Step outside \mathbb{Q} into a bigger number field — say \mathbb{Q}(\sqrt 5) or \mathbb{Q}(i) — and a question that seemed settled forever comes back to life: which of these new numbers deserve to be called "integers"? Inside \mathbb{Q} the answer was obvious — the whole numbers, full stop. Inside a bigger field, full of numbers like \sqrt 5 and \tfrac{1+\sqrt5}{2}, "obvious" runs out fast.

We need a rule that (a) picks out exactly the ordinary integers when applied back inside \mathbb{Q}, and (b) is precise enough to work in any number field, however exotic. The rule that does the job is the algebraic integer, and it is one of the cleverest small definitions in all of mathematics: get it slightly wrong, and every later theorem about arithmetic in number fields collapses.

It's tempting to guess that "integer" should just mean "not a fraction" — but that guess turns out to be far too crude once irrational numbers are in the mix. We need a test built entirely out of equations, one that can be applied uniformly to any number in any number field, however unfamiliar it looks. That is exactly what the definition below delivers.

The definition

Every ordinary integer k qualifies trivially — it is the root of x - k = 0, which is monic with integer coefficients. So the new definition never throws away anything we already had; it only asks whether it can be extended to catch new numbers living in bigger fields. Restricted back to \mathbb{Q}, the algebraic integers turn out to be exactly \mathbb{Z} — no more, no fewer. That is the check that tells us the definition is the right generalisation, not just a plausible-looking one.

Worked example: is \sqrt 2 an algebraic integer?

To show a number is an algebraic integer, all we have to do is exhibit one monic integer polynomial that it satisfies — we don't need to check every polynomial, just find one witness.

Take \alpha = \sqrt 2. Square both sides of \alpha = \sqrt 2 to clear the root:

The polynomial x^2 - 2 is monic (leading coefficient 1) and every coefficient (1, 0, -2) is an ordinary integer. Since \sqrt 2 is a root of it, \sqrt 2 is an algebraic integer — even though it is famously irrational. Being an algebraic integer has nothing to do with being a "nice" decimal; it is about which polynomial equations a number can satisfy.

Worked example: the golden ratio

Now try something that looks far stranger at first glance: the golden ratio \varphi = \tfrac{1+\sqrt5}{2} \approx 1.618, the number the ancient Greeks prized for its uniquely balanced proportions. Is it an algebraic integer?

Start from \varphi = \tfrac{1+\sqrt5}{2}, so 2\varphi - 1 = \sqrt 5. Square both sides:

Divide through by 4 to make it monic:

The polynomial x^2 - x - 1 is monic with integer coefficients (1, -1, -1), and \varphi is a root. So the golden ratio — a number built from a square root divided by two — is, perhaps surprisingly, an algebraic integer. This is the first hint that the "integers" of a number field can be a richer, stranger set than the naive guess a + b\sqrt5 with a, b \in \mathbb{Z} would suggest.

More generally, whenever d \equiv 1 \pmod 4, the field \mathbb{Q}(\sqrt d) hides these extra "half-integer" algebraic integers of the form \tfrac{a + b\sqrt d}{2} (with a, b both odd). For d = 5 the full ring of integers is

strictly larger than \mathbb{Z}[\sqrt 5]. Guess the ring of integers wrong — miss this hidden half — and every later arithmetic argument in \mathbb{Q}(\sqrt 5) breaks.

Worked example: why \tfrac12 is NOT an algebraic integer

Contrast this with an ordinary fraction. Could \tfrac12 be a root of some monic integer polynomial? Suppose it were — suppose

Multiply through by 2^{n} to clear denominators:

Every term after the leading 1 is a multiple of 2, so the whole left-hand side is 1 plus an even number — always odd, and therefore never equal to 0. No such polynomial can exist. This is really a special case of the rational root theorem: any rational root of a monic integer polynomial must itself be an integer, because the leading coefficient 1 leaves the denominator nowhere to hide. \tfrac12 is not an integer, so it fails — it simply isn't "whole" enough to belong.

Beyond square roots: any degree works

Nothing in the definition limits us to quadratic examples. Take \alpha = \sqrt[3]{2}, the real cube root of 2. Cubing both sides of \alpha = \sqrt[3]2 gives

and x^3 - 2 is monic with integer coefficients, so \sqrt[3]2 is an algebraic integer living in the degree-3 field \mathbb{Q}(\sqrt[3]2). The same idea scales to any degree at all: a root of any monic x^n + c_{n-1}x^{n-1} + \cdots + c_0 with integer c_i is an algebraic integer, whatever n is. The smallest such n for a given \alpha — the degree of its minimal polynomial — is called the degree of the algebraic integer; \sqrt 2 has degree 2, \sqrt[3]2 has degree 3, and every ordinary integer has degree 1.

They form a ring

A single algebraic integer is a curiosity; the real power shows up once you collect all of them inside a number field K. Remarkably, they are closed under both addition and multiplication — add or multiply two algebraic integers of K and you land on another one. (This is not obvious at all: it takes real work to show that a sum of two roots of possibly different monic polynomials is again a root of some monic integer polynomial.) That closure means the algebraic integers of K form the ring of integers \mathcal{O}_K — the proper stage on which all arithmetic in K is done. In \mathbb{Q}(i) this ring is the familiar Gaussian integers; in \mathbb{Q}(\sqrt 5) it is the golden-ratio ring \mathbb{Z}[\varphi] from above.

Once you have a well-behaved ring of "integers" inside every number field, the next natural question is whether they arithmetic like the ordinary integers — above all, whether every element factors uniquely into primes. The honest answer is a genuine shock: sometimes it does not, and fixing that is the story of ideals and unique factorisation, next.