Remark

Under the assumptions of the corollary, let $q$ be an element of Q such that we have $\lambda (p\otimes q) = 0$ for every $p\in P$. We then have $\tau (q) = 0$, that is, $q= 0$. Likewise, if $p$ is an element of P such that we have $\lambda (p\otimes q) = 0$ for every $q\in Q$, then $p= 0$.