Quadratic Integers

The Gaussian integers $a + bi$ are one ring in an infinite family. Pick any integer $d$ that is not a square. The field $\mathbb{Q}(\sqrt d)$ has a ring of integers $\mathcal{O}_d$: the numbers $x + y\omega$ with $x, y \in \mathbb{Z}$, where

$$\omega = \begin{cases} \dfrac{-1 + \sqrt d}{2} & d \equiv 1 \pmod 4, \\[1ex] \sqrt d & \text{otherwise.} \end{cases}$$

For $d = -1$ this is $\mathbb{Z}[i]$. For $d = -3$ it is the Eisenstein integers, whose $\omega = e^{2\pi i/3}$ is a cube root of unity. A $d$ with square factors names the same field as its squarefree part, so $\mathbb{Q}(\sqrt 8) = \mathbb{Q}(\sqrt 2)$.

Each element is a point of a lattice, and the lattice is where the arithmetic is easiest to see. Every plot on this page is a Show of one LatticeTiles layer, its tiles colored by ColorRules. Drag to pan; ⌘/Ctrl-scroll to zoom; hover a tile to read its factorization.

The norm

Multiplying by the conjugate gives the norm:

$$N(x + y\omega) = (x + y\omega)(x + y\bar\omega) = x^2 + sxy - ry^2, \qquad \omega^2 = s\omega + r.$$

The norm is a rational integer, and it is multiplicative: $N(\alpha\beta) = N(\alpha)N(\beta)$. A factorization of $\alpha$ therefore gives a factorization of $N(\alpha)$. So an element whose norm is a rational prime cannot factor. The units are the elements of norm $\pm 1$.

#Gaussian primes, colored by norm

source

<Show GridLines="[10, 10]" Axes ImageSize="[Automatic, 440]">
  <LatticeTiles
    ColorRules='[IsPrime -> ColorData(["Dusk", [0, 10], "Reflected"])(Sqrt(Abs(Norm))), IsUnit -> White]'
    BoundaryStyle="[IsZero -> White]"
  >
    GaussianIntegers
  </LatticeTiles>
</Show>

A prime's color comes from its norm, through a band. The gradient runs from its first color at $\sqrt{|N|} = 0$ to its last at $\sqrt{|N|} = 10$, back to its first at $20$, and so on. The band is the grid step, so the colors turn at the grid's own spacing; the legend says where. A band is a choice of presentation, not of mathematics: pick another scheme from the legend's bar, or have it wrap or stop (↺) instead of reflecting back. For $\mathbb{Z}[i]$, $\sqrt{|N|}$ is the distance from 0, so the bands are rings. True scale (AspectRatio -> Automatic) makes no difference here, because $\mathbb{Z}[i]$ is already square.

Square and hexagonal

When $d \equiv 1 \pmod 4$, $\omega$ sits halfway along. The lattice is then centered, and each element owns a hexagon. The Eisenstein integers tile the plane with regular hexagons, and their primes have six-fold symmetry because there are six units.

#Eisenstein primes

source

<Show GridLines="[10, 10]" Axes ImageSize="[Automatic, 440]">
  <LatticeTiles
    ColorRules='[IsPrime -> ColorData(["Viridis", [0, 10], "Reflected"])(Sqrt(Abs(Norm))), IsUnit -> White]'
    BoundaryStyle="[IsZero -> White]"
  >
    EisensteinIntegers
  </LatticeTiles>
</Show>

By default the plot measures the non-real axis in units of $\omega$'s own height. Every ring then gets square or regular-hexagonal tiles, and fields can be compared side by side. True scale, AspectRatio -> Automatic, draws $x + y\sqrt d$ at $(x, y\sqrt{|d|})$ instead. For $d < 0$ that is the complex plane itself, so the lattice lines up with a complex plot of the same view.

When factorization is not unique

In $\mathbb{Z}[\sqrt{-5}]$,

$$6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}),$$

and none of the four factors splits any further. They are irreducible, but none of them is prime: $2$ divides the product $(1 + \sqrt{-5})(1 - \sqrt{-5})$ without dividing either factor.

Ideals repair this. The ideal $(\alpha)$ always factors uniquely into prime ideals; here

$$(6) = \mathfrak{p}_2^2\, \mathfrak{p}_3\, \mathfrak{p}_3', \qquad \mathfrak{p}_2 = (2, 1 + \sqrt{-5}),\ \mathfrak{p}_3 = (3, 1 + \sqrt{-5}),\ \mathfrak{p}_3' = (3, -1 + \sqrt{-5}).$$

None of these is principal. The elements $2$, $3$ and $1 \pm \sqrt{-5}$ are four different ways of grouping them into principal products.

The plot paints irreducibles that are not prime teal, a color that sits apart from every color of the gradient, so an irreducible never reads as a prime of some norm. $6$ is selected, and every irreducible divisor of the selection lights up, from both factorizations at once; hover it to read both factorizations and its ideals. Shift-click to select more.

#ℤ[√−5]: primes by norm, and irreducibles that are not prime

source

<Show
  Variables="[_s -> Variable(Automatic, [(6, 0)])]"
  Selection="_s"
  GridLines="[10, 10]"
  Axes
  ImageSize="[Automatic, 440]"
>
  <LatticeTiles
    ColorRules='[
      IsPrime -> ColorData(["Dusk", [0, 10], "Reflected"])(Sqrt(Abs(Norm))),
      IsIrreducible && !IsPrime -> Teal,
      IsUnit -> White,
      IrreducibleFactors(Selected) -> Opacity(0.4, White)
    ]'
    ColorMixing='"Screen"'
    BoundaryStyle="[IsZero -> White, Selected -> Directive(White, AbsoluteThickness(2.5))]"
  >
    <QuadraticIntegers>-5</QuadraticIntegers>
  </LatticeTiles>
</Show>

How badly unique factorization fails is measured by the class group, the ideals modulo the principal ones. Its order $h$ is the class number. An element is irreducible exactly when no proper sub-product of its prime ideals is principal. That is how the plot decides, with ideal classes computed as reduced binary quadratic forms. When $h = 1$ (for $d < 0$, only $d = -1, -2, -3, -7, -11, -19, -43, -67, -163$), irreducible and prime coincide and that color disappears.

Real fields

For $d > 0$ the norm $x^2 - dy^2$ takes both signs, and the units form an infinite group $\pm\varepsilon^k$ around a fundamental unit $\varepsilon$, from the continued fraction of $\omega$. A prime's associates run off along the hyperbolas $N = \pm p$, so the primes arrange themselves along those curves.

#ℚ(√229): a real field with class number 3

source

<Show GridLines="[10, 10]" Axes ImageSize="[Automatic, 440]">
  <LatticeTiles
    ColorRules="[
      Splits -> Teal,
      Inert -> Gold,
      Ramified -> Red,
      IsIrreducible && !IsPrime -> Purple,
      IsUnit -> White
    ]"
    BoundaryStyle="[IsZero -> White]"
  >
    <QuadraticIntegers>229</QuadraticIntegers>
  </LatticeTiles>
</Show>

Each prime is colored by how the rational prime $p$ below it splits: teal when it splits, gold when it stays inert, red when it ramifies. A prime $p$ either splits into two conjugate ideals (norm $\pm p$), stays inert (the prime is $p$ itself, norm $p^2$) or ramifies (it divides the discriminant). The Kronecker symbol $\left(\frac{D}{p}\right)$ decides which.

The logarithmic embedding

A real field has two real embeddings, $\sigma_1$ and $\sigma_2$, sending $\sqrt d$ to $\pm\sqrt d$. Plot each element at $(\log|\sigma_1(\alpha)|, \log|\sigma_2(\alpha)|)$ and the multiplicative structure turns additive. The norm is $\sigma_1\sigma_2$, so the elements of norm $\pm n$ lie on the line $x + y = \log n$, and the units lie on the antidiagonal. Multiplying by the fundamental unit $\varepsilon$ is a translation along it, by $(\log\varepsilon, -\log\varepsilon)$: every norm line repeats with period the regulator $\log\varepsilon$.

#ℚ(√5), logarithmically

source

<Show GridLines="[1, 1]" Axes ImageSize="[Automatic, 440]">
  <LatticeTiles
    Embedding="Logarithmic"
    ColorRules='[
      IsUnit -> White,
      IsPrime -> ColorData(["Dusk", [0, 4], "Reflected"])(Log(Abs(Norm))),
      IsZero -> White
    ]'
  >
    <QuadraticIntegers>5</QuadraticIntegers>
  </LatticeTiles>
</Show>

The units march down the antidiagonal at steps of $\log\varphi \approx 0.48$, $\varphi$ the golden ratio. Each prime's associates do the same on its own line. Only the elements of norm up to 200 are drawn, from a finite box of the lattice, so the lines thin out toward their ends.

Orders

$\mathcal{O}_d$ is the largest of the rings in $\mathbb{Q}(\sqrt d)$. The others are its orders $\mathbb{Z} + f\mathcal{O}_d$, named by their discriminant $f^2 D_K$: QuadraticOrder(-12) is $\mathbb{Z}[\sqrt{-3}]$, of conductor 2 in the Eisenstein integers. It has the square lattice of $\sqrt{-3}$, and loses unique factorization at 2: the purple tiles are irreducible but not prime.

#ℤ[√−3], the order of conductor 2

source

<Show GridLines="[10, 10]" Axes ImageSize="[Automatic, 440]">
  <LatticeTiles
    ColorRules="[
      IsIrreducible && !IsPrime -> Purple,
      Splits -> Teal,
      Inert -> Gold,
      Ramified -> Red,
      IsUnit -> White
    ]"
    BoundaryStyle="[IsZero -> White]"
  >
    <QuadraticOrder>-12</QuadraticOrder>
  </LatticeTiles>
</Show>

Higher degrees, and orders in general, are in Orders and Algebraic Integers.

Sweeping fields

The quadratic primes explorer has the same component, with a stepper and a random button to move through fields; any figure here goes full-window from its corner button.

The same lattices carry numbers written in a complex base: base $-1 + i$ with digits $0$ and $1$ writes every Gaussian integer and tiles a fractal. That is the Complex bases guide.