If $a$ belongs to an extension field of $F$ and $f(a)$ is algebraic over $F$, then $a$ is algebraic over $F$.

Question

I know since $f(x)$ is nonconstant of $F[x]$ there exists a splitting field for $f(x)$ over $F$. Then since $f(a)$ is algebraic over $F$,it means $f(a)$ as a whole is a zero of some other nonzero polynomial in $F[x]$. This is where I get stumped, does this mean "$a$" is a zero of $f(x) - f(a) ∊ F(f(a))[x] $?

Answer 1

If $f(a)$ is algebraic over $F$ there is some non constant polynomial $g\in F[X]$ such that $g(f(a))=0$. Note that $h:=g\circ f\in F[X]$ is non constant and that $h(a)=0$.