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 $p^e$, $e\ge 1$. 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:

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

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

Theorem 1.

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

Proof.

Given elements $x,y\in GF(p^{\mathrm{\infty }})$ 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 $GF(p^{\mathrm{\infty }})$. The properties of a field are thus inherited and we have that $GF(p^{\mathrm{\infty }})$ is a field. Furthermore, for any $e\ge 1$, $GF(p^e)$ is contained in $GF(p^{e!})$ as $e|e!$, and so $GF(p^e)$ is contained in $GF(p^{\mathrm{\infty }})$.

Now given $p(x)$ a polynomial over $GF(p^{\mathrm{\infty }})$ 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 $GF(p^{\mathrm{\infty }})$. Therefore $GF(p^{\mathrm{\infty }})$ is algebraically closed. ∎

We say $GF(p^{\mathrm{\infty }})$ 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.

Corollary 2.

The algebraic closure of a finite field is countable.

Proof.

By construction the algebraic closure is a countable union of finite sets so it is countable. ∎

Title	algebraic closure of a finite field
Canonical name	AlgebraicClosureOfAFiniteField
Date of creation	2013-03-22 16:40:51
Last modified on	2013-03-22 16:40:51
Owner	Algeboy (12884)
Last modified by	Algeboy (12884)
Numerical id	5
Author	Algeboy (12884)
Entry type	Derivation
Classification	msc 12F05
Related topic	FiniteField
Related topic	FiniteFieldCannotBeAlgebraicallyClosed