Editing Algebraic extension (section)

==Some properties==
All transcendental extensions are of infinite [[degree of a field extension|degree]]. This in turn implies that all finite extensions are algebraic.<ref>See also Hazewinkel et al. (2004), p. 3.</ref> The [[converse (logic)|converse]] is not true however: there are infinite extensions which are algebraic.<ref>Fraleigh (2014), Theorem 31.18, p. 288.</ref>  For instance, the field of all [[algebraic number]]s is an infinite algebraic extension of the rational numbers.<ref>Fraleigh (2014), Corollary 31.13, p. 287.</ref>

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 <math>K(a).</math> 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, <math>K(a)=K[a]</math>, the smallest [[ring (mathematics)|ring]] containing {{math|''K''}} and {{mvar|a}}. In this case, <math>K(a)</math> is a finite extension of {{mvar|K}} and all its elements are algebraic over {{mvar|K}}.  In particular, <math>K(a)</math> is a {{mvar|K}}-vector space with basis <math>\{1,a,...,a^{d-1}\}</math>, where ''d'' is the degree of the [[Minimal polynomial (field theory)|minimal polynomial]] of {{mvar|a}}.<ref>Fraleigh (2014), Theorem 30.23, p. 280.</ref> These properties do not hold if {{mvar|a}} is not algebraic. For example, <math>\Q(\pi)\neq \Q[\pi],</math> and they are both infinite dimensional vector spaces over <math>\Q.</math><ref>Fraleigh (2014), Example 29.8, p. 268.</ref>

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

An extension ''L''/''K'' is algebraic [[if and only if]] every sub ''K''-[[Algebra over a field|algebra]] of ''L'' is a field.