$X$ is algebraic over $E$

Question

Let $X$ be transcendental over a field $F$, and let $E$ be a subfield of $F(X)$ properly containing $F$. Prove that $X$ is algebraic over $E$.

Could we maybe use also the following??

Let $f(x) \in E \Rightarrow f(x) \in F(X) \Rightarrow f(x)=\frac{h(X)}{g(X)}$, where $g(X) \neq 0, h,g \in F[x]$.

Or doesn't this help ??

Answer 1

Hint: With a change of notation write $t = f(X)$ and let $t = \frac{h(X)}{g(X)}$ with $h(X), g(X) \in \ F(X), g(X) \neq 0, \text{gcd}(h(X), g(X)) = 1.$ Now consider the polynomial $h(Y) - tg(Y)$ in variable $Y$ and coefficients in $F(t).$ Show that this polynomial is irreducible over $F(t)$ and $X$ is a root. Use this to show that $[F(X) : F(t)] < \infty $

EDIT: To show that $h(Y) - tg(Y)$ is irreducible over $F(t),$ it is sufficient to show that it is irreducible in $(F[t])[Y].$ We have $(F[t])[Y] = (F[Y])[t].$ If we have a factorization of $h(Y) - tg(Y)$ in $(F[t])[Y]$ then it will induce a factorization of the same polynomial in $(F[Y])[t],$ but $h(Y) - tg(Y)$ is irreducible in $(F[Y])[t].$ Now note that $n := \text{deg}(h(Y) - tg(Y)) = \text{max} \{\text{deg}(h(X)), \text{deg}(g(X))\}$ as a polynomial of $Y$ over $F(t).$ So $[F(X) : F(t)] = n < \infty$