Exercise 7.2.15 (Algebraic extensions)

Solution: By definition of an algebraic extension, we have that if $M:K$ is algebraic then $\forall <!--\; FUNCTION\; APPLICATION\; -->\alpha \in M\exists <!--\; FUNCTION\; APPLICATION\; -->f\ne 0\in K[t]:f(\alpha )=0$. Since we have that $M$ is a field extension of $L$ it must contain all of $L$, therefore we have that $\forall <!--\; FUNCTION\; APPLICATION\; -->\alpha \in L\exists <!--\; FUNCTION\; APPLICATION\; -->f\ne 0\in K[t]:f(\alpha )=0$, hence $L:K$ is algebraic. Similarly, since $L$ extends $K$ any $f\ne 0\in K[t]$ must exist in $L[t]$, hence we that $\forall <!--\; FUNCTION\; APPLICATION\; -->\alpha \in M\exists <!--\; FUNCTION\; APPLICATION\; -->f\ne 0\in L[t]:f(\alpha )=0$, hence $M:L$ is algebraic.