$A$ algebraic over $B$, $B$ algebraic over $C$. $A$ algebraic over $C$?

Question

Let $A/B/C$ be field extensions, with $A$ algebraic over $B$, $B$ algebraic over $C$. Must $A$ be algebraic over $C$?

I think the answer is yes, but I don't know how to prove it.

Answer 1

I assume you know that a field extension $E/F$ is finite if and only if it is finitely generated and algebraic, if and only if it is finitely generated by algebraic elements. I will use that for your exercise.

Let $\alpha\in A$. We want to show that it is algebraic over $C$. We know that it is algebraic over $B$, hence there is a polynomial $p(x)=\sum_{i=0}^n b_ix^i\in B[x]$ such that $p(\alpha)=0$. Since the extension $B/C$ is algebraic we know that the elements $b_0,...,b_n$ are algebraic over $C$. Hence $[C(b_0,...,b_n):C]<\infty$. Now note that $p(x)$ is actually a polynomial in $C(b_0,...,b_n)[x]$ and hence $\alpha$ is an algebraic element over $C(b_0,...,b_n)$. Hence $[C(b_0,...,b_n,\alpha):C(b_0,...,b_n)]<\infty$. Since extension degree is multiplicative we conclude that $[C(b_0,...,b_n,\alpha):C]<\infty$, hence $C(b_0,...,b_n,\alpha)/C$ is an algebraic extension. Since $\alpha$ is an element in it we know it is algebraic over $C$.