Algebraic number: Difference between revisions

{{Short description|Complex number that is a root of a non-zero polynomial in one variable with rational coefficients}}

In [[mathematics]], an '''algebraic number''' is a number that is a [[root of a function|root]] of a non-zero [[polynomial]] in one variable with [[integer]] (or, equivalently, [[Rational number|rational]]) coefficients.  For example, the [[golden ratio]] <math>(1 + \sqrt{5})/2</math> is an algebraic number, because it is a root of the polynomial <math>X^2 - X - 1</math>, i.e., a solution of the equation <math>x^2 - x - 1 = 0</math>, and the [[complex number]] <math>1 + i</math> is algebraic as a root of <math>X^4 + 4</math>. Algebraic numbers include all [[integer]]s, [[rational number]]s, and [[nth root|''n''-th roots of integers]].

Numbers which are not algebraic are called [[transcendental number|transcendental]] and include [[pi|{{pi}}]] and {{mvar|[[e (mathematical constant)|e]]}}. There are [[countable set|countably many]] algebraic numbers, hence [[almost all]] real (or complex) numbers (in the sense of [[Lebesgue measure]]) are transcendental.

For any {{math|&alpha;}}, the [[simple extension]] of the rationals by {{math|&alpha;}}, denoted by <math>\Q(\alpha)</math> (whose elements are the <math>f(\alpha)</math> for <math>f</math> a [[rational function]] with rational coefficients which is defined at <math>\alpha</math>), is of finite [[Degree of a field extension|degree]] if and only if {{math|&alpha;}} is an algebraic number.

The condition of finite degree means that there is a finite set <math>\{a_i | 1\le i\le k\}</math> in <math>\Q(\alpha)</math> such that <math>\Q(\alpha) = \sum_{i=1}^k a_i \Q</math>; that is, every member in <math>\Q(\alpha)</math> can be written as <math>\sum_{i=1}^k a_i q_i</math> for some rational numbers <math>\{q_i | 1\le i\le k\}</math> (note that the set <math>\{a_i\}</math> is fixed).

Indeed, since the <math>a_i-s</math> are themselves members of <math>\Q(\alpha)</math>, each can be expressed as sums of products of rational numbers and powers of {{math|&alpha;}}, and therefore this condition is equivalent to the requirement that for some finite <math>n</math>, <math>\Q(\alpha) = \{\sum_{i=-n}^n \alpha^{i} q_i | q_i\in \Q\}</math>.

The latter condition is equivalent to <math>\alpha^{n+1}</math>, itself a member of <math>\Q(\alpha)</math>, being expressible as <math>\sum_{i=-n}^n \alpha^i q_i</math> for some rationals <math>\{q_i\}</math>, so <math>\alpha^{2n+1} = \sum_{i=0}^{2n} \alpha^i q_{i-n}</math> or, equivalently, {{math|&alpha;}} is a root of <math>x^{2n+1}-\sum_{i=0}^{2n} x^i q_{i-n}</math>; that is, an algebraic number with a minimal polynomial of degree not larger than <math>2n+1</math>.

For any two algebraic numbers {{math|&alpha;}}, {{math|&beta;}}, this follows directly from the fact that the [[simple extension]] <math>\Q(\gamma)</math>, for <math>\gamma</math> being either <math>\alpha+\beta</math>, <math>\alpha-\beta</math>, <math>\alpha\beta</math> or (for <math>\beta\ne 0</math>) <math>\alpha/\beta</math>, is a [[linear subspace]] of the finite-[[Degree of a field extension|degree]] field extension <math>\Q(\alpha,\beta)</math>, and therefore has a finite degree itself, from which it follows (as shown [[#Degree of simple extensions of the rationals as a criterion to algebraicity|above]]) that <math>\gamma</math> is algebraic.

That the field of algebraic numbers is algebraically closed can be proven as follows: Let {{math|&beta;}} be a root of a polynomial <math> \alpha_0 + \alpha_1 x + \alpha_2 x^2 ... +\alpha_n x^n</math> with coefficients that are algebraic numbers <math>\alpha_0</math>, <math>\alpha_1</math>, <math>\alpha_2</math>... <math>\alpha_n</math>. The field extension <math>\Q^\prime \equiv \Q(\alpha_1, \alpha_2, ... \alpha_n)</math> then has a finite degree with respect to <math>\Q</math>. The simple extension <math>\Q^\prime(\beta)</math> then has a finite degree with respect to <math>\Q^\prime</math> (since all powers of {{math|&beta;}} can be expressed by powers of up to <math>\beta^{n-1}</math>). Therefore, <math>\Q^\prime(\beta) = \Q(\beta, \alpha_1, \alpha_2, ... \alpha_n)</math> also has a finite degree with respect to <math>\Q</math>. Since <math>\Q(\beta)</math> is a linear subspace of <math>\Q^\prime(\beta)</math>, it must also have a finite degree with respect to <math>\Q</math>, so {{math|&beta;}} must be an algebraic number.

  | last = Garibaldi | first = Skip

*{{citation |last1=Hardy |first1=Godfrey Harold |author-link1=G. H. Hardy |last2=Wright |first2=Edward M. |author-link2=E. M. Wright |date=1972 |title=An introduction to the theory of numbers |edition=5th  |location=Oxford |publisher=Clarendon|isbn=0-19-853171-0}}