Remark

If $A$ is any commutative ring and $R$ an alternating square matrix of odd order n over $A$, one has det $R = 0$. This results from Cor. 3 when $A$ is a field. A direct proof in the case where $A$ is a field of characteristic $\neq 2$ is as follows: since ${}^tR = -R$, one has det $R = \det {}^tR = (-1)^n \det R$, whence $2 \det {}^tR = 0$. Now, since the determinant of an alternating matrix $(\alpha_{ij})$ is a polynomial with integer coefficients with respect to the $\alpha_{ij}$ such that $i < j$, the principle of extension of algebraic identities (Chapter IV, § 2, No. 5, Scholium) shows that our assertion is then true for an arbitrary commutative ring $A$.