# $\mathbb{F}$ contains a unique subfield of cardinality $p^n$

I'm referring to thepage 197 of the book Field extension and Galois theory by Julio R. Bastida.

let $\mathbb{P}$ be a prime field with prime characteristic, $p$ and $\mathbb{F}$ be an algebraic closure of $\mathbb{P}$.
Then if $n$ is a positive integer, $\mathbb{F}$ contains a unique sub field of cardinality $p^n$.

Can somebody please explain why this happens? * also notice that being the algebraic closure we can prove $\mathbb{F}$ is infinite and Galois over $\mathbb{K}$

The field with $p^n$ elements is made up of the roots of $X^{p^n}-X$ which happen to be distinct.
This gives also uniqueness because any element $x$ in a field with $p^n$ elements must satisfy the relation $x^{p^n}=x$.
• Thank you Andrea. To make it clear, is it because of the fact that $\mathbb{F}$ is algebraically closed, we consider the above polynomial is seperable (hence distinct roots)? – gune Jan 19 '17 at 16:52
• @CharithJeewantha: exactly. Since $\Bbb F$ is algebraically closed it must contain all the roots of that polynomial. That makes it. – Andrea Mori Jan 19 '17 at 16:54