The integers a field contains

The rational numbers contain the integers, and nobody needs telling which they are. A field built by adjoining $\sqrt 5$ to the rationals — every $a + b\sqrt5$ with rational $a$ and $b$ — contains an obvious candidate for its integers: the numbers with whole $a$ and $b$, a ring written $\mathbb{Z}[\sqrt5]$.

The candidate is wrong, and the golden ratio is the proof. $\varphi = (1 + \sqrt5)/2$ has a half in both coordinates, so it is not in $\mathbb{Z}[\sqrt5]$. But it satisfies

$$\varphi^2 = \varphi + 1,$$

a polynomial equation with whole-number coefficients and leading coefficient 1, and in every respect that matters for arithmetic it behaves as an integer does. The seven-powers argument found that $\sqrt2 + \sqrt[3]{3}$ also satisfies a polynomial with leading coefficient 1, and hinted that this was not an accident. This essay is about what that property is, which numbers have it, and why the numbers that have it are closed under addition — a fact that has no reason to be true and a very short reason why it is.

A definition that does not use coordinates #

A number is an algebraic integer when it is a root of a monic polynomial with whole-number coefficients — one whose leading coefficient is 1. The definition says nothing about how the number is written, and that is its point: coordinates depend on a choice of basis, and integrality should not.

The definition gives the right answer for the rationals. A rational root $p/q$ in lowest terms of a polynomial with whole coefficients has numerator dividing the constant term and denominator dividing the leading coefficient; if the polynomial is monic, the denominator divides 1.

So a rational number is an algebraic integer exactly when it is an ordinary integer. The new definition extends the old one without changing it, and that is the minimum a good definition of integer must do.

For numbers in $\mathbb{Q}(\sqrt5)$ it is easy to test. The number $(a + b\sqrt5)/2$, with $b \neq 0$, has minimal polynomial

$$x^2 - a\,x + \frac{a^2 - 5b^2}{4},$$

because its conjugate is $(a - b\sqrt5)/2$, the two add to $a$ and multiply to $(a^2 - 5b^2)/4$. The polynomial is already monic, so the number is an algebraic integer exactly when that constant term is a whole number.

$a^2 - 5b^2$ is divisible by 4 exactly when $a$ and $b$ have the same parity, since $5 \equiv 1$ modulo 4 and an odd square is 1 modulo 4. So the algebraic integers of $\mathbb{Q}(\sqrt5)$ are the numbers $(a + b\sqrt5)/2$ with $a \equiv b \pmod 2$, and those are precisely the whole combinations $m + n\varphi$. The integers of the field are $\mathbb{Z}[\varphi]$, and $\mathbb{Z}[\sqrt5]$ — the even–even squares of the checkerboard — is half of them.

A lattice, and its index #

The integers of a quadratic field have a geometric shape that makes the relationship between the two rings visible.

Place each number of the field at the point (number, conjugate) in the plane. The integers become a lattice — a grid, sheared and stretched, of points spaced evenly in two independent directions. The basic cell spanned by $1$ and $\varphi$ has area $|\varphi - \varphi'| = \sqrt5$. The cell of $\mathbb{Z}[\sqrt5]$, spanned by $1$ and $\sqrt5$, has area $2\sqrt5$: twice as large, so the smaller ring contains exactly half the points.

The square of that cell area — 5 for $\mathbb{Z}[\varphi]$, 20 for $\mathbb{Z}[\sqrt5]$ — is the discriminant, and it is the first invariant anyone computes for a ring of integers. For $\mathbb{Q}(\sqrt d)$ with $d$ square-free, the integers are $\mathbb{Z}[\sqrt d]$ when $d \equiv 2$ or $3 \pmod 4$ and $\mathbb{Z}[(1 + \sqrt d)/2]$ when $d \equiv 1 \pmod 4$, and the discriminant is $4d$ or $d$ accordingly. The golden ratio’s field is the second case, and so is the field of $\sqrt{-3}$, whose integers are the Eisenstein integers with their hexagonal lattice.

The same mistake, of taking whole coordinates for integers, happens in the quaternions, where the integers turn out to include the points with a half in every coordinate, for the same reason: a monic equation is satisfied by more than the grid that coordinates suggest.

Why the sum of two integers is an integer #

The definition raises a question with no obvious answer. If $\alpha$ satisfies one monic polynomial and $\beta$ another, why should $\alpha + \beta$ satisfy a monic polynomial at all? The dimension count shows it satisfies some polynomial, found as a dependency among its powers — but a dependency found by rational elimination has no reason to come out monic.

The argument that works replaces a vector space with a lattice and a dependency with a determinant.

Consider all whole combinations of $\alpha^i \beta^j$ for $i$ below the degree of $\alpha$ and $j$ below the degree of $\beta$. Because the minimal polynomials are monic, any higher power of $\alpha$ can be rewritten as a whole combination of lower ones — no division is ever needed — and the same for $\beta$. So multiplying any element of this finite list by $\alpha$ or by $\beta$ lands in whole combinations of the list again. Multiplication by $\alpha + \beta$ is therefore a square matrix $M$ of whole numbers acting on the list’s coefficients.

Now the determinant does the work. $(\alpha + \beta)$ times the vector of list elements equals $M$ times that vector, so $(\alpha + \beta)\,I - M$ kills a non-zero vector, so its determinant is nought:

$$\det\big((\alpha+\beta)\,I - M\big) = 0.$$

That determinant is the characteristic polynomial of $M$ evaluated at $\alpha + \beta$, and the characteristic polynomial of a whole-number matrix is monic with whole coefficients. So $\alpha + \beta$ is an algebraic integer, and the same argument with $\alpha\beta$ in place of $\alpha + \beta$ handles products. The algebraic integers form a ring. The argument is sometimes called the determinant trick, and it is the Cayley–Hamilton theorem used as a tool.

The seven-powers table has a second reading in this light. The minimal polynomial it found for $\sqrt2 + \sqrt[3]{3}$ came out monic with whole coefficients, and that was not luck: $\sqrt 2$ and $\sqrt[3]{3}$ are algebraic integers, so their sum is, and the minimal polynomial of an algebraic integer is always monic with whole coefficients — a theorem of Gauss’s about factoring monic polynomials, which says a monic whole-number polynomial cannot factor into monic pieces with fractional coefficients. The exact elimination had no way to know the answer should be integral, and produced integers because the number was one.

The squares and the hexagons #

Two other rings of integers are already familiar, and the monic definition picks them out without fuss.

For $\mathbb{Q}(i)$, with $i^2 = -1$, the field is $\mathbb{Q}(\sqrt{-1})$ and $-1 \equiv 3 \pmod 4$, so the integers are just the whole combinations $a + bi$: the Gaussian integers, a square lattice. A number $(a + bi)/2$ with $a$ and $b$ odd has norm $(a^2 + b^2)/4$, and $a^2 + b^2 \equiv 2 \pmod 4$ for odd $a, b$, so the norm is not a whole number and the number is not an integer. The square grid is exactly right, which is why a prime as a sum of two squares can be read off lattice points of the plain grid.

For $\mathbb{Q}(\sqrt{-3})$, with $-3 \equiv 1 \pmod 4$, the half-integers come back: $\omega = (-1 + \sqrt{-3})/2$ satisfies $\omega^2 + \omega + 1 = 0$, a monic equation, so it is an integer, and the ring of integers is $\mathbb{Z}[\omega]$ — the Eisenstein integers. Drawn in the complex plane they form a hexagonal lattice, the densest arrangement of discs in the plane, and the grid of whole combinations $a + b\sqrt{-3}$ is a sublattice of index two inside it, exactly as $\mathbb{Z}[\sqrt5]$ is inside $\mathbb{Z}[\varphi]$. The same congruence modulo 4 decides both, and in both it is the difference between a lattice with the right arithmetic and one missing half its points.

Integers stay on the lattice #

There is a quick test that separates integers from non-integers without computing any polynomial, and it is what the lattice picture suggests.

If $\alpha$ is an algebraic integer of degree $d$, every power $\alpha^k$ is a whole combination of $1, \alpha, \dots, \alpha^{d-1}$, because the monic polynomial rewrites high powers without division. So all the powers of an integer lie on one finitely generated lattice. The converse holds too: if all the powers of $\alpha$ lie in some lattice of finite rank, then multiplication by $\alpha$ is a whole-number matrix on that lattice, and the determinant trick makes $\alpha$ an integer.

The powers of $\varphi$ are Fibonacci combinations, because $\varphi^{k+1} = \varphi^k + \varphi^{k-1}$ is the Fibonacci recurrence read in the coordinates $1, \varphi$. The powers of $\varphi/2$ satisfy $4x^2 = 2x + 1$, which cannot be made monic without fractions, and their coordinates need ever larger powers of two in the denominator — no lattice holds them all. That is the same fact that makes the powers of the point $(3 + 4i)/5$ on the unit circle wander for ever instead of repeating: it is not an algebraic integer, its powers need a denominator of $5^k$, and no lattice holds them; whereas an algebraic integer whose conjugates all lie on the unit circle has powers trapped among finitely many lattice points, and must come back round as a root of unity.

Units, and a Pell equation #

In the ordinary integers only $1$ and $-1$ have reciprocals that are integers. In $\mathbb{Z}[\varphi]$ there are infinitely many, and they are all visible in one table.

The norm of $\alpha$ is $\alpha$ times its conjugate, a whole number for an integer, and it multiplies: the norm of a product is the product of the norms. An integer has an integer reciprocal exactly when its norm is $\pm1$, since then the conjugate divided by the norm is the reciprocal. The norm of $\varphi$ is $\varphi\varphi' = -1$, so $\varphi$ is a unit, and so is every power of it, positive or negative. Dirichlet’s unit theorem says that for a real quadratic field there is nothing else: every unit is $\pm\varphi^k$.

In $\mathbb{Z}[\sqrt5]$ only a third of those units survive. A unit $x + y\sqrt5$ with whole $x, y$ is a solution of $x^2 - 5y^2 = \pm1$, and the smallest, $2 + \sqrt5$, is $\varphi^3$. That is the Pell equation, generated by one solution, and the table shows its solutions sitting inside a larger group generated by $\varphi$, of which the Pell solutions are the cubes. The larger ring has the simpler unit group, as it has the simpler lattice, the smaller discriminant and — most importantly — the better arithmetic.

Why the larger ring is the right one #

The reason to insist on $\mathbb{Z}[\varphi]$ rather than $\mathbb{Z}[\sqrt5]$ is not taste. It is that factorisation works in one and fails in the other.

In $\mathbb{Z}[\sqrt5]$ the number 4 factors as $2 \cdot 2$ and also as $(1 + \sqrt5)(-1 + \sqrt5)$, and neither $2$ nor $1 + \sqrt5$ can be broken further within that ring — two genuinely different factorisations into irreducibles. In $\mathbb{Z}[\varphi]$ the second factorisation dissolves, because $1 + \sqrt5 = 2\varphi$ and $-1 + \sqrt5 = 2\varphi^{-1}$, each twice a unit, so both factorisations are $2 \cdot 2$ up to units. $\mathbb{Z}[\varphi]$ has unique factorisation, and $\mathbb{Z}[\sqrt5]$ does not.

This is the general pattern, and it is why the ring of all algebraic integers in a field — the maximal order — is the one number theory uses. Smaller rings like $\mathbb{Z}[\sqrt5]$ are missing elements that the monic definition supplies, and the missing elements are exactly what is needed to repair factorisation. Even the maximal order can fail to factor uniquely — $\mathbb{Z}[\sqrt{-5}]$ is the standard example, where $6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5})$ and no larger ring of integers exists to fix it — and Dedekind’s response was to factor ideals instead of numbers, which restores uniqueness in every ring of integers without exception.

The failure is measured by a finite group, the class group, whose size is the class number: 1 when factorisation is unique, 2 for $\mathbb{Z}[\sqrt{-5}]$. And the class group is the same object Gauss had already met from the other side, as the set of genuinely different quadratic forms $ax^2 + bxy + cy^2$ of a given discriminant, composed by a rule he found without any ideals at all. That correspondence is why which primes a form takes is a question about rings of integers in disguise: a prime is represented by the form $x^2 + 5y^2$ when it factors into principal ideals in $\mathbb{Z}[\sqrt{-5}]$, and by the other form of that discriminant, $2x^2 + 2xy + 3y^2$, when its factors are not principal. Two forms, class number two, and a prime’s behaviour decided by which class its factors fall in.

For $\mathbb{Q}(\sqrt5)$ itself the group is trivial, which is why nothing about $\mathbb{Z}[\varphi]$ above has needed ideals. Every non-zero element of $\mathbb{Z}[\varphi]$ that is not a unit is a product of irreducibles in one way only, up to multiplying the factors by powers of $\varphi$ and by $-1$. That short list of units is the whole of the ambiguity.

What the lattices and tables cannot show #

The lattice figure plots points with small coordinates and the cells that tile the plane with them. That the whole of $\mathbb{Z}[\varphi]$ is this lattice, and nothing else, rests on the parity argument, which the grid confirms for 169 numbers and proves for none.

The determinant trick is shown for one pair, $\sqrt2$ and $\varphi$, where the matrices are four by four. For a pair of degrees $m$ and $n$ the matrix is $mn$ by $mn$ and cannot be drawn for anything interesting, and the argument’s force is that its size does not matter: a whole-number matrix of any size has a monic whole-number characteristic polynomial. The polynomial it produces is also not always the minimal one — for $\sqrt2 + \varphi$ it happens to be, at degree four, but for a sum whose parts are related the determinant still gives a monic polynomial of the full size, of which the minimal polynomial is only a factor. The trick proves integrality; it does not find the smallest equation.

And the claim about unique factorisation is stated, not drawn. Nothing in the figures factors anything; the counterexample in $\mathbb{Z}[\sqrt5]$ and its repair in $\mathbb{Z}[\varphi]$ are checked by the arithmetic in the text, and the general theorem that $\mathbb{Z}[\varphi]$ factors uniquely rests on its being a Euclidean ring, which is a separate argument.

Still open: how many real quadratic fields factor uniquely #

The quadratic fields whose integers factor uniquely are the fields of class number one. For imaginary fields — $\mathbb{Q}(\sqrt{-d})$ — the question is completely settled: there are exactly nine, the largest being $d = 163$, a list Gauss conjectured and Heegner, Baker and Stark proved complete in the twentieth century.

For real fields like $\mathbb{Q}(\sqrt5)$ the situation is the opposite. Class number one appears to be common — computations find it for a large majority of prime discriminants — and Gauss’s conjecture that infinitely many real quadratic fields have class number one is still unproved. The difference is the unit group. Imaginary fields have only finitely many units, and class numbers grow; real fields have infinitely many, generated by one fundamental unit like $\varphi$, and a large fundamental unit can keep the class number small. Controlling how large those units get is the obstacle, and it is the same unit whose powers fill the table above.

An integer is something a monic equation holds #

The picture of integers as the points with whole coordinates is the one everybody starts with, and it is the wrong picture as soon as the coordinates are not the integers’ own. The right picture is a lattice defined by a property — satisfying a monic equation with whole coefficients — and that property turns out to be closed under addition and multiplication by a determinant, to give back the ordinary integers when applied to rational numbers, and to supply exactly the extra elements that make arithmetic in a field behave.

The golden ratio, with its half in both coordinates, is the smallest example of an integer the coordinates miss. Its powers stay on the lattice as Fibonacci numbers, its norm is $-1$, and it generates every unit its field has — none of which $\mathbb{Z}[\sqrt5]$, without it, can say.