Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
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

2 Answers 2

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.

share|improve this answer
    
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 $$\mathbb{F}_{p^m}\subseteq\mathbb{F}_{q^m}\subseteq\mathbb{F}_{p^{nm}}.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.