The class number is a degree

Units that form a lattice ended at a formula in which the class number and the regulator of a field appear only as a product, so that the hardest thing to compute about a field’s units hides the hardest thing to compute about its ideals. For the fields $\mathbb{Q}(\sqrt{-d})$ there are no units to hide behind — the regulator is 1 — and the class number can be counted directly. Counting the classes that break factorisation did that: every positive form $ax^2 + bxy + cy^2$ of discriminant $D = b^2 - 4ac$ is a point $\tau = (-b + \sqrt{D})/2a$ of the upper half-plane, the reduced forms $|b| \le a \le c$ are exactly the points in one small region, and the class number $h$ is how many there are.

That makes $h$ a count. This essay is about a stranger description of the same number, which turns it into a degree. There is a single function on the upper half-plane whose value at the $h$ reduced points are $h$ numbers that belong together as the roots of one polynomial with whole-number coefficients — a polynomial of degree exactly $h$, and irreducible, so that each of its roots is an algebraic number of degree $h$. A field with one class gets a root of degree one, which is to say a whole number, and the whole number leaves a fingerprint visible to anybody with a calculator.

A function that cannot tell equivalent forms apart #

Two forms are equivalent when a change of variables of determinant one turns one into the other, and on the points $\tau$ such a change acts as $\tau \mapsto (p\tau + q)/(r\tau + s)$ with whole $p, q, r, s$ and $ps - qr = 1$ — the modular group. A function on the half-plane that does not change under any of these moves takes one value on each class of forms, so it is a function of the class rather than of the form.

The simplest such function is Klein’s $j$. Write $q = e^{2\pi i \tau}$, which is small when $\tau$ is high up the half-plane, and set

$$E_4 = 1 + 240\sum_{n \ge 1} \sigma_3(n)\, q^n, \qquad \Delta = q \prod_{n \ge 1} (1 - q^n)^{24}, \qquad j = \frac{E_4^{\,3}}{\Delta},$$

where $\sigma_3(n)$ is the sum of the cubes of the divisors of $n$. Both $E_4$ and $\Delta$ change under the modular group, but by factors that cancel in the ratio, and $j$ is unchanged. Expanded in powers of $q$ it begins

$$j = \frac{1}{q} + 744 + 196\,884\, q + 21\,493\,760\, q^2 + 864\,299\,970\, q^3 + \cdots,$$

with whole-number coefficients throughout. Everything below is computed from these series in exact fixed-point arithmetic, carried to 140 decimal places: the exponential, sine and cosine that give $q$, then $E_4$ and the product, then $j$.

For $D = -23$ there are three reduced forms, $x^2 + xy + 6y^2$ and the pair $2x^2 \pm xy + 3y^2$, and the figure evaluates $j$ at the three points. The first, high up at $\tau = (-1 + \sqrt{-23})/2$ where $q$ is about $-3 \times 10^{-7}$, gives $j = -3{,}493{,}225.69997\ldots$, real and negative. The other two, lower down where $q$ is larger, give a conjugate pair $737.84998 \pm 1764.01894\,i$. None of the three looks special. Multiply out $(X - j_1)(X - j_2)(X - j_3)$, though, and the product is

$$X^3 + 3{,}491{,}750\,X^2 - 5{,}151{,}296{,}875\,X + 12{,}771{,}880{,}859{,}375,$$

with every coefficient within $10^{-113}$ of a whole number before rounding. The three values are the three roots of one cubic with whole coefficients. Each is an algebraic integer of degree three, and the three are conjugates of one another in exactly the way $\sqrt 2$ and $-\sqrt 2$ are.

Which values come out real is itself readable from the picture. The forms $2x^2 + xy + 3y^2$ and $2x^2 - xy + 3y^2$ are mirror images across the imaginary axis — their points are $\tau$ and $-\bar\tau$ — and because every coefficient of $j$’s expansion is a real whole number, $j(-\bar\tau)$ is the complex conjugate of $j(\tau)$. So mirror-image forms give a conjugate pair, and a form that is its own mirror image, with $b = 0$ or $|b| = a$ or $a = c$, gives a real value. Those self-mirrored forms are exactly the classes equal to their own inverse under composition, the classes of order one or two, and Gauss’s genus theory counts them: there are $2^{t-1}$, where $t$ is the number of primes dividing $D$. So the number of real roots of the class polynomial is $2^{t-1}$: one for $-23$ and for $-167$, whose discriminants are prime, two for $-15 = -3 \cdot 5$, where both roots are real. The geometry of the half-plane, the genus theory of forms and the reality of an algebraic number are one fact read three ways.

One degree for each class #

The cubic is no accident of $-23$. For every negative discriminant $D$, the product of $X - j(\tau)$ over the reduced forms is a polynomial with whole-number coefficients, called Hilbert’s class polynomial $H_D$, and it is irreducible — it cannot be factored over the rationals — so its degree, the number of reduced forms, is the degree of each of its roots. The table computes it for nine discriminants, with class numbers 1, 2, 3, 4, 5, 7 and 11. For $D = -15$, with two classes, the two values of $j$ are $(-191{,}025 \pm 85{,}995\sqrt5)/2$, and their polynomial is $X^2 + 191{,}025X - 121{,}287{,}375$. For $D = -167$, with eleven classes, the polynomial has degree eleven and a largest coefficient of 68 digits; before rounding, every one of its twelve coefficients was within $10^{-38}$ of a whole number, which is a whole number recovered from eleven transcendental-looking series by multiplication alone.

The theory that predicts this is complex multiplication, the part of number theory Leopold Kronecker called the dearest dream of his youth. The values $j(\tau)$ at the points of a field’s forms are called singular moduli, and their properties go beyond being algebraic. Adjoining any one of them to $\mathbb{Q}(\sqrt{D})$ produces a field whose symmetries over $\mathbb{Q}(\sqrt{D})$ are exactly the class group — the classes of forms, multiplied by Gauss’s composition, permute the values of $j$ among themselves. That field is the Hilbert class field, the largest extension in which every prime ideal of the base becomes principal, and $j$ writes it down explicitly. In the language of the tower of degrees, the extension has degree $h$ over $\mathbb{Q}(\sqrt{D})$ and $2h$ over the rationals, and the class group is its Galois group.

The coefficients grow because the root at the top form dominates. With $a = 1$ the top point has $|q| = e^{-\pi\sqrt{|D|}}$, so $|j|$ there is about $e^{\pi\sqrt{|D|}}$, and the constant term of $H_D$ is a product of all $h$ values. That is why computing $H_D$ this way needs precision that grows with $\sqrt{|D|}$ times $h$, and why 140 places are plenty for $-167$ and would not be for a discriminant with a thousand classes.

When the degree is one #

A field with one class has one reduced form, and its class polynomial has degree one: $j(\tau)$ is itself a whole number. The nine such fields give

$$0,\; 1728,\; -15^3,\; 20^3,\; -32^3,\; -96^3,\; -960^3,\; -5280^3,\; -640320^3,$$

for $D = -3, -4, -7, -8, -11, -19, -43, -67, -163$, and the figure finds each by evaluating the series and rounding — the computed value was within $10^{-89}$ of the whole number even for 163, where it has eighteen digits. That they are all perfect cubes is a theorem of Weber’s: $j$ has a cube root, $\gamma_2$, which is itself a modular function for a smaller group and takes algebraic-integer values at these points, so for class number one that cube root is a whole number too. That $j - 1728$ is $\pm|D|$ times a square has the same kind of explanation, through a square root of $j - 1728$.

The factorisations hide a further surprise. Every prime appearing in the table is smaller than the discriminant: $640320 = 2^6 \cdot 3 \cdot 5 \cdot 23 \cdot 29$ and $40{,}133{,}016 = 2^3 \cdot 3^3 \cdot 7 \cdot 11 \cdot 19 \cdot 127$, all well below 163. A random eighteen-digit number almost always has a prime factor in the millions. Benedict Gross and Don Zagier explained it in 1985: the difference of two singular moduli, here $j(\tau) - j(i) = j(\tau) - 1728$ and $j(\tau) - j(\omega) = j(\tau) - 0$, is divisible only by primes that divide numbers of the form $(D_1 D_2 - x^2)/4$, which are bounded by the discriminants. Their paper on singular moduli was the arithmetic preparation for the theorem with Goldfeld that later made the class number problem effective.

The cubic modulo p #

A polynomial with whole coefficients can be reduced modulo any prime, and how a polynomial breaks modulo the primes found that the way it factors is a census of its symmetries. For the class polynomial the census has a meaning in terms of forms. The figure takes every prime from 29 to 599, counts the roots of $H_{-23}$ modulo $p$ by trying every residue, and separately searches for whole numbers $x, y$ with one of the three reduced forms equal to $p$. The cubic has three roots exactly when $x^2 + xy + 6y^2$ takes the value $p$, none when $2x^2 + xy + 3y^2$ does, and one when no form of discriminant $-23$ takes $p$ at all — on all 100 primes, without exception.

That is a statement which primes a form takes could not make with congruences. For $x^2 + y^2$ the primes it takes are the ones leaving remainder 1 on division by 4, a condition on $p$ alone. For $x^2 + xy + 6y^2$ no condition of that kind exists, because the class group has order three and is not a group of residues; a prime such as 59 is taken by $x^2 + xy + 6y^2$ while 31 is taken by $2x^2 + xy + 3y^2$, and no list of residues, for any modulus, sorts all the primes of one kind from the other. What decides it is a polynomial: $p$ is taken by the principal form exactly when $H_{-23}$ splits completely modulo $p$. That is the general shape of the answer, and the class polynomial is what supplies it for every $D$.

The census matches the symmetries, too. The splitting field of $H_{-23}$ has the symmetries of a triangle, the same group as $x^3 - 2$, and the shares in the figure — 14, 36 and 50 out of 100 for three, none and one root — are on their way to $\tfrac16$, $\tfrac13$ and $\tfrac12$, the shares of the identity, the rotations and the flips. And a much smaller cubic does the same job. Modulo every one of these primes, $x^3 - x - 1$ has as many roots as $H_{-23}$, because its real root, the plastic number $1.3247\ldots$, generates the same cubic field as $j(\tau_1)$; the class polynomial is simply the most extravagant description of that field.

How near is nearly whole #

Now the near-integer. At the point $\tau = (1 + \sqrt{-d})/2$ of the reduced form $x^2 + xy + \tfrac{1+d}{4}y^2$, the variable $q = e^{2\pi i\tau}$ equals $-e^{-\pi\sqrt d}$, real, negative and tiny. The expansion of $j$ becomes

$$j = -e^{\pi\sqrt d} + 744 - 196\,884\, e^{-\pi\sqrt d} + 21\,493\,760\, e^{-2\pi\sqrt d} - \cdots.$$

If $d$ has one class, the left side is a whole number, $-j_0^3$ say, and rearranging,

$$e^{\pi\sqrt d} = j_0^3 + 744 - 196\,884\, e^{-\pi\sqrt d} + \cdots.$$

So $e^{\pi\sqrt d}$ is a whole number minus a correction, and the size of the correction is the question. The figure plots each term of the expansion on a logarithmic scale for the four largest $d$ with one class. The first term is $e^{\pi\sqrt d}$ itself, the second is 744, and from then on each term is a coefficient of $j$ — which grows, roughly like $e^{4\pi\sqrt n}$ — times $e^{-\pi\sqrt d\, n}$, which shrinks much faster. At $d = 163$ the first correction is $7.5 \times 10^{-13}$ and each later one is about fifteen powers of ten smaller still, so the whole tail is one number, $196{,}884\,e^{-\pi\sqrt{163}}$, to all the precision anyone could want. At $d = 19$ the first correction is a fifth and the later ones are not negligible, which is why $e^{\pi\sqrt{19}} = 885{,}479.78$ is an unimpressive near-miss.

Twelve nines #

The arithmetic gives

$$e^{\pi\sqrt{163}} = 262\,537\,412\,640\,768\,743.99999999999925007\ldots,$$

twelve nines after the decimal point, short of $640320^3 + 744$ by $7.4993 \times 10^{-13}$, and the predicted correction $196{,}884\,e^{-\pi\sqrt{163}}$ agrees with the measured shortfall to every digit the figure prints. Charles Hermite noticed the near-miss in 1859. Martin Gardner announced in his April column of 1975 that Ramanujan had proved $e^{\pi\sqrt{163}}$ to be a whole number, and a good many readers believed him for a month, which is a measure of how implausible the true explanation sounds: a transcendental number that misses being whole because a quadratic field happens to factor uniquely.

The other three are weaker versions of the same event. $e^{\pi\sqrt{67}}$ falls short of $5280^3 + 744$ by $1.3 \times 10^{-6}$, and $e^{\pi\sqrt{43}}$ of $960^3 + 744$ by $2.2 \times 10^{-4}$, each shortfall again $196{,}884\,e^{-\pi\sqrt d}$. There is no fifth, because there is no tenth field of class number one; the essay that counted the classes traced the history of that theorem, and Kurt Heegner’s original proof of it in 1952 went through modular functions of exactly this kind: he showed that a tenth field would force a singular modulus with properties no whole number can have.

The coefficient 196,884 has its own story, and it is the one connection here that nobody predicted. John McKay noticed in 1978 that it is one more than 196,883, the smallest dimension in which the Monster — the largest sporadic finite simple group, with about $8 \times 10^{53}$ elements — can act by rotations. The next coefficient, 21,493,760, is $1 + 196{,}883 + 21{,}296{,}876$, a sum of the Monster’s first three dimensions, and the pattern continues. John Conway and Simon Norton named it monstrous moonshine in 1979, and Richard Borcherds proved it in 1992. So the number that holds $e^{\pi\sqrt{163}}$ a trillionth short of whole is, minus one, the size of the Monster’s smallest stage.

Where the class polynomial does work #

The class polynomial is not only a curiosity; it is how a particular kind of elliptic curve is built. Every value of $j$ is the $j$-invariant of an elliptic curve $y^2 = x^3 + ax + b$ — two curves are the same over the complex numbers exactly when their $j$-invariants agree — and the singular moduli are the $j$-invariants of the curves with complex multiplication, whose lattices of periods are the ideals of an imaginary quadratic field. Reduce $H_D$ modulo a prime $p$, find a root, and the curve with that $j$-invariant over the field of $p$ elements has a number of points that is known in advance: if $4p = t^2 + |D|u^2$, it has $p + 1 \pm t$ points, the numbers give or take twice the square root found always lie within $2\sqrt p$ of $p + 1$.

That is the complex multiplication method, and it runs in the other direction from the usual problem. Counting the points of a random curve modulo a large prime is a hard computation; choosing $D$, solving $4p = t^2 + |D|u^2$, and taking a root of $H_D$ produces a curve whose count is known before it is written down. It is how curves with a prescribed number of points are generated for cryptography, and in the reverse direction it is the engine of Atkin and Morain’s elliptic curve primality proving, which certifies primes of many thousands of digits by building such curves modulo the candidate. The precision needed to compute $H_D$ — growing with $h\sqrt{|D|}$ — is the practical limit on which discriminants the method can use, which is why the coefficient sizes in the table matter outside this essay.

What the computations show and what they assume #

The coefficients are measured, not quoted. Every value of $j$ is computed to 140 decimal places from the defining series, the products are expanded in the same arithmetic, and the distance of each coefficient from the nearest whole number is reported: $10^{-113}$ for $-23$, $10^{-38}$ for $-167$, where eleven roots and the growth of the coefficients use up the precision. That the coefficients are exactly whole is the theorem of complex multiplication, which no finite computation proves; the computation shows that they are whole to far more places than any coincidence could supply.

The census modulo primes is exact. Roots modulo $p$ are found by trying every residue, the forms’ values by searching every small pair, and the agreement is checked on every prime from 29 to 599. It is evidence for the statement about all primes, and the theorem behind it — that the principal form takes $p$ exactly when $H_D$ splits completely modulo $p$, for primes not dividing the polynomial’s discriminant — is class field theory.

And the picture shows only imaginary fields. A real quadratic field $\mathbb{Q}(\sqrt d)$ has indefinite forms, whose points lie on the real line rather than in the upper half-plane, and $j$ has nothing to say there; real fields that factorise uniquely counted their classes by cycles of forms instead, with no single number whose degree is the class number.

Still open: the same dream for the real fields #

Kronecker’s dream was to generate every abelian extension of a field by special values of explicit functions. For the rationals, the exponential does it: the roots of unity, values of $e^{2\pi i x}$ at rational $x$, generate every abelian extension of $\mathbb{Q}$, which is the Kronecker–Weber theorem. For an imaginary quadratic field, $j$ and its companions do it, as here. David Hilbert asked in 1900, as the twelfth of his problems, for the functions that would do the same for every number field.

For real quadratic fields, no analogue of $j$ is known. Harold Stark conjectured in the 1970s that the values of certain derivatives of $L$-series are logarithms of units generating the right extensions, and the conjecture has been checked numerically in many cases; Samit Dasgupta and Mahesh Kakde gave a $p$-adic answer for totally real fields in 2021, through units constructed $p$-adically rather than by a complex function. A description as concrete as “evaluate this function at the points of the reduced forms” is still missing for $\mathbb{Q}(\sqrt{5})$, and the question of what replaces the upper half-plane and the modular group there is one of the central open problems connecting number theory to analysis.

A count that is also a degree #

The class number of $\mathbb{Q}(\sqrt{-d})$ is a count of reduced forms and also the degree of an algebraic number: the modular function $j$ at the reduced points gives $h$ conjugate values whose polynomial has whole-number coefficients. For $-23$ the three values make a cubic whose roots modulo each prime say which form takes that prime; for $-167$ eleven values make a polynomial of degree eleven; and for the nine fields with one class the value is a whole number, always a cube, built from small primes only. At $d = 163$ that whole number sits next to $e^{\pi\sqrt{163}}$, separated by a correction of 196,884 times a trillionth of a trillionth of its size — a coefficient that, less one, counts the dimensions of the Monster’s smallest representation.