Purely transcendental extensions do not affect the minimal polynomial.

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I've been studying for qualiying exams and recently came across this problem.

Let $L/M/K$ be a field extensions with $[L:M]<\infty$. Let $A$ be a subfield consisting of all elements of $L$ that are algebraic over $K$. Suppose $M\cap A=K$.

(a) Show that if $\alpha\in A\backslash K$ and $m_{\alpha,A}\in M[x]$ is the minimal polynomial of $\alpha$ over $M$. Then $m_{\alpha,A}\in K[x]$.

(b) Show that if $char(K)=0$ and $K\subseteq B\subseteq A$ is an intermediary extension with $[B:K]<\infty$, then $[B:K]\leq [L:M]$.

(c) Show that if $char(K)=0$ then $[A:K]\leq [L:M]$.

Attempts

For $(a)$ I've deduced that it's enough to show that $m_{\alpha,K}(x)$ is irreducible in $M[x]$ where $m_{\alpha,K}(x)$ is the irreducible polynomial of $\alpha$ over $K$. I think there is something to be said for the fact that $(x-\alpha)\notin M[x]$ and so any factorisation $m_{\alpha,K}(x)=g(x)h(x)$ with $g(x),h(x)\in M[x]\backslash(M\cup K[x])$ will have $\deg(h(x))>1$ and $\deg(g(x))>1$. I suspect I need to incorporate the hypothesis $[L:M]<\infty$.

For (b) and (c) I think the characteristic zero assumption has something to do with separability. I know finite extensions in characteristic zero are separable but I can't see how this will help with obtaining bounds on the degree.