# Uniqueness of finite field [duplicate]

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?

## marked as duplicate by Dietrich Burde abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 28 '18 at 12:51

• 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. – Saucy O'Path Jul 28 '18 at 12:42
• @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. – Upc Jul 28 '18 at 14:12
HINT: Prove that every element of such extension is a root of $x^{p^n} - x$.
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 $X^{p^n - 1} - 1$.