Let $F$ be a field, and let $c$ be transcendental over $F$. Prove that $F(c)$ is the field of quotients of $\{a(c): a(x) \in F[x]\}$, and is isomorphic to $F(x)$, the field of quotients of $F[x]$.
Here is the hint/solution from the book:
Let $Q$ denote the field of quotients of $\{a(c) : a(x) ∈ F[x]\}$. Since $F(c)$ is a field which contains $F$ and $c$, it follows that $F(c)$ contains every $a(c) = a_0 + a_1c + \cdots + a_nc^n$ where $a_0, \dots, a_n \in F$. Thus, $Q \subseteq F(c)$. Conversely, by definition $F(c)$ is contained in any field containing $F$ and $c$; hence $F(c) \subseteq Q$. Complete the solution.
However, it seems to me $F(c)$ doesn't just contain every $a(c) = a_0 + a_1c + \cdots + a_nc^n$ where $a_0, \dots, a_n \in F$, but also every quotient in the form of $a(c)/b(c)$ with $b(c)\ne 0$.
So here is my attempt, borrowing from the above argument:
Let $Q_c$ denote the field of quotients of $\{a(c) : a(x) \in F[x]\}$. Since $F(c)$ is a field which contains $F$ and $c$, it follows that $F(c)$ contains every $a(c)/b(c)$, where $a(x), b(x)\in F[x]$ and $b(x)\ne 0$. Thus, $Q_c \subseteq F(c)$. Conversely, by definition $F(c)$ is contained in any field containing $F$ and $c$; hence $F(c) \subseteq Q_c$.
Let $F(x)$ be the field of quotients of $F[x]$. Let $\sigma_c: F(x) \rightarrow F(c)$ such that $\sigma_c\left(a(x)/b(x)\right) = a(c)/b(c)$ for $a(x), b(x)\in F[x]$ and $b(x)\ne 0$. Note $\sigma_c$ is a homomorphism from $F(x)$ to $F_c$. It is surjective by definition of $\sigma_c$. It is injective for the kernel is trivial, given $c$ is transcendental. Hence $F(x)$ and $F(c)$ are isomorphic.
Is this reasonable, or did I over complicate?