Showing that all elements of a Field Extension are transcendental 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 $q(\alpha) = 0$.
My Concerns: 


*

*$\beta$ 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?
 A: Proffering the use of uniqueness of factorization of polynomials as an alternative.
Assume that contrariwise that $\beta=r(\alpha)/s(\alpha), \beta\notin F,$ is algebraic over $F$.
Here $r(\alpha),s(\alpha)\in F[\alpha]$. W.l.o.g. we can assume that:


*

*$\gcd(r(\alpha),s(\alpha))=1$ because we can simply cancel any common factor, and 

*at least one of $r(\alpha), s(\alpha)$ is not a constant. For otherwise $\beta\in F$.


Let $p(x)=a_0+a_1x+\cdots+a_nx^n\in F[x]$ be the minimal polynomial of $\beta$ over $F$. Because $\beta\notin F$ we know that $p(x)$ has degree at least two. Furthermore, $p(x)$ is irreducible in $F[x]$, so we can assume that $a_n=1$ and $a_0\neq0$.
Let's plug in $x=\beta$. We arrive at
$$
a_0+a_1\frac{r(\alpha)}{s(\alpha)}+\cdots+\left(\frac{r(\alpha)}{s(\alpha)}\right)^n=0.
$$
Using a common denominator allows a rewrite
$$
0=\frac{a_0s(\alpha)^n+a_1s(\alpha)^{n-1}r(\alpha)+a_2s(\alpha)^{n-2}r(\alpha)^2+
\cdots+a_{n-1}s(\alpha)r(\alpha)^{n-1}+r(\alpha)^n}{s(\alpha)^n}.
$$
Here the numerator must be the zero polynomial in the ring $F[\alpha]$. But, the two bullets above imply that there exists an irreducible non-constant polynomial $m(\alpha)\in F[\alpha]$ such that $m(\alpha)$ is a factor of exactly one of $r(\alpha), s(\alpha)$ (and hence not a factor of the other). 
Look at the numerator again. If $m(\alpha)\mid r(\alpha)$ and $m(\alpha)\nmid s(\alpha)$, then all the terms in the numerator with the sole exception of the first are divisible by $m(\alpha)$. Implying that the numerator is not divisible by $m(\alpha)$. This is a contradiction because the numerator is supposed to be zero.
If $m(\alpha)\mid s(\alpha)$ and $m(\alpha)\nmid r(\alpha)$ then a similar problem occurs. This time the last term of the numerator is the only one not divisible by $m(\alpha)$.
Unique factorization of polynomials was used in the step when we deduced that a power of a polynomial not divisible by $m(\alpha)$ could not itself be divisible by $m(\alpha)$ either.
A: I would use the multiplicativity of the degree in the tower of extensions $F \subseteq F(\beta) \subseteq F(\alpha)$. Since $\beta \in F(\alpha)$, then $\beta$ is a rational function in $\alpha$, i.e., $\beta = \frac{f(\alpha)}{g(\alpha)}$ for some polynomials $f,g \in F[x]$, $g \neq 0$. Then $\alpha$ is a root of $\beta g(x) - f(x) \in F(\beta)[x]$. Assume $\beta \in F(\alpha) \setminus F$. Then at least one of $f$ and $g$ is nonconstant, and since $F[x]$ is a PID we may take $f$ and $g$ relatively prime.  Then $\beta g(x) - f(x)$ is not the zero polynomial (since $\beta g \in F(\beta)[x] \setminus F[x]$ while $f \in F[x]$), so $\alpha$ is algebraic over $F(\beta)$, hence $[F(\alpha) : F(\beta)] < \infty$.
For contradiction, assume $\beta$ is algebraic over $F$. Then $[F(\beta) : F] < \infty$, so by multiplicativity we have
$$
[F(\alpha) : F] = [F(\alpha) : F(\beta)] [F(\beta) : F] < \infty \, ,
$$
contradicting the fact that $\alpha$ is transcendental over $F$.
