Is a vector space with countable basis is countable?

Question

Let $F$ be countable field and the vector space $V$ spaned by $\{1,\alpha\}$. For example, for $F=\mathbb Q$, if $\alpha$ is like $\sqrt2$, then

$$f:V\to\mathbb Q\times\mathbb Q\\ p+q\alpha\mapsto(p,q)$$

is an isomorphism. From this we can conclude that $V=\mathbb Q(\alpha)$ is also countable. But what can we say if $\alpha$ is like $\pi$. It seems that

$$f:V\to\mathbb Q\times\mathbb Q\times\mathbb Q\times\cdots\\ p+q\alpha+r\alpha^2+\cdots\mapsto(p,q,r,\cdots)$$

is an ismorphism too, but I didn't see somewhere that if a countable product of countable sets $\mathbb Q\times\mathbb Q\times\mathbb Q\times\cdots$ is countable. Also I read somewhere that $\mathbb Q(\alpha)$ is the set of elements like $\frac{p_1+p_2\alpha+\cdots p_n\alpha^{n-1}}{q_1+q_2\alpha+\cdots q_n\alpha^{n-1}}$. So I'm a bit confused about that, since I got any element of $V$ like $p+q\alpha+r\alpha^2+\cdots$. Could you please help me?

Answer 1

Hint:

The set of sums $p_0 + p_1\alpha$ is countable.

The set of sums $p_0 + p_1\alpha + p_2\alpha^2$ is countable.

$\cdots$

Answer 2

The field $\mathbb{Q}(\alpha)$ is the smallest field containing $\mathbb{Q}$ and $\alpha$. Therefore it should contain any element reachable by finitely many products, sums, inversions built from $\alpha$ and rational numbers. This does not include series which require infinitely many steps.

From the fact that there are countably many fractions $\dfrac{P(\alpha)}{Q(\alpha)}$ with $\deg P\leq n$ and $\deg Q\leq n$, you can conclude that $\mathbb{Q}(\alpha)$ is countable.