Prove all elements of $A$ is algebraic over $C$, if all elements of $A$ are algebraic over $B$ and $B$ are algebraic over $C$

Question

Let there be 3 fields $A$, $B$ and $C$.

If all elements of $A$ are algebraic over $B$ and all elements of $B$ are algebraic over $C$, prove that this implies that all elements of $A$ is algebraic over $C$.

Answer 1

Let it be that element $\alpha\in A$ has minimal polynomial $\sum_{i=0}^{n}\beta_{i}X^{i}$ with $\beta_{i}\in B$.

Then $C\subset B_{0}:=C\left(\beta_{0},\dots,\beta_{n}\right)$ and $B_{0}\subset B_{0}\left(\alpha\right)$ are finite extensions.

Consequently $C\subset B_{0}\left(\alpha\right)$ is a finite (hence algebraic) extension.

This tells us that $\alpha$ is algebraic over $C$.

---

edit:

Used are the following lemma's:

• If $K\subset L$ and $L\subset M$ are finite extensions then $K\subset M$ is a finite extension.

• Every finite extension is algebraic. (A proof of that can be found here).

Answer 2

If you have an element $\alpha \in A$ which is algebraic over $B$, what can you say about the form of $\alpha$? Similarly, if you have an element $\beta \in B$ which is algebraic over $C$, what can you say about the form of $\beta$? Can you somehow write $\alpha \in A$ as an algebraic element over $C$?