Let $E=F(x)$ to be the fraction field of the ring $F[x] $, show that $x\in E$ is transcendental over $F$
let $x\in E$ then, it exists the function $f(y)=y-x$ such that $f(x)=0$
, thus $x$ is algebraic in $E$
clearly, it doesn't exist the function $f(y)=y-x \in F$ because $x$ doesn't exists in $F$.
Assume that it exist $f(y)\neq0 \in F$ such that $f(x)=0$ how can I show this is a contradiction.
Specially, when, for example $\sqrt{2} \notin \mathbb{Q}$ but $\sqrt{2}$ is algebraic over $\mathbb{Q}$ I don't know, this doesn't make a lot of sense.
Also, I found this, cant say I understand it, seemed too much for something simple. I thought it must be an easier way.
Edit: The following answer works if we assume that $F(x)$ is isomorphic to the field of fractions of $F[X]$, where $F[X]$ is the ring of polynomials over $F$. Otherwhise as @lulu pointed out in the comments, we do not have enough information to give a proper answer!
Let $F$ be a field and $F(x)$ is isomorphic to $Quot(F[X])$. Let us call this isomorphism $\iota: Quot(F[X])\rightarrow F(x)$, $X\mapsto x$ Since it is an isomorphism, it is injective. From this we get that $\iota':F[X]\rightarrow F(x)$, $X\mapsto x$ is also injective, so it has trivial kernel. But this means there exist no polynomial $P$ in $F[X]$ (besides the zero polynomial) such that $\iota'(P) = 0$, hence $x$ is transcendental.