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Algebraic Integer — Definition, Formula & Examples

An algebraic integer is a complex number that is a root of some monic polynomial (leading coefficient 1) whose coefficients are all ordinary integers.

A complex number α\alpha is an algebraic integer if there exists a monic polynomial p(x)=xn+an−1xn−1+⋯+a1x+a0p(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 with ai∈Za_i \in \mathbb{Z} for all ii, such that p(α)=0p(\alpha) = 0. The set of all algebraic integers forms a ring, often denoted Z‾\overline{\mathbb{Z}}.

Key Formula

p(α)=αn+an−1αn−1+⋯+a1α+a0=0,ai∈Zp(\alpha) = \alpha^n + a_{n-1}\alpha^{n-1} + \cdots + a_1\alpha + a_0 = 0, \quad a_i \in \mathbb{Z}
Where:
  • α\alpha = The complex number being tested as an algebraic integer
  • nn = The degree of the monic polynomial
  • aia_i = Integer coefficients of the polynomial (the leading coefficient is implicitly 1)

How It Works

To show a number is an algebraic integer, you need to find a monic polynomial with integer coefficients that has that number as a root. Every ordinary integer nn qualifies because it is a root of x−n=0x - n = 0. Square roots like 5\sqrt{5} also qualify since x2−5=0x^2 - 5 = 0 is monic with integer coefficients. However, the fraction 12\frac{1}{2} is not an algebraic integer: its minimal polynomial is 2x−1=02x - 1 = 0, which is not monic when written with integer coefficients. The key requirement is that the leading coefficient must be exactly 1.

Worked Example

Problem: Show that the golden ratio φ=1+52\varphi = \frac{1 + \sqrt{5}}{2} is an algebraic integer.
Step 1: Write the equation that defines the golden ratio and rearrange it into a polynomial equation.
φ=1+52  ⟹  2φ−1=5  ⟹  (2φ−1)2=5\varphi = \frac{1 + \sqrt{5}}{2} \implies 2\varphi - 1 = \sqrt{5} \implies (2\varphi - 1)^2 = 5
Step 2: Expand and simplify to get a monic polynomial with integer coefficients equal to zero.
4φ2−4φ+1=5  ⟹  4φ2−4φ−4=0  ⟹  φ2−φ−1=04\varphi^2 - 4\varphi + 1 = 5 \implies 4\varphi^2 - 4\varphi - 4 = 0 \implies \varphi^2 - \varphi - 1 = 0
Step 3: The polynomial x2−x−1x^2 - x - 1 is monic (leading coefficient 1) with all integer coefficients, and φ\varphi is a root.
p(x)=x2−x−1,p(φ)=0p(x) = x^2 - x - 1, \quad p(\varphi) = 0
Answer: Since φ\varphi is a root of the monic integer-coefficient polynomial x2−x−1x^2 - x - 1, the golden ratio is an algebraic integer.

Why It Matters

Algebraic integers are central to algebraic number theory, where they generalize the ordinary integers to number fields like Q(−5)\mathbb{Q}(\sqrt{-5}). Unique factorization can fail in rings of algebraic integers, motivating the theory of ideals that underpins modern algebra and cryptographic applications.

Common Mistakes

Mistake: Confusing algebraic integers with algebraic numbers. For instance, claiming 12\frac{1}{2} is an algebraic integer because it satisfies 2x−1=02x - 1 = 0.
Correction: Every algebraic integer is an algebraic number, but not vice versa. The polynomial must be monic — its leading coefficient must be 1. Since the minimal polynomial of 12\frac{1}{2} over Z\mathbb{Z} is 2x−12x - 1, which is not monic, 12\frac{1}{2} is an algebraic number but not an algebraic integer.