A = E\oplus Ej$. We have $s^2=$ Id$_E$, so the element $j^2$ of A belongs to the center of A; hence there exists an element $\gamma$ of $K^*$ such that $j^2=\gamma$.

When $p\not= 2$, there exist an element $i$ of E and an element $\alpha \in K^*$ such that $E = K(i)$ and $i^2=\alpha$ (V, §11, No. 9, p. 93, Example 3); in this case, A is isomorphic to the quaternion algebra of type $(\alpha ,0, \gamma )$. When $p= 2$, there exist an element $i$ of E and an element $\alpha$ of K such that $E = K(i)$ and $i^2=i+\alpha$ (V, §11, No. 9, p. 93, Example 2), so that A is isomorphic to the quaternion algebra of type $(\alpha ,1, \gamma )$.

If K is finite, then the algebra A is isomorphic to a matrix algebra $\mathbf{M}_n(K)$ (VIII, p. 357, Corollary 2) and therefore contains elements of degree $n$ over K; the assumption implies $n= 2$ and therefore the result in this case (VIII, p. 362, Example). Suppose that the field K is infinite. Let L be a maximal étale subalgebra of A. By V, §7, No. 4, p. 41, Proposition 7, there exists an element $x$ of A such that the K-algebra L is equal to $K[x]$, hence by assumption has degree $\leqslant 2$. Since we have $[A : K] = [L : K]^2$ (VIII, p. 264, Proposition 4 and p. 262, Proposition 3), we conclude that [A : K] = 4. Corollary 1 then follows from Proposition 4.

Every element $u$ of E satisfies $u^2-T_E(u)u+ N_E(u) = 0$, so the K-algebra E is isomorphic to a quaternion algebra (Corollary 1).

(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.

Denote the center of D by Z, and let L be a maximal commutative subfield of D. We have $[D : Z] = [L : Z]^2$ by VIII, p. 265, Corollary 2; we also have $[L : R]\leqslant 2$ because C is an algebraic closure of R. There are consequently three possible cases:

a) We have R = Z = L, so [D : Z] = 1 and D = R.

b) We have $R\not= Z$ and Z = L, so [D : Z] = 1 and D = L. In this case, D is isomorphic to C.

c) We have R = Z and [L : R] = 2, so [D : R] = 4. By Proposition 4 of VIII, p. 364, the R-algebra D is isomorphic to a quaternion algebra of type $(\alpha ,0, \gamma )$, where $\alpha$ and $\gamma$ are nonzero elements of R. Let $i\in D$- Z be such that $i^2=\alpha$. We have $\alpha \not= 0$. If $\alpha >0$, then there exists an $a\in R$ such that $a^2=\alpha$ (VI, §2, No. 6, p. 26, Theorem 3); we then have $(a-i)(a+i) = 0$, which is absurd because D is a field. So we have $\alpha <0$. The inequality $\gamma <0$ is shown analogously. There then exist elements $a$ and $c$ of $R^*$ such that $\alpha =-a^2$ and $\gamma =-c^2$ (loc. cit.). The algebra D is therefore isomorphic to the quaternion algebra of type $(-1,0,-1)$ (VIII, p. 362, Remark 2), that is, to H.