Proposition. minimal polynomial for $\alpha$ over $F$ [1405]

    Let $K / F$ be a field extension and $\alpha \in K$ be an element algebraic over $F$. There exists a unique monic irreducible polynomial $m_{\alpha, F}(x) \in F[x]$ such that $m_{\alpha, F}(\alpha) = 0$.

    We call the polynomial $m_{\alpha, F}(x)$ the minimal polynomial for $\alpha$ over $F$, and call $\deg m_{\alpha, F}(x)$ the degree of $\alpha$ over $F$.

    Proof. sketch [1405A]

      Existence: let $p(x)$ be the very polynomial in the proof of Proposition 1403. Let $a$ be the leading coefficient of $p(x)$. Then $m_{\alpha, F}(x) = a^{-1} p(x)$. It is obviously monic, and the irreducibility is inherited from $p(x)$.

      Uniqueness: use the same notation as in the proof of Proposition 1403. Then $\ker \varphi = \left( p(x) \right) = \left( m_{\alpha, F}(x) \right)$. Suppose there is another monic irreducible polynomial $m_{\alpha, F}'(x) \in F[x]$ such that $m_{\alpha, F}'(\alpha) = 0$. Then $m_{\alpha, F}'(x) \in \left( m_{\alpha, F}(x) \right)$, hence $m_{\alpha, F}'(x) = m_{\alpha, F}(x) q(x)$ for some $q(x) \in F[x]$. Since $m_{\alpha, F}'(x), m_{\alpha, F}(x)$ are monic, $q(x)$ should also be monic. If $\deg q > 0$, then it contradicts irreducibility of $m_{\alpha, F}'(x)$. So $q(x) = 1$, which implies $m_{\alpha, F}' = m_{\alpha, F}$.