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In the following MathOverflow question, it has been pointed out that $\overline{\mathbb{F}_p}$ is an uncountable set. Whereas according to http://press.princeton.edu/chapters/s9103.pdf (see page 4 theorem 1.2.1) the closure $\overline{\mathbb{F}_p}$ is $\cup_{n=1}^{\infty}\mathbb{F}_{p^n}$, which I think is a countable union of finite sets and hence countable. Where am I going wrong in this?

Also, in the same document before the same theorem its mentioned that if $\mathbb{F}_q$ has characteristic $p$ then its closure is same as that of $\mathbb{F}_p$ but I think that the set $\cup_{n=1}^{\infty}\mathbb{F}_{q^n}$ is a proper subset of $\cup_{n=1}^{\infty}\mathbb{F}_{p^n}$ since $q$ is a power of $p$, thus they are not the same. Again where is the fault in my reasoning?

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    Oh yes. I misunderstood. Thanks.2012-06-22

2 Answers 2

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The field discussed on MO is $\mathbb{F}_p((t))$, the field of formal Laurent series over $\mathbb{F}_p$. This has uncountable algebraic closure. The algebraic closure of $\mathbb{F}_p$ is countable, as you have correctly stated in the question.

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    I get that. Thanks.2012-06-22
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Your reasoning for the first question is correct.

Hint for the second question: If $q=p^n$, then $\mathbb{F}_{p^m}\subseteq\mathbb{F}_{q^m}\subseteq\mathbb{F}_{p^{nm}}.$