Proposition. $f(\alpha)=0$ if and only if $m_{\alpha, F} \mid f$ [1406]

    Let $K / F$ be a field extension and $\alpha \in K$. Suppose that $f(x) \in F[x]$. Then $f(\alpha) = 0$ if and only if $m_{\alpha, F}(x) \mid f(x)$ in $F[x]$.

    In this case, $f(x)$ is monic and irreducible if and only if $f(x) = m_{\alpha, F}(x)$.

    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 $f(\alpha) = 0$.