transfinite arithmetic, cardinal arithmetic, ordinal arithmetic
prime field, p-adic integer, p-adic rational number, p-adic complex number
arithmetic geometry, function field analogy
An algebraic number is a root of a polynomial with integer coefficients (or, equivalently, with rational coeffients). Equivalently, an element $\alpha$ of a field extension $K$ of the rational numbers $\mathbb{Q}$ is algebraic if the subfield $\mathbb{Q}(\alpha)$ is a finite degree extension, i.e., is finite-dimensional as a vector space over $\mathbb{Q}$.
Since the rational numbers are a subfield of the complex numbers, and since the complex numbers are an algebraically closed field, algebraic numbers are naturally regarded as a sub-field of complex numbers
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Visualisation of the (countable) field of algebraic numbers in the complex plane. Colours indicate degree of the polynomial the number is a root of (red = linear, i.e. the rationals, green = quadratic, blue = cubic, yellow = quartic…). Points becomes smaller as the integer polynomial coefficients become larger. View shows integers 0,1 and 2 at bottom right, $+i$ near top.
due to Stephen J. Brooks here
But the collection of all algebraic numbers forms itself already an algebraically closed field, typically denoted $\overline{\mathbb{Q}}$, as this is the algebraic closure of the field $\mathbb{Q}$ of rational numbers. This also follows easily from the equivalent definition of algebraic numbers in terms of finite degree extensions. The absolute Galois group $Gal(\overline{\mathbb{Q}}, \mathbb{Q})$ is peculiar, see there.
A number (especially a complex number) which is not algebraic is called transcendental; famous examples are the base ($\mathrm{e} = 2.7\ldots$) and period ($2 \pi \mathrm{i} = 6.28\ldots \mathrm{i}$, or equivalently $\pi = 3.14\ldots$) of the natural logarithm.
Given a field $k$, an algebraic number field $K$ over $k$ is a finite-degree extension of $k$. By default, the term “algebraic number field” means an algebraic number field over the rational numbers. If $\alpha$ is an algebraic number over $\mathbb{Q}$ then $\mathbb{Q}[\alpha]$ is a number field, however the field of all algebraic numbers is not a number field.
An algebraic integer is a root of a monic polynomial with integer coefficients. Equivalently, an element $\alpha$ of a field extension $K$ of $\mathbb{Q}$ is an algebraic integer if the ring $\mathbb{Z}[\alpha]$ is of finite rank as a $\mathbb{Z}$-module. It follows easily from this characterization that the collection of all algebraic integers forms a commutative ring.
algebraic number, algebraic integer
See also
Last revised on October 1, 2018 at 23:40:21. See the history of this page for a list of all contributions to it.