# algebraic closure of a finite field

Fix a prime $p$ in $\mathbb{Z}$. Then the Galois fields $GF(p^{e})$ denotes the finite field of order $p^{e}$, $e\geq 1$. This can be concretely constructed as the splitting field  of the polynomials  $x^{p^{e}}-x$ over $\mathbb{Z}_{p}$. In so doing we have $GF(p^{e})\subseteq GF(p^{f})$ whenever $e|f$. In particular, we have an infinite chain:

 $GF(p^{1!})\subseteq GF(p^{2!})\subseteq GF(p^{3!})\subseteq\cdots\subseteq GF(% p^{n!})\subseteq\cdots.$

So we define $\displaystyle GF(p^{\infty})=\bigcup_{n=1}^{\infty}GF(p^{n!})$.

###### Theorem 1.

$GF(p^{\infty})$ is an algebraically closed field of characteristic  $p$. Furthermore, $GF(p^{e})$ is a contained in $GF(p^{\infty})$ for all $e\geq 1$. Finally, $GF(p^{\infty})$ is the algebraic closure  of $GF(p^{e})$ for any $e\geq 1$.

###### Proof.

Given elements $x,y\in GF(p^{\infty})$ then there exists some $n$ such that $x,y\in GF(p^{n!})$. So $x+y$ and $xy$ are contained in $GF(p^{n!})$ and also in $GF(p^{\infty})$. The properties of a field are thus inherited and we have that $GF(p^{\infty})$ is a field. Furthermore, for any $e\geq 1$, $GF(p^{e})$ is contained in $GF(p^{e!})$ as $e|e!$, and so $GF(p^{e})$ is contained in $GF(p^{\infty})$.

Now given $p(x)$ a polynomial over $GF(p^{\infty})$ then there exists some $n$ such that $p(x)$ is a polynomial over $GF(p^{n!})$. As the splitting field of $p(x)$ is a finite extension  of $GF(p^{n!})$, so it is a finite field $GF(p^{e})$ for some $e$, and hence contained in $GF(p^{\infty})$. Therefore $GF(p^{\infty})$ is algebraically closed. ∎

We say $GF(p^{\infty})$ is the algebraic closure indicating that up to field isomorphisms, there is only one algebraic closure of a field. The actual objects and constructions may vary.

## References

Title algebraic closure of a finite field AlgebraicClosureOfAFiniteField 2013-03-22 16:40:51 2013-03-22 16:40:51 Algeboy (12884) Algeboy (12884) 5 Algeboy (12884) Derivation msc 12F05 FiniteField FiniteFieldCannotBeAlgebraicallyClosed