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In abstract algebra, a field extension L/K is called algebraic if every element of L is algebraic over K, i.e. if every element of L is a root of some non-zero polynomial with coefficients in K. Field extensions that are not algebraic, i.e. which contain transcendental elements, are called transcendental.

For example, the field extension R/Q, that is the field of real numbers as an extension of the field of rational numbers, is transcendental, while the field extensions C/R and Q(√2)/Q are algebraic, where C is the field of complex numbers.

All transcendental extensions are of infinite degree. This in turn implies that all finite extensions are algebraic.[1] The converse is not true however: there are infinite extensions which are algebraic. For instance, the field of all algebraic numbers is an infinite algebraic extension of the rational numbers.

If a is algebraic over K, then K[a], the set of all polynomials in a with coefficients in K, is not only a ring but a field: an algebraic extension of K which has finite degree over K. The converse is true as well, if K[a] is a field, then a is algebraic over K. In the special case where K = Q is the field of rational numbers, Q[a] is an example of an algebraic number field.

A field with no nontrivial algebraic extensions is called algebraically closed. 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.

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


The class of algebraic extensions forms a distinguished class of field extensions, that is, the following three properties hold:[2]

  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:

  1. 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 article: Substructure

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

\left\{y\in N\Big|p(y)\right\}

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.

See also[edit]


  1. ^ See also Hazewinkel et al. (2004), p. 3.
  2. ^ Lang (2002) p.228


Original courtesy of Wikipedia: http://en.wikipedia.org/wiki/Algebraic_extension — Please support Wikipedia.
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Tue, 12 Mar 2013 11:16:25 -0700

+Asahiko Matsuda Because the concept of "constructible number" is a precise technical mathematical term. It refers to a number that belongs to an "algebraic extension of the rational numbers that can be built via a sequence of simple extensions of ...

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