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. ∎
References
- 1 McDonald, Bernard R., Finite rings with identity^{}, Pure and Applied Mathematics, Vol. 28, Marcel Dekker Inc., New York, 1974, p. 48.
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 |