algebraic extension

All transcendental extensions are of infinite degree. This in turn implies that all finite extensions are algebraic.See also Hazewinkel et al. (2004), p. 3. The converse is not true however: there are infinite extensions which are algebraic.Fraleigh (2014), Theorem 31.18, p. 288. For instance, the field of all algebraic numbers is an infinite algebraic extension of the rational numbers.Fraleigh (2014), Corollary 31.13, p. 287.

Let {{math|E}} be an extension field of {{math|K}}, and {{math|a ∈ E}}. The smallest subfield of {{math|E}} that contains {{math|K}} and {{mvar|a}} is commonly denoted $K(a).$ If {{mvar|a}} is algebraic over {{math|K}}, then the elements of {{math|K(a)}} can be expressed as polynomials in {{mvar|a}} with coefficients in K; that is, $K(a)=K[a]$, the smallest ring containing {{math|K}} and {{mvar|a}}. In this case, $K(a)$ is a finite extension of {{mvar|K}} and all its elements are algebraic over {{mvar|K}}. In particular, $K(a)$ is a {{mvar|K}}-vector space with basis $\backslash \{1,a,...,a^\{d-1\}\backslash \}$, where d is the degree of the minimal polynomial of {{mvar|a}}.Fraleigh (2014), Theorem 30.23, p. 280. These properties do not hold if {{mvar|a}} is not algebraic. For example, $\backslash Q(\backslash pi)\backslash neq\; \backslash Q[\backslash pi],$ and they are both infinite dimensional vector spaces over $\backslash Q.$Fraleigh (2014), Example 29.8, p. 268.

An algebraically closed field F has no proper algebraic extensions, that is, no algebraic extensions E with F < E.Fraleigh (2014), Corollary 31.16, p. 287. An example is the field of complex numbers. Every field has an algebraic extension which is algebraically closed (called its algebraic closure), but proving this in general requires some form of the axiom of choice.Fraleigh (2014), Theorem 31.22, p. 290.

An extension L/K is algebraic if and only if every sub K-algebra of L is a field.

The following three properties hold:Lang (2002) p.228

1. If E is an algebraic extension of F and F is an algebraic extension of K then E is an algebraic extension of K.
2. If E and F are algebraic extensions of K in a common overfield C, then the compositum EF is an algebraic extension of K.
3. If E is an algebraic extension of F and E > K > F then E is an algebraic extension of K.

These finitary results can be generalized using transfinite induction:

{{ordered list|start=4

| The union of any chain of algebraic extensions over a base field is itself an algebraic extension over the same base field.

}}

This fact, together with Zorn's lemma (applied to an appropriately chosen poset), establishes the existence of algebraic closures.

{{Main|Substructure (mathematics)}}

Model theory generalizes the notion of algebraic extension to arbitrary theories: an embedding of M into N is called an algebraic extension if for every x in N there is a formula p with parameters in M, such that p(x) is true and the set

:$\backslash left\backslash \{y\backslash in\; N\; \backslash mid\; p(y)\backslash right\backslash \}$

is finite. It turns out that applying this definition to the theory of fields gives the usual definition of algebraic extension. The Galois group of N over M can again be defined as the group of automorphisms, and it turns out that most of the theory of Galois groups can be developed for the general case.

Given a field k and a field K containing k, one defines the relative algebraic closure of k in K to be the subfield of K consisting of all elements of K that are algebraic over k, that is all elements of K that are a root of some nonzero polynomial with coefficients in k.

• Integral element
• Lüroth's theorem
• Galois extension
• Separable extension
• Normal extension

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Category:Field extensions