Question

If $L$is a field such that $F\subseteq K\subseteq L$and $u\in L$is algebraic over $F$show that $u$is algebraic over $K$.

Short answer

It is proved that$u$ is algebraic over $F$.

Step-by-step solution

Definition of an algebraic function.

If$K$ is an extension field of $F$,then$K$ is said to be an algebraic extension of$F$ if every element of $K$algebraic over$F$ .

Showing that u is algebraic over  F 

Since$u\in L$ , it is algebraic over$K$ . Therefore, there exists a polynomial$p\left(x\right)\in F\left[x\right]$ that can be considered a polynomial over $K$.

Therefore, the same polynomial$p\left(x\right)\in F\left[x\right]$ serves as a polynomial over$K$ such that$p\left(u\right)=0$ .

Thus, there exists a polynomial $p\left(x\right)\in F\left[x\right]$such that $p\left(u\right)=0$. So, $u$is algebraic over $K$.