How do we know "Simple extension of a countable field is also countable?" This intuitively makes sense but I'm not sure how to rigorously prove this. I want to use this to prove that there does not exist a simple extension from $\mathbb{Q}$ to $\mathbb{R}$. Since $\mathbb{Q}$ is countable, and any simple extension from $\mathbb{Q}$ is countable, but $\mathbb{R}$ is uncountable so there does not exist such extension...
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If you understand why $K(x)$ is countable whenever $K$ is, that's a good step in the right direction. – Jan 04 '16 at 01:54
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@user hmmm that's exactly what my question is -- I'm not sure how to rigorously prove, using definition of countableness and field operations. – nekodesu Jan 04 '16 at 01:57
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Start by showing that $K[x]$ is countable, and then compare $K[x] \times K[x]$ with $K(x)$ (as a Cartesian product, not direct product). – Jan 04 '16 at 02:00
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If you can show that the field $K(x)$ is countable whenever $K$ is, then that should give you enough to finish solving the more general problem. To this end, complete two steps:
If $K$ is countable then the polynomial ring $K[x]$ is countable. This is a simple application of the fact that the set of polynomials of degree $n$ is clearly countable, and the union of such sets is the entire ring.
If a ring is countable, so is its field of fractions. To prove this, follow the ideas used in proving that the field of fractions of $\mathbb{Z}$ is countable.
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Would it not be sufficient to note that after showing $K[x]$ is countable for a countable $K$, then since $K(x) \subseteq K[x]$, $K(x)$ must also be countable? – Natasha Apr 26 '20 at 19:08
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