0
$\begingroup$

This question already has an answer here:

Assume $L$ is the algebraic closure of $\mathbb{F}_p$. Show there exists a unique finite field of cardinality $p^n$ containing $\mathbb{F}_p$. The existence is easy just have to define the splitting field of $X^{p^n}-X$. But what about uniqueness?

$\endgroup$

marked as duplicate by Dietrich Burde abstract-algebra Jul 28 '18 at 12:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Well, there are at most $p^n$ solutions in $\overline{\mathbb F}_p$ to the equation $x^{p^n}-x=0$. On the other hand, all the elements of a field of $p^n$ elements must satisfy the equation $x^{p^n}-x=0$. So the elements of such a field are exactly those $p^n$ solutions. $\endgroup$ – Saucy O'Path Jul 28 '18 at 12:42
  • $\begingroup$ @SaucyO'Path: Why the element of that field with $p^n$ must satisfy the equation? How can I deduce it without mentioning the fact that every finite field has cyclic multiplicative group. $\endgroup$ – Upc Jul 28 '18 at 14:12
  • $\begingroup$ You don't need it to be cyclic. You just need it to be a group (which it is by definition). $\endgroup$ – Saucy O'Path Jul 28 '18 at 14:14
2
$\begingroup$

HINT: Prove that every element of such extension is a root of $x^{p^n} - x$.

$\endgroup$
1
$\begingroup$

The multiplicative subgroup of nonzero elements of a field with $p^n$ elements is an abelian group of $p^n - 1$ elements; every element is a root of $X^{p^n - 1} - 1$.

$\endgroup$

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