My question is whether my strategy for this proof is correct. I'll put the fact to be proven, my strategy, and why I am hesitant about my strategy.
Thing to be proven: I want to show that if $E$ is an extension field of $F$, and $\alpha \in E$ is transcendental over $F$ then every element of $F(\alpha)$ (field of quotients of $F[\alpha]$) is also transcendental over $F$.
My Proof Strategy: Assume $\beta \in F(\alpha)$ and $\beta$ is algebraic. Then $\exists p(x) \in F[x]$ such that $p(\beta) = 0$. But since $\beta$ is a polynomial in $\alpha$, that means that some polynomial in $\alpha$ with coefficients in $F$ is equal to $0$ so $\exists q(x) \in F[x]$ such that $q(\alpha) = 0$.
$\beta$ is in a field of quotients of $F[\alpha]$, but this can be circumvented because in order for that to be $0$ we just need the "numerator" which is in $F[\alpha]$ to have a root.
Showing rigorously that $p(\beta)$ is a polynomial in $\alpha$ seems like a pain. It makes sense, but doing a general solution seems hard.
This makes me think that an easier solution might be to somehow employ theorems about the fact that $\langle p(x)\rangle$ must be a maximal ideal.
So my question is, which route would yield an easier and more elegant solution?