Cardinality of the quotient field of power series ring

Question

Let $k$ be a field which is countable and let $x$ be an indeterminate over $k$. I have hard time to prove $$\operatorname{card} k((x)) = \operatorname{card}\mathbb R.$$ Thank you.

Answer 1

When your field is $\mathbb{Z}_{2}$ Define a bijection between $k[[x]]$ and the set of numbers between 0 and 1 which can have only 0 and 1 as their digits. For other fields you should be more carious. If for example your field is something as $\mathbb{C}$, you can see easily that as $$k[[x]]\cong\prod_{i\in\mathbb{N}\cup\{0\}}k$$ so its cardinal is again $|\mathbb{R}|=c$ . For fields which their cardinal is less than $c$. As you can define a one-to-one function from $\mathbb{Z}_2[[x]]$ to $k[[x]]$ and a one-to-one function from $k[[x]]$ to $\mathbb{C}[[x]]$, so you have $c=|\mathbb{Z}_2[[x]]|\leq|k[[x]]|\leq|\mathbb{C}[[x]]|=c$ and therefore $|k[[x]]|=c$ . And now as you assumed that $k$ is countable so its cardinal is less than $c$ and the problem is solved.

Now pay attention that $k[[x]]\subset k((x))$ and you can look at $k((x))$ as a subset of $k[[x]]\times k[[x]]$ so $c\leq|k((x))|\leq c^2=c$ .