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I suspect that this is a very simple question, but I need to ask.

My question is

How do the fields of characteristic $p$ look like?

If $K$ is a finite field of order $p^n$, then $K$ has characteristic $p$ ($p$ prime). We can take the algebraic closure of $K$ and we get $$ \bar{K} = \bigcup_n K^n. $$ Then $K$ has characteristic $p$ as well.

Are all the algebraically closed (hence infinite) fields of characteristic $p$ algebraic closures of a union of finite fields in this way?

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Nope; every element of a finite field is algebraic over $\mathbf{F}_p$, but $\mathbf{F}_p(x)$ contains a element transcendental over $\mathbf{F}_p$... and thus so does its algebraic closure.

However, every field of characteristic $p$ can be written as a union of finite fields and copies of $\mathbf{F}_p(x)$. The easiest way is to just throw in a new copy of $\mathbf{F}_p(x)$ for every transcendental element of your field, obtained by mapping $x$ to that element.

Also, if $K$ is any field at all, the algebraic closure of $K$ can be written as a union of finite extensions of $K$. (which won't be finite fields, of course, unless $K$ is finite itself)

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  • $\begingroup$ Can you make the second paragraph more explicit? $\endgroup$ – user87952 Apr 29 '14 at 18:39
  • $\begingroup$ @user: Did you ask that before or after my edit? If after, what part of it do you need explained? $\endgroup$ – user14972 Apr 29 '14 at 18:40
  • $\begingroup$ I think I did it after your edit. So could you make the "every field of char. $p$ can be written ..." more explicit. $\endgroup$ – user87952 Apr 29 '14 at 18:41
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There are algebraically closed fields of characteristic $p$ of any infinite cardinality. An uncountable algebraically closed field of characteristic $p$ cannot be isomorphic to a countable union of finite fields.

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  • $\begingroup$ Can you give a classification of sorts of the algebraically closed fields of characteristic $p$? $\endgroup$ – user87952 Apr 29 '14 at 18:38
  • $\begingroup$ You can classify the countable ones by transcendence degree over $\mathbb{Z}_p$. For any uncountable cardinal $\kappa$, there is up to isomorphism only one algebraically closed field of characteristic $p$. $\endgroup$ – André Nicolas Apr 29 '14 at 18:40

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