The Gaussian integers showed the power of enlarging the integers with a new number, i. That same trick works for the rationals, too: take \mathbb{Q} and adjoin a new number like \sqrt 2 or i, and you get a bigger universe of numbers — closed under addition, subtraction, multiplication, and division just like \mathbb{Q} itself — called a number field.

Algebraic number theory makes this systematic: adjoin a root of any polynomial and study the arithmetic of the resulting world. Once you see it this way, ordinary integer arithmetic stops looking like the whole story and starts looking like the smallest, simplest case of something much bigger — the ground floor of a building with many more floors above it.

Algebraic numbers

A number is algebraic if it is a root of a polynomial with rational coefficients. Familiar examples include \sqrt{2} (root of x^2 - 2), i (root of x^2 + 1), and the golden ratio. Numbers that are not algebraic — like \pi and e — are called transcendental. Almost all real numbers are transcendental, yet algebraic numbers are the ones number theory can grasp and reason about precisely.

Worked example. Is 1 + \sqrt2 algebraic? Let x = 1 + \sqrt2, so x - 1 = \sqrt2. Squaring both sides gives x^2 - 2x + 1 = 2, i.e. x^2 - 2x - 1 = 0 — a polynomial with rational (in fact integer) coefficients that 1+\sqrt2 satisfies exactly. So yes: algebraic, even though it isn't itself a "nice" square root. This little trick — manipulate the expression until the irrational parts cancel and a polynomial equation falls out — is the standard way to confirm a new number belongs to the algebraic world at all.

What a number field is

A number field is what you get by adjoining an algebraic number \alpha to the rationals — the smallest field containing both \mathbb{Q} and \alpha, written \mathbb{Q}(\alpha). The simplest beyond \mathbb{Q} itself are the quadratic fields:

Worked example — closure. "Field" is a strong claim: it says you can add, subtract, multiply, and divide (by anything nonzero) and never leave the set. Check it on two elements. Take 1 + 2\sqrt2 and 3 - \sqrt2:

Both answers are still of the form a + b\sqrt2 with a, b \in \mathbb{Q} — nothing escaped. That's not a coincidence: because (\sqrt2)^2 = 2 is rational, every product of two elements of this form collapses back down to the same form. Division works too (multiply top and bottom by a - b\sqrt2 to clear the square root from the denominator), which is exactly what makes \mathbb{Q}(\sqrt2) a genuine field, not just a ring.

The Gaussian rationals

Adjoining i instead of \sqrt2 gives another quadratic field, \mathbb{Q}(i) = \{\,a + bi : a, b \in \mathbb{Q}\,\}, sometimes called the Gaussian rationals. It sits directly above the Gaussian integers \mathbb{Z}[i]: every Gaussian integer is a Gaussian rational (just take a, b to be whole numbers), but a Gaussian rational like \tfrac12 + \tfrac13 i is not a Gaussian integer at all — it has no whole-number coordinates.

Worked example. Multiply two Gaussian rationals the same way you'd multiply any complex numbers:

Here both factors happened to be Gaussian integers, and the product landed back on a Gaussian integer too — integer-times-integer always stays inside the smaller lattice. But mix in a fraction, like \left(\tfrac12 + i\right)(2 - i) = 1 - \tfrac12 i + 2i - i^2 = 2 + \tfrac32 i, and the result only has to land back in the full field \mathbb{Q}(i), not in the integer lattice — which is exactly the distinction the next section is about.

A field is bigger than its integers

Below, the plane is the complex numbers. Step through: first the tidy lattice of Gaussian integers — whole-number coordinates only — then a handful of other points that also belong to the field \mathbb{Q}(i) but land strictly between the lattice points, because their coordinates are fractions.

The degree of a number field

Both \mathbb{Q}(\sqrt2) and \mathbb{Q}(i) are finite-dimensional vector spaces over \mathbb{Q} — the idea of a basis and dimension from linear algebra, just with \mathbb{Q} playing the role of the scalars. Every element of \mathbb{Q}(\sqrt2) is built from exactly two "coordinates" against the basis \{1, \sqrt2\}; every element of \mathbb{Q}(i) from two coordinates against \{1, i\}. That count — 2 in both cases — is called the field's degree, written [\mathbb{Q}(\sqrt2):\mathbb{Q}] = 2.

Worked example — a bigger degree. Adjoin the real cube root of 2 instead: \mathbb{Q}(\sqrt[3]{2}). Now you need three basis elements to reach every number in the field — \{1, \sqrt[3]{2}, \sqrt[3]{4}\} — because \sqrt[3]{2} is a root of the degree-3 polynomial x^3 - 2, and cubing it again just produces \sqrt[3]{4}, still expressible in that basis, rather than collapsing back to something rational. So [\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3 — number fields can have any finite degree at all, not just two.

That word finite is doing real work. However exotic the algebraic number you adjoin, the resulting field never blows up to infinite dimension — it is always some finite-dimensional \mathbb{Q}-vector space, pinned down by one whole number, its degree. That finiteness is exactly what makes number fields tame enough to build a whole theory of "integers," "primes," and factorisation inside them — an infinite-dimensional mess would offer nowhere near as much structure to hold onto.

Why generalise this far

Many integer problems become transparent in the right number field. Sums of two squares live naturally in \mathbb{Q}(i): the identity 5 = 1^2 + 2^2 is really the factorisation 5 = (1+2i)(1-2i) in disguise, and questions about which numbers split as a sum of two squares become questions about which numbers factor a certain way in this bigger field. Pell's equation lives in \mathbb{Q}(\sqrt d), where its solutions turn out to be the "units" of a certain ring sitting inside the field; Fermat's equation was attacked in fields built from roots of unity. In each case, a hard problem about ordinary whole numbers turns into a more tractable problem about factoring inside a larger, friendlier number system — the same move, again and again, at the heart of algebraic number theory.

But to do arithmetic in this larger system — to talk about primes, factorisation, and divisibility, not just addition and multiplication — we first need the right notion of "integer" inside a number field: the algebraic integers, next.