Editing Algebraic number (section)

==Field==
[[File:Algebraic number in the complex plane.png|thumb|Algebraic numbers colored by degree (blue = 4, cyan = 3, red = 2, green = 1). The unit circle is black.{{explain|reason=What does this figure tell us about algebraic numbers? Can we get some insight out of it, or it this just mathematical art?|date=July 2024}}]]
The sum, difference, product, and quotient (if the denominator is nonzero) of two algebraic numbers is again algebraic:

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.

An alternative way of showing this is constructively, by using the [[resultant]].

Algebraic numbers thus form a [[field (mathematics)|field]]{{sfn|Niven|1956|p=92}} <math>\overline{\mathbb{Q}}</math> (sometimes denoted by <math>\mathbb A</math>, but that usually denotes the [[adele ring]]).

===Algebraic closure===
Every root of a polynomial equation whose coefficients are ''algebraic numbers'' is again algebraic. That can be rephrased by saying that the field of algebraic numbers is [[algebraically closed field|algebraically closed]]. In fact, it is the smallest algebraically closed field containing the rationals and so it is called the [[algebraic closure]] of the rationals.

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.