Algebraic extension

The finite algebraic extensions of the field ${\displaystyle \mathbb{Q}}$ of the rational numbers are called algebraic number fields and are the main objects of study of algebraic number theory. Another example of a common algebraic extension is the extension ${\displaystyle \mathbb{C}/\mathbb{R}}$ of the real numbers by the complex numbers.

Some properties

All transcendental extensions are of infinite degree. This in turn implies that all finite extensions are algebraic. 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.

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

An algebraically closed field F has no proper algebraic extensions, that is, no algebraic extensions E with F < E. 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.