On simple extension Show that $\mathbb{R}$
  is not a simple extension of $\mathbb{Q}$
  as follow:
a. $\mathbb{Q}$
  is countable.
b. Any simple extension of a countable field is countable.
c. $\mathbb{R}$
  is not countable.
I 've done a. and c. Can anyone help me a hint to prove b.?
 A: Let $F$ be a countable field, then the collection of all polynomials of degree $\leq n$ is countable. Hence, $F[x]$ is countable, being the countable union of countable sets. Hence, $F[x] \times F[x]\setminus\{0\}$ is countable. There is a surjective function $F[x]\times F[x]\setminus\{0\} \to F(x)$ by
$$
(f(x), g(x)) \mapsto \frac{f(x)}{g(x)}
$$
Hence, $F(x)$ is countable. If $a$ is transcendental over $F$, then $F(x) \cong F(a)$, which is thus countable.
A: To simplify things let us divide in two cases:
The extension of $\;\Bbb Q\;$ is algebraic: Then we have $\;\Bbb Q(\alpha)/\Bbb Q\;$  of degree $\;n\;$ , and every element in the extension is of the form $\;a_0+a_1\alpha+\ldots+a_{n-1}\alpha^{n-1}\;,\;\;a_i\in\Bbb Q\;$ ...
The extension of $\;\Bbb Q\;$ is transcendental: Then $\;\Bbb Q(\omega)/\Bbb Q\;$ isn't finite, but in fact $\;\Bbb Q(\omega)\;$ is the fractions field of $\;\Bbb Q[\omega]\cong\Bbb Q[x]=\;$the polynomial ring in one variable over the rationals, and then we're almost done since
$$\left|\Bbb Q(\omega)\right|:=\left|\left\{\frac ab\;;\;a,b\in\Bbb Q[\omega]\;,\;b\neq 0\right\}\right|\le\left|\Bbb Q[x]\times\Bbb Q[x]\right|=\aleph_0\cdot\aleph_0=\aleph_0$$
A: Here is a hint: Let $F$ be a countable field, $F(a)$ a simple extension. Then every element in $F(a)$ can be written as a quotient $\frac{p(a)}{q(a)}$, where $p$ and $q$ are polynomials. Now, combine known results on countability.
