Algebraic element
In mathematics, if L is a field extension of K, then an element a of L is called an algebraic element over K, or just algebraic over K, if there exists some non-zero polynomial g(x) with coefficients in K such that g(a) = 0. Elements of L which are not algebraic over K are called transcendental over K.
These notions generalize the algebraic numbers and the transcendental numbers (where the field extension is C/Q, C being the field of complex numbers and Q being the field of rational numbers).
Examples
- The square root of 2 is algebraic over Q, since it is the root of the polynomial g(x) = x2 − 2 whose coefficients are rational.
- Pi is transcendental over Q but algebraic over the field of real numbers R: it is the root of g(x) = x − π, whose coefficients (1 and −π) are both real, but not of any polynomial with only rational coefficients. (The definition of the term transcendental number uses C/Q, not C/R.)
Properties
The following conditions are equivalent for an element [math]\displaystyle{ a }[/math] of [math]\displaystyle{ L }[/math]:
- [math]\displaystyle{ a }[/math] is algebraic over [math]\displaystyle{ K }[/math],
- the field extension [math]\displaystyle{ K(a)/K }[/math] is algebraic, i.e. every element of [math]\displaystyle{ K(a) }[/math] is algebraic over [math]\displaystyle{ K }[/math] (here [math]\displaystyle{ K(a) }[/math] denotes the smallest subfield of [math]\displaystyle{ L }[/math] containing [math]\displaystyle{ K }[/math] and [math]\displaystyle{ a }[/math]),
- the field extension [math]\displaystyle{ K(a)/K }[/math] has finite degree, i.e. the dimension of [math]\displaystyle{ K(a) }[/math] as a [math]\displaystyle{ K }[/math]-vector space is finite,
- [math]\displaystyle{ K[a] = K(a) }[/math], where [math]\displaystyle{ K[a] }[/math] is the set of all elements of [math]\displaystyle{ L }[/math] that can be written in the form [math]\displaystyle{ g(a) }[/math] with a polynomial [math]\displaystyle{ g }[/math] whose coefficients lie in [math]\displaystyle{ K }[/math].
To make this more explicit, consider the polynomial evaluation [math]\displaystyle{ \varepsilon_a: K[X] \rightarrow K(a),\, P \mapsto P(a) }[/math]. This is a homomorphism and its kernel is [math]\displaystyle{ \{P \in K[X] \mid P(a) = 0 \} }[/math]. If [math]\displaystyle{ a }[/math] is algebraic, this ideal contains non-zero polynomials, but as [math]\displaystyle{ K[X] }[/math] is a euclidean domain, it contains a unique polynomial [math]\displaystyle{ p }[/math] with minimal degree and leading coefficient [math]\displaystyle{ 1 }[/math], which then also generates the ideal and must be irreducible. The polynomial [math]\displaystyle{ p }[/math] is called the minimal polynomial of [math]\displaystyle{ a }[/math] and it encodes many important properties of [math]\displaystyle{ a }[/math]. Hence the ring isomorphism [math]\displaystyle{ K[X]/(p) \rightarrow \mathrm{im}(\varepsilon_a) }[/math] obtained by the homomorphism theorem is an isomorphism of fields, where we can then observe that [math]\displaystyle{ \mathrm{im}(\varepsilon_a) = K(a) }[/math]. Otherwise, [math]\displaystyle{ \varepsilon_a }[/math] is injective and hence we obtain a field isomorphism [math]\displaystyle{ K(X) \rightarrow K(a) }[/math], where [math]\displaystyle{ K(X) }[/math] is the field of fractions of [math]\displaystyle{ K[X] }[/math], i.e. the field of rational functions on [math]\displaystyle{ K }[/math], by the universal property of the field of fractions. We can conclude that in any case, we find an isomorphism [math]\displaystyle{ K(a) \cong K[X]/(p) }[/math] or [math]\displaystyle{ K(a) \cong K(X) }[/math]. Investigating this construction yields the desired results.
This characterization can be used to show that the sum, difference, product and quotient of algebraic elements over [math]\displaystyle{ K }[/math] are again algebraic over [math]\displaystyle{ K }[/math]. For if [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] are both algebraic, then [math]\displaystyle{ (K(a))(b) }[/math] is finite. As it contains the aforementioned combinations of [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math], adjoining one of them to [math]\displaystyle{ K }[/math] also yields a finite extension, and therefore these elements are algebraic as well. Thus set of all elements of [math]\displaystyle{ L }[/math] which are algebraic over [math]\displaystyle{ K }[/math] is a field that sits in between [math]\displaystyle{ L }[/math] and [math]\displaystyle{ K }[/math].
Fields that do not allow any algebraic elements over them (except their own elements) are called algebraically closed. The field of complex numbers is an example. If [math]\displaystyle{ L }[/math] is algebraically closed, then the field of algebraic elements of [math]\displaystyle{ L }[/math] over [math]\displaystyle{ K }[/math] is algebraically closed, which can again be directly shown using the characterisation of simple algebraic extensions above. An example for this is the field of algebraic numbers.
See also
References
- Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4
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