# What is an algebraically closed field of characteristic $p$?

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?

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)

• Can you make the second paragraph more explicit? – user87952 Apr 29 '14 at 18:39
• @user: Did you ask that before or after my edit? If after, what part of it do you need explained? – user14972 Apr 29 '14 at 18:40
• I think I did it after your edit. So could you make the "every field of char. $p$ can be written ..." more explicit. – user87952 Apr 29 '14 at 18:41

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.

• Can you give a classification of sorts of the algebraically closed fields of characteristic $p$? – user87952 Apr 29 '14 at 18:38
• 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 $p$. – André Nicolas Apr 29 '14 at 18:40