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?
Best Answer
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)