If $F \subseteq L$ is an algebraic field extension and $\phi:L \to L$ is an $F$-monomorphism, then $\phi(L)=L$.

Question

Let $F \subseteq L$ be an algebraic field extension and let $\phi:L \to L$ be an $F$-monomorphism. I want to show that $\phi(L)=L$. The question is from Martin Isaacs' Algebra.

I have a hint: A polynomial $f\in F[x]$ must have as many roots in $\phi(L)$ as it does in $L$. But I can not use the hint.

Answer 1

Let $u \in L$. Then, $u$ is algebraic over $F$.

Consider now the minimal polynomial $f$ of $u$. Use the hint provided: Since $\phi(L) \subset L$ and $f$ has as many roots in $\phi(L)$ as in $L$, then all the roots of $f$ in $L$ belong to $\phi(L)$.

Since $u$ is one of the roots, the above shows that $u \in \phi(L)$.

Answer 2

Assume there's $a \in L$ such that $a$ is not in the image of $\phi$ then since the extension is algebraic there is an irreducible polynomial $f \in F[x]$ such that $f(a) =0$. Let $S$ denote the set of zero's of $f$ in $L$ then from your hint $\phi$ idnuces a self-bijection of $S$.