Algebraic closure

Examples

• The fundamental theorem of algebra states that the algebraic closure of the field of real numbers is the field of complex numbers.
• The algebraic closure of the field of rational numbers is the field of algebraic numbers.
• There are many countable algebraically closed fields within the complex numbers, and strictly containing the field of algebraic numbers; these are the algebraic closures of transcendental extensions of the rational numbers, e.g. the algebraic closure of \mathbf{Q} (\pi).
• For a finite field of prime power order q, the algebraic closure is a countably infinite field that contains a copy of the field of order q^n for each positive integer n (and is in fact the union of these copies).

Existence of an algebraic closure and splitting fields

Let S = { f_{\lambda} \mid \lambda \in \Lambda} be the set of all monic irreducible polynomials in K[x]. For each f_{\lambda} \in S, introduce new variables u_{\lambda,1},\ldots,u_{\lambda,d} where d = {\rm degree}(f_{\lambda}). Let R be the polynomial ring over K generated by u_{\lambda,i} for all \lambda \in \Lambda and all i \leq {\rm degree}(f_{\lambda}). Write : f_{\lambda} - \prod_{i=1}^d (x-u_{\lambda,i}) = \sum_{j=0}^{d-1} r_{\lambda,j} \cdot x^j \in R[x]

with r_{\lambda,j} \in R. Let I be the ideal in R generated by the r_{\lambda,j}. Since I is strictly smaller than R, Zorn's lemma implies that there exists a maximal ideal M in R that contains I. The field K_1=R/M has the property that every polynomial f_{\lambda} with coefficients in K splits as the product of x-(u_{\lambda,i} + M), and hence has all roots in K_1. In the same way, an extension K_2 of K_1 can be constructed, etc. The union of all these extensions is the algebraic closure of K, because any polynomial with coefficients in this new field has its coefficients in some K_n with sufficiently large n, and then its roots are in K_{n+1}, and hence in the union itself.

It can be shown along the same lines that for any subset S of K[x], there exists a splitting field of S over K.

Separable closure

An algebraic closure K^{\text{alg}}of K contains a unique separable extension K^{\text{sep}} of K containing all (algebraic) separable extensions of K within K^{\text{alg}}. This subextension is called a separable closure of K. Since a separable extension of a separable extension is again separable, there are no finite separable extensions of K^{\text{sep}}, of degree 1. Saying this another way, K is contained in a separably-closed algebraic extension field. It is unique (up to isomorphism).

The separable closure is the full algebraic closure if and only if K is a perfect field. For example, if K is a field of characteristic p and if X is transcendental over K, K(X)(\sqrt[p]{X}) \supset K(X) is a non-separable algebraic field extension.

In general, the absolute Galois group of K is the Galois group of K^{\text{sep}} over K.

References

References

1. McCarthy (1991) p.21
2. [[Michael Atiyah. M. F. Atiyah]] and [[I. G. Macdonald]] (1969). ''Introduction to commutative algebra''. Addison-Wesley publishing Company. pp. 11–12.
3. Kaplansky (1972) pp.74-76
4. [https://mathoverflow.net/questions/46566/is-the-statement-that-every-field-has-an-algebraic-closure-known-to-be-equivalent Mathoverflow discussion]
5. (1989). "Infinite Algebraic Extensions of Finite Fields". [[American Mathematical Society]].
6. McCarthy (1991) p.22
7. (2008). "Field arithmetic". [[Springer-Verlag]].