Algebraic Numbers and Integers — Mathematics Knowledge System

Overview

An algebraic number is any complex number that is a root of some nonzero polynomial with rational coefficients; an algebraic integer is a root of a monic polynomial with integer coefficients. Every algebraic integer is an algebraic number, but not conversely (1/2 is algebraic — root of 2x−1 — but not an algebraic integer). The algebraic integers form a ring, generalizing ℤ, and this ring inside a number field is the central object of algebraic number theory, playing the role that ℤ plays inside ℚ. Numbers that are not algebraic at all, like π and e, are called transcendental, and their existence (proved by Liouville in 1844 and Hermite/Lindemann later in the 19th century) settled ancient questions like squaring the circle.

Intuition

Algebraic numbers are exactly the numbers that arise as solutions to polynomial equations with rational coefficients — this includes every rational number itself (root of qx−p), every root of an integer, and much more exotic combinations, but it excludes numbers like π that satisfy no polynomial relation whatsoever. Restricting further to MONIC integer polynomials singles out the algebraic integers, and this monic condition is exactly analogous to how an ordinary integer n is a root of the monic polynomial x−n, while a fraction like 1/2 is only a root of the non-monic 2x−1. The key surprising fact is that the algebraic integers form a ring (closed under addition and multiplication) even though this is not obvious from the definition — it requires showing that a sum or product of roots of monic integer polynomials is again a root of some monic integer polynomial, which uses the theory of finitely generated ℤ-modules.

Formal Definition

Definition

Let α ∈ ℂ. Define:

\alpha \text{ is algebraic} \iff \exists\, p(x) \in \mathbb{Q}[x],\ p \neq 0,\ p(\alpha) = 0

Algebraic number

\alpha \text{ is an algebraic integer} \iff \exists\, p(x) = x^n + c_{n-1}x^{n-1} + \cdots + c_0 \in \mathbb{Z}[x] \ \text{(monic)},\ p(\alpha)=0

Algebraic integer (monic, integer coefficients)

\overline{\mathbb{Q}} = \{\alpha \in \mathbb{C} : \alpha \text{ algebraic over } \mathbb{Q}\}, \qquad \mathcal{O}_{\overline{\mathbb{Q}}} = \{\alpha \in \mathbb{C} : \alpha \text{ an algebraic integer}\}

The field of algebraic numbers and the ring of algebraic integers

\mathcal{O}_{\overline{\mathbb{Q}}} \cap \mathbb{Q} = \mathbb{Z}

Rational algebraic integers are exactly the ordinary integers

Properties

Closure under +, ×, and field ops

\text{The algebraic numbers } \overline{\mathbb{Q}} \text{ form a field: closed under } +, -, \times, \div \text{ (by nonzero elements).}

Algebraic integers form a ring

\text{The algebraic integers form a subring of } \mathbb{C}\text{, closed under } + \text{ and } \times \text{ (but not division).}

Rational algebraic integers are ordinary integers

\alpha \in \mathbb{Q} \text{ and } \alpha \text{ an algebraic integer} \implies \alpha \in \mathbb{Z}.

Minimal polynomial characterization

\alpha \text{ algebraic} \implies \text{it has a unique monic minimal polynomial } m_\alpha(x) \in \mathbb{Q}[x] \text{ of least degree with } m_\alpha(\alpha)=0.

Countability

\overline{\mathbb{Q}} \text{ is countable (a countable union of finite root-sets over countably many polynomials), so almost all complex numbers are transcendental.}

Worked Examples

1/2 is a root of the rational polynomial 2x−1, so it is algebraic.
$2\left(\tfrac12\right) - 1 = 0$
For 1/2 to be an algebraic integer, it would need to be a root of SOME monic integer polynomial x^n+c_{n-1}x^{n-1}+\cdots+c_0. But by the Rational Root Theorem, any rational root p/q (in lowest terms) of a monic integer polynomial must have q=1, i.e. must be an integer.
$\alpha = p/q \in \mathbb{Q}, \ \alpha \text{ alg. integer} \implies q \mid 1$
Since 1/2 has denominator 2 ≠ 1 in lowest terms, it cannot be a root of any monic integer polynomial.
$\tfrac12 \notin \mathbb{Z} \implies \tfrac12 \text{ is not an algebraic integer}$

Answer: 1/2 is algebraic (root of 2x−1 ∈ ℚ[x]) but not an algebraic integer, since the Rational Root Theorem forces any rational algebraic integer to be a genuine integer.

Practice Problems

Difficulty 5/10

Is √5 an algebraic integer? Justify by exhibiting a monic integer polynomial it satisfies.

Difficulty 6/10

Find the minimal polynomial of α = 1 + √2 over ℚ, and verify it is monic with integer coefficients.

Difficulty 7/10

Explain why π is not algebraic (i.e. is transcendental), and what famous classical problem this settles.

Common Mistakes

Common Mistake

Assuming 'algebraic integer' just means 'algebraic number that happens to be real' or 'algebraic number that is an integer when simplified'.

It specifically means a root of a MONIC polynomial with INTEGER coefficients; e.g. 1/2 is algebraic but not an algebraic integer despite being a real, simple rational number.

Common Mistake

Thinking irrational and transcendental are the same concept.

√2 is irrational but algebraic (root of x²-2); transcendental is strictly stronger, meaning not algebraic at all — every transcendental number is irrational, but not conversely.

Quiz

Summary

An algebraic number is a root of some nonzero polynomial in ℚ[x]; an algebraic integer is a root of a monic polynomial in ℤ[x].
Every algebraic integer that happens to be rational is an ordinary integer — the Rational Root Theorem forces this.
The algebraic numbers form a field; the algebraic integers form a subring (closed under + and × but not division).
√2+√3 has minimal polynomial x⁴−10x²+1 and 1+√2 has minimal polynomial x²−2x−1 — both monic with integer coefficients, hence algebraic integers.
Numbers satisfying no polynomial at all, like π and e, are transcendental; π's transcendence (Lindemann, 1882) proved squaring the circle impossible.

References

WebsiteWikipedia — Algebraic number
WebsiteWikipedia — Algebraic integer
WebsiteWikipedia — Transcendental number