Remark 6

Soient $K$ a commutative field of characteristic 2, $A$ a separable quadratic extension of $K$; one has $A = K(\theta)$, where $\theta$ is root of an irreducible polynomial $X^2 + X + \beta$ of $K[X]$ and the $K$-automorphism $J$ of $A$, distinct from the identity, is such that $\theta^J = \theta + 1$ (Chapter V, § 11, Exercise 8). Now prove that if $E$ and $G$ are vector spaces over $A$, $\Phi$ a sesquilinear mapping (for $J$) from $E \times E$ into $G$, one has, putting $Q(x) = \Phi(x, x)$,