algebraic closure of a finite field
Fix a prime in . Then the Galois fields denotes thefinite field of order , . This can be concretely constructed asthe splitting field of the polynomials
over . In so doing wehave whenever . In particular, we have aninfinite chain:
So we define .
Theorem 1.
is an algebraically closed field of characteristic .Furthermore, is a contained in for all .Finally, is the algebraic closure
of for any .
Proof.
Given elements then there exists some such that. So and are contained in and alsoin . The properties of a field are thus inherited and we havethat is a field. Furthermore, for any , iscontained in as , and so is contained in .
Now given a polynomial over then there exists some such that is a polynomial over . As the splitting fieldof is a finite extension of , so it is a finite field for some , and hence contained in . Therefore is algebraically closed.∎
We say is the algebraic closure indicating that up to fieldisomorphisms, there is only one algebraic closure of a field. The actual objectsand 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 soit 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.