Proof. sketch [1406A]

$\implies$: Just use the same argument in the proof of Proposition 1403. We have $f(x) \in \ker \varphi = \left( m_{\alpha, F}(x) \right)$. So $m_{\alpha, F}(x) \mid f(x)$ in $F[x]$.

$\impliedby$: Similarly, $m_{\alpha, F}(x) \mid f(x) \implies f(x) \in \left( m_{\alpha, F}(x) \right) = \ker\varphi$ and therefore $f(\alpha) = 0$.