If $K$ be an algebraic extension of $E$ and $E$ be an algebraic extension of $F$ then $K$ is an algebraic extension of $F$. Let $K$ be an algebraic extension of $E$ and $E$ be an algebraic extension of $F$ then $K$ is an algebraic extension of $F$
Proof-
Let $a \in K$,then since $a$ is algebraic over $E$, so there exist an irreducible polynomial in $E[x]$ say $p(x)$ such that $p(a) = 0$.
Let
$p(x) = b_{0} + b_{1}x +... +b_{n}x^n$
where $b_{0},b_{1},...,b_{n} \in E$
We need to show that $a$ is in some finite extension of $F$
How do we construct such a finite extension.
In the remaining proof it adds element by element to the field $F$ but I am not sure why are we adding the elements $b_{i}$, $0 \leq i \leq n$ in order to show the existence of such a finite extension?.
Any help is great!
 A: This one is not difficult and pretty standard. Let $a\in K$ then $a$ is algebraic over $E$ so we have polynomial $f(x) \in E[x] $ such that $f(a) =0$. Let $b_{i} $ be coefficients of $f$. Each $b_i\in E$ and hence is algebraic over $F$ so that $L=F(b_{0},b_1,b_2,\dots b_n) $ is a finite extension of $F$. Now $f(x) $ is also a polynomial in $L[x] $ and $f(a) =0$ so that $a$ is algebraic over $L$ and hence $L(a) $ is a finite extension of $L$. It is now clear that $$[L(a) :F] =[L(a) :L] [L:F] $$ so that $L(a) $ is a finite extension of $F$. Thus $a$ is algebraic over $F$. Since $a$ was an arbitrary element of $K$ it follows that $K$ is an algebraic extension of $F$.

Notice that the proof hinges on the following results:


*

*If $K$ is an extension of field $F$ then any element $a\in K$ is algebraic over $F$ if and only if $F(a) $ is a finite extension of $F$.

*If $K$ is a finite extension of $E$ and $E$ is a finite extension of $F$ then $K$ is a finite extension of $F$ and $$[K:F] =[K:E] [E:F] $$

*If $K$ is a finite extension of $F$ then every element of $K$ is algebraic over $F$ so that finite extensions are always algebraic.
In the proof given above, the first result is used to deduce that $L$ is a finite extension of $F$. The last result is used at the end to show that $a$ is algebraic over $F$. 
