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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 (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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In the MO thread I see that Pete Clark mentions the algebraic closure of $\mathbb F_p((t))$. Where is it claimed that $\overline{\mathbb F}_p$ is uncountable? – Dylan Moreland Jun 22 '12 at 21:12
Oh yes. I misunderstood. Thanks. – Abhishek Gupta Jun 22 '12 at 21:22
up vote 7 down vote accepted

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. – Abhishek Gupta Jun 22 '12 at 21:18

Your reasoning for the first question is correct.

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

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