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 $x^{p^n}=x$.

| cite | improve this answer | |
  • $\begingroup$ 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)? $\endgroup$ – gune Jan 19 '17 at 16:52
  • 1
    $\begingroup$ @CharithJeewantha: exactly. Since $\Bbb F$ is algebraically closed it must contain all the roots of that polynomial. That makes it. $\endgroup$ – Andrea Mori Jan 19 '17 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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