Proposition 4

Suppose that $A$ is a field, and let $\Phi$ be a bilinear form (resp. sesquilinear form for $J$) on $E \times F$; for $E/F^0$ to be of finite dimension, it is necessary and sufficient that $F/E^0$ be of finite dimension, and these dimensions are then equal.

Indeed, let $\Phi_1$ be the non-degenerate form associated with $\Phi$, on $(E/F^0) \times (F/E^0)$ (No. 3). Suppose that $E/F^0$ is of finite dimension $n$; since the linear mapping $d_{\Phi_1}$ from $F/E^0$ (resp. $(F/E^0)^J$) into $(E/F^0)^*$ is injective, $F/E^0$ is of finite dimension $n' \leq n$; considering $s_{\Phi_1}$, we see similarly that $n \leq n'$.