Corollary 4

The hypotheses and notation being those of Cor. 3, suppose moreover that $E_1$ and $E_2$ are orthogonal, and transport to $C(Q_1) \otimes C(Q_2)$, by means of the bijection $p$, the algebra structure of $C(Q)$. If $a_i$ and $b_i$ are even or odd elements of $C(Q_i)$ ($i = 1,\ 2$), then $a\ (a_1 \otimes a_2)(b_1 \otimes b_2) = \varepsilon(a_1b_1) \otimes (a_2b_2)$, with $\varepsilon = 1$ unless $a_2$ and $b_1$ are odd, in which case $\varepsilon = -1$.

It suffices indeed to prove that $p_2(a_2)p_1(b_1) = \varepsilon p_1(b_1)p_2(a_2)$, and for this purpose we may suppose that $p_2(a_2)$ (resp. $p_1(b_1)$) is a product $x_1 \ldots x_h$ (resp. $y_1 \ldots y_k$) of elements of $\rho_Q(E_2)$ (resp. $\rho_Q(E_1)$). Since $E_1$ and $E_2$ are orthogonal, we have

whence

The conclusions of cor. 3 and 4 remain true if we omit the hypothesis that $E_1$ and $E_2$ are free modules (cf. exerc. 1).