Degree of a field extension by a transcendental element

Question

Let $F$ be a field, and let $F(x)$ be the field of fractions of the polynomial ring $F[x]$. I'm interested in the degree of the field extension $[F(x) : F]$. Obviously it is infinite, but what exactly is its cardinality? Is it $\aleph_0$? Does it depend on the field $F$?

Answer 1

The natural $F$-basis of $F(x)$ is $$\{ x^k, k\ge 0\} \cup \{ x^l/h^m, m\ge 1,l<\deg(h), h \in F[x]\text{ monic irreducible}\}$$ Thus (for $F$ infinite) the cardinality of the basis is comprised between that of $F$ and $F[x]^2$, ie. it is the same as $F$.

Answer 2

For any infinite field $F$, $F[x] = \oplus_{n \geq 0} F (x^n)$ is of equal cardinality to $F$, and there is a surjective mapping $F[x] \times (F[x])^* \rightarrow F(x)$ given by $(p(x), q(x)) \mapsto \frac{p(x)}{q(x)}$ (where $(F[x])^* = F[x] \setminus \{ 0 \}$). Since $F[x] \times (F[x])^*$ is of equal cardinality to $F[x]$, the result follows.

If $F$ is finite, $F[x]$ is countably infinite, and by the same logic as above, $F(x)$ is countably infinite as well.