How to show that this simple extension avoids adding more elements in $L$?

Question

$F\subset L \subset K$ is a field extension, $α \in K$ is algebraic element over $F$ , whose minimal polynomial $p(x)$ over $F$ is irreducible over $L$, show that $F(α)\cap L = F$

I don't know how to do..thanks for any help..

Answer 1

$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$Recall that every element of $F(\alpha)$ can be written uniquely as $q(\alpha)$, where $q(x) \in F[x]$ is either zero, or has degree less than the degree of $p(x)$.

Also, since $p(x)$ is irreducible over $L$, it is the minimal polynomial of $\alpha$ not only over $F$, but also over $L$

Suppose by way of contradiction that $F(\alpha) \cap L \supsetneq F$. Then there is a non-constant polynomial $q(x) \in F[x]$, of degree smaller than the degree of $p(x)$, such that $q(\alpha) = b \in L$. But then $\alpha$ is a root of $0 \ne q(x) - b \in L[x]$. This is a contradiction, as $q(x) - b$ has degree less than $p(x)$, which was assumed to be the minimal polynomial of $\alpha$ over $L$.