Element in field of quotients is transcendental

Let $F\subseteq E$ be fields, and let $c\in E$. Let $F(c)$ be the field of quotients containing $F$ and $c$. Suppose $c$ is transcendental over $F$. Prove that every element in $F(c)$ but not in $F$ is transcendental over $F$.

An element in $x\in F(c)$ can be written as $x=\dfrac{a_0+a_1c+\ldots+a_mc^m}{b_0+b_1c+\ldots+b_nc^n}$ for some $a_0,\cdots,a_m,b_0,\cdots,b_n\in F$. Suppose $x\not\in F$, and suppose it were algebraic over $F$, i.e. it satisfies a polynomial equation $d_0+d_1x+\ldots+d_kx^k=0$ for some $d_0,\ldots,d_k\in F$.

Expanding everything in this last equation, we get a polynomial equation in $c$, which should show that $c$ is algebraic, yielding the desired contradiction.

My question: How do we know that this polynomial equation in $c$ won't be $0=0$?

• Consider the highest terms of $x$. – Martin Brandenburg Jan 17 '14 at 16:33
• @MartinBrandenburg But still, we need to plug in the expression of $x$, which is a quotient of two polynomials in $c$. – JJ Beck Jan 17 '14 at 16:35

Write $x = \frac{p(c)}{q(c)}$, where $p$ and $q$ are relatively prime polynomials. Then we get $$d_0 q^k + d_1 p q^{k-1} + \cdots + d_k p^k = 0, \quad k \geq 1, \quad d_0, d_k \ne 0.$$
Then $q | p^k$, so $q$ is constant. Likewise $p | q^k$, so $p$ is constant. This shows $x = \frac{p}{q} \in F$, a contradiction.
Longer answer: For any $x \in F(c) \ F$, there are polynomials $p$,$q$ such that $x = \frac{p(c)}{q(c)}$ and $1$ is a $\gcd$ of $p$,$q$ in $F[x]$, which is a PID because $F$ is a field, and hence it should work out.