Let \(L/F\) be a field extension, and let \(\alpha\) be algebraic over \(F\). Let \(m_F(x) = \operatorname{irr}(\alpha, F) \in F[x]\) and \(m_L(x) = \operatorname{irr}(\alpha, L) \in L[x]\) be the minimal polynomials of \(\alpha\) over \(F\) and over \(L\), respectively. Prove that \(m_L(x)\) divides \(m_F(x)\) in the polynomial ring \(L[x]\).
Solution
Because \(F\subseteq L\), we may regard \[ m_F(x)\in F[x] \] as a polynomial in \(L[x]\). Divide it by the minimal polynomial \(m_L(x)\): \[ m_F(x)=q(x)m_L(x)+r(x), \qquad \deg r<\deg m_L. \] Evaluating at \(\alpha\) gives \[ 0=m_F(\alpha)=q(\alpha)m_L(\alpha)+r(\alpha)=r(\alpha). \] If \(r\ne0\), this would be a nonzero polynomial in \(L[x]\) of degree smaller than \(m_L\) that annihilates \(\alpha\), contradicting minimality of \(m_L\). Hence \(r=0\), so \[ \boxed{m_L(x)\mid m_F(x)\text{ in }L[x]}. \]