Let \(F\subseteq L\subseteq\Omega\) be fields and let \(\alpha\in\Omega\) be algebraic over \(F\). Show that the minimal polynomial of \(\alpha\) over \(L\) divides the minimal polynomial of \(\alpha\) over \(F\) in \(L[x]\).
Solution
Assume \(F\subseteq L\subseteq\Omega\) are fields and \(\alpha\in\Omega\) is algebraic over \(F\). Then \(\alpha\) is also algebraic over \(L\), since its minimal polynomial over \(F\) lies in \(L[x]\) and annihilates \(\alpha\).
Let \[ m_F=m_{\alpha,F},\qquad m_L=m_{\alpha,L}. \] Divide \(m_F\) by \(m_L\) in \(L[x]\): \[ m_F=q\,m_L+r,\qquad \deg r<\deg m_L. \] Evaluating at \(\alpha\) gives \[ 0=m_F(\alpha)=r(\alpha), \] because \(m_L(\alpha)=0\). By minimality of \(m_L\), the only polynomial in \(L[x]\) of degree smaller than \(\deg m_L\) that annihilates \(\alpha\) is the zero polynomial. Hence \(r=0\).
Therefore \[ \boxed{m_{\alpha,L}\mid m_{\alpha,F}\quad\text{in }L[x].} \]