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I'm working on a question saying $A \subseteq B \subseteq C$ fields, and $a \in C$ algebraic over $A$, either prove $[A(a):A] ≥ [B(a):B]$ or give a counter example.

I think it's true and used the following equation: $[B(a):A(a)][A(a):A] = [B(a):B][B:A]$, and I think we can show that $[B(a):A(a)] < [B:A]$ by using the classic definition regarding basis of field extensions, but I'm not sure how to do so since I don't even know if it's a finite extension.

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    $\begingroup$ If $f$ is the minimal polynomial of $a$ over $A$, then this is not necessary irreducible over $B$. (I hope you know that $\deg f=[A(a):A]$.) $\endgroup$
    – user26857
    Dec 6, 2015 at 8:46
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    $\begingroup$ Also, try to use the latex here, it's much more readable that way $\endgroup$
    – Zelos Malum
    Dec 6, 2015 at 9:04

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