# Binary quadratic forms over the rational integers and class numbers of quadratic ﬁelds.

August 23, 2014 by . 3 comments

I wrote an article with Irving Kaplansky on indefinite binary quadratic forms, integral coefficients. At the time, I believe I used high-precision continued fractions or similar. It took me years to realize that the right way to solve Pell’s equation, or find out the “minimum” of an indefinite form (and other small primitively represented values), or the period of its continued fraction, was the method of “reduced” forms in cycles/chains, due to Lagrange, Legendre, Gauss. It is also the cheapest way to find the class number and group multiplication for ideals in real quadratic fields, this probably due to Dirichlet. For imaginary quadratic fields, we have easier “reduced” positive forms.

A binary quadratic form, with integer coefficients, is some $$f(x,y) = A x^2 + B xy + C y^2.$$ The discriminant is $$\Delta = B^2 – 4 A C.$$ We will abbreviate this by $$\langle A,B,C \rangle.$$ It is primitive if $${\gcd(A,B,C)=1. }$$ Standard fact, hard to discover but easy to check: $$(A x^2 + B x y + C D y^2 ) (C z^2 + B z w + A D w^2 ) = A C X^2 + B X Y + D Y^2,$$ where $${ X = x z – D yw, \; Y = A xw + C yz + B yw. }$$ This gives us Dirichlet’s definition of “composition” of quadratic forms of the same discriminant, $$\langle A,B,CD \rangle \circ \langle C,B,AD \rangle = \langle AC,B,D \rangle.$$ In particular, if this $${D=1,}$$ the result represents $${1}$$ and is ($${SL_2 \mathbb Z}$$) equivalent to the “principal” form for this discriminant. Oh, duplication or squaring in the group; if $${\gcd(A,B)=1,}$$ $$\langle A,B,AD \rangle^2 = \langle A^2,B,D \rangle.$$ This comes up with positive forms: $${ \langle A,B,C \rangle \circ \langle A,-B,C \rangle = \langle 1,B,AC \rangle }$$ is principal, the group identity. Probably should display some $${SL_2 \mathbb Z}$$ equivalence rules, these are how we calculate when things are not quite right for Dirichlet’s rule: $$\langle A,B,C \rangle \cong \langle C,-B,A \rangle,$$ $$\langle A,B,C \rangle \cong \langle A, B + 2 A, A + B +C \rangle,$$ $$\langle A,B,C \rangle \cong \langle A, B – 2 A, A – B +C \rangle.$$

Imaginary first. Suppose we want to know about $${\mathbb Q(\sqrt {-47}).}$$ Reduced positive forms $${ \langle A,B,C \rangle }$$ obey $${|B| \leq A \leq C}$$ and $${B \neq -A,}$$ plus whenever $${A = C,}$$ we have $${B \geq 0.}$$ Our group of binary forms is

$$\begin{array}{rrrrr} \langle & 1, & 1, & 12 & \rangle \\ \langle & 2, & -1, & 6 & \rangle \\ \langle & 2, & 1, & 6 & \rangle \\ \langle & 3, & -1, & 4 & \rangle \\ \langle & 3, & 1, & 4 & \rangle \end{array}$$

This is an abelian group in any case, so it is cyclic of order 5. These are also the five elements in the ring of integers of $${\mathbb Q(\sqrt {-47}).}$$ Here is the mapping from forms to ideals: given $${ \langle A,B,C \rangle, }$$ drop the letter $${C.}$$ That’s it. $$\langle A,B,C \rangle \mapsto \left[ A, \frac{B + \sqrt \Delta}{2} \right].$$ Oh, why is this an ideal, rather than just some $${\mathbb Z}$$-lattice? Because, given $${\alpha,\beta}$$ rational integers, $$\left[ \alpha, \frac{\beta + \sqrt \Delta}{2} \right]$$ is an ideal if and only if $$4 \alpha | ( \Delta – \beta^2 ).$$

Group: we already see how to do $$\langle 2,1,6 \rangle^2 \cong \langle 4,1,3 \rangle \cong \langle 3,-1,4 \rangle;$$ $$\langle 2,1,6 \rangle \circ \langle 3,-1,4 \rangle \cong \langle 2,5,9 \rangle \circ \langle 3,5,6 \rangle \cong \langle 6,5,3 \rangle \cong \langle 3,-5,6 \rangle \cong \langle 3,1,4 \rangle;$$ $$\langle 2,1,6 \rangle \circ \langle 3,1,4 \rangle \cong \langle 6,1,2 \rangle \cong \langle 2,-1,6 \rangle.$$ $$\langle 2,1,6 \rangle \circ \langle 2,-1,6 \rangle \cong \langle 1,1,12 \rangle$$ in any case.

Oh, not at all by the way, if we want $${\mathbb Q(\sqrt {K})}$$ for squarefree $${K \equiv 2,3 \pmod 4,}$$ we actually use quadratic forms of discriminant $${\Delta = 4K.}$$ Not my fault. Indeed, below, in the Pell solution for $${61,}$$ the discriminant is actually $${244,}$$ there is a single cycle of discriminant $${61}$$ which looks a bit different, mainly due to the odd numbers in the middle coefficient.

There are a few extra tricks for indefinite forms. The right way to calculate things is to use “reduced” forms. The definition of $${ \langle A,B,C \rangle }$$ being reduced is $$0 < B < \sqrt \Delta, \; \; \mbox{and} \; \; \sqrt \Delta -B < 2 |A| < \sqrt \Delta + B,$$ this being equivalent to $$0 < B < \sqrt \Delta, \; \; \mbox{and} \; \; \sqrt \Delta -B < 2 |C| < \sqrt \Delta + B.$$

If you actually calculate enough Conway topographs, you realize eventually that

THEOREM $${ \langle A,B,C \rangle }$$ is reduced if and only if $$AC < 0 \; \; \mbox{and} \; \; B > |A+C|$$ I have never seen this in print, it should be attributed to Conway or to Jagy or to Marty Weissman of UC Santa Cruz and Singapore; Conway knows much more than he writes down. With the required relation $$B^2 – 4 AC = \Delta$$ and evident inequalities, this gives a quick way to find all reduced forms.

Now, different reduced forms may be equivalent, which sounds bad, but the cycle method quickly decides these relationships. No decimal accuracy or “cycle detection” is ever required. A single calculation is done, $${\left\lfloor \sqrt \Delta \right\rfloor }$$ and remembered forever. I actually wrote Newton’s square root method with integer arithmetic. Everything else, forever, is integer arithmetic. Given a form $${ \langle A,B,C \rangle }$$ we want to find its “right neighbor.” First, we find a nonzero integer I like to call $${\delta,}$$ with $$\delta C > 0 \; \; \mbox{and} \; \; |\delta| = \; \left\lfloor \; \frac{B + \left\lfloor \sqrt \Delta \right\rfloor }{2|C|} \; \right\rfloor .$$ The absolute values of the $${\delta}$$’s are the partial quotients in the continued fraction of, well, something. Note that $${B}$$ and $${\left\lfloor \sqrt \Delta \right\rfloor }$$ are positive.

Alright, given a $${\delta,}$$ the form and its right neighbor are $$\langle A,B,C \rangle \rightarrow \langle C,-B + 2 C \delta, A – B \delta +C \delta^2 \rangle.$$ For those who know what a Gram matrix is, call it $${G,}$$ the Gram matrix of the right neighbor is $${P^T G P,}$$ where $$P \; = \; \left( \begin{array}{rr} 0 & -1 \\ 1 & \delta \end{array} \right).$$

Keeping a cumulative product of these $${P}$$ matrices is how we solve Pell, quickly and without tears.

$$\begin{array}{rrrrrrrr} 0. & \langle & 1, & 14, & -12 & \rangle & \delta= & -1 \\ 1. & \langle & -12, & 10, & 3 & \rangle & \delta= & 4 \\ 2. & \langle & 3, & 14, & -4 & \rangle & \delta= & -3 \\ 3. & \langle & -4, & 10, & 9 & \rangle & \delta= & 1 \\ 4. & \langle & 9, & 8, & -5 & \rangle & \delta= & -2 \\ 5. & \langle & -5, & 12, & 5 & \rangle & \delta= & 2 \\ 6. & \langle & 5, & 8, & -9 & \rangle & \delta= & -1 \\ 7. & \langle & -9, & 10, & 4 & \rangle & \delta= & 3 \\ 8. & \langle & 4, & 14, & -3 & \rangle & \delta= & -4 \\ 9. & \langle & -3, & 10, & 12 & \rangle & \delta= & 1 \\ 10. & \langle & 12, & 14, & -1 & \rangle & \delta= & -14 \\ 11. & \langle & -1, & 14, & 12 & \rangle & \delta= & 1 \\ 12. & \langle & 12, & 10, & -3 & \rangle & \delta= & -4 \\ 13. & \langle & -3, & 14, & 4 & \rangle & \delta= & 3 \\ 14. & \langle & 4, & 10, & -9 & \rangle & \delta= & -1 \\ 15. & \langle & -9, & 8, & -5 & \rangle & \delta= & 2 \\ 16. & \langle & 5, & 12, & -5 & \rangle & \delta= & -2 \\ 17. & \langle & -5, & 8, & 9 & \rangle & \delta= & 1 \\ 18. & \langle & 9, & 10, & -4 & \rangle & \delta= & -3 \\ 19. & \langle & -4, & 14, & 3 & \rangle & \delta= & 4 \\ 20. & \langle & 3, & 10, & -12 & \rangle & \delta= & -1 \\ 21. & \langle & -12, & 14, & 1 & \rangle & \delta= & 14 \\ 22. & \langle & 1, & 14, & -12 & \rangle & & \end{array}$$

Automorph, written on right of Gram matrix: $$\left( \begin{array}{rr} 183241189 & 2713847760 \\ 226153980 & 3349396909 \end{array} \right).$$

Pell automorph $$\left( \begin{array}{rr} 1766319049 & 13795392780 \\ 226153980 & 1766319049 \end{array} \right).$$

Pell unit $$1766319049^2 – 61 \cdot 226153980^2 = 1$$

Pell NEGATIVE $$29718^2 – 61 \cdot 3805^2 = -1$$

4 PRIMITIVE $$1523^2 – 61 \cdot 195^2 = 4$$

-4 PRIMITIVE $$39^2 – 61 \cdot 5^2 = -4$$

Here we are using $$\langle 1,0,-61 \rangle \cong \langle 1,14,-12 \rangle.$$ It begins with $${\langle 1,14,-12 \rangle}$$ and does not end until it gets $${\langle 1,14,-12 \rangle}$$ again, this being guaranteed. No floating point reals anywhere, no decimal accuracy, no doubts. Think Gwen Stefani. Note that “Automorph” is a traditional word for the matrix $${P}$$ when $${P^T G P = G.}$$

Ideals; with positive prime discriminant $${p \equiv 1 \pmod 4,}$$ the principal form always represents both $${1}$$ and $${-1,}$$ and, just as with positive forms, the mapping to ideals is a bijection a group isomorphism. I did some examples with class number above one; note that the first computer they gave me was called phoebus after Phoebus Apollo, this was intended to be phoebusjunior but the spelling went sideways. Also, I ran out of steam putting angle brackets $${\langle \rangle}$$ around the coefficient triples for binary forms, I hope you can adjust:

jagy@phobeusjunior:
jagy@phobeusjunior: ./indefCycle_All_Reduced 229
1. 1 15 -1 cycle length 2
2. 3 13 -5 cycle length 6
3. 5 13 -3 cycle length 6
form class number is 3
jagy@phobeusjunior: ./indefCycle_All_Reduced 257
1. 1 15 -8 cycle length 6
2. 2 15 -4 cycle length 6
3. 4 15 -2 cycle length 6
form class number is 3
jagy@phobeusjunior: ./indefCycle_All_Reduced 401
1. 1 19 -10 cycle length 6
2. 2 19 -5 cycle length 6
3. 5 19 -2 cycle length 6
4. 4 17 -7 cycle length 10
5. 7 17 -4 cycle length 10
form class number is 5
jagy@phobeusjunior:

Now, if the discriminant is divisible by any prime $${q \equiv 3 \pmod 4,}$$ or we just have bad luck with the prime factors $${p \equiv 1 \pmod 4,}$$ then the principal form does not represent $${-1}$$ and the mapping from forms to ideals becomes two to one. I really thought there was a big mystery, but no. Here is an example $${\Delta = 5 \cdot 13 \cdot 17 \cdot 41 = 45305.}$$ There are 316 reduced forms collected into 16 cycles:
jagy@phobeusjunior: ./indefCycle_All_Reduced 45305
45305 factored 5 * 13 * 17 * 41
1. 1 211 -196 cycle length 12
2. -1 211 196 cycle length 12
3. 2 211 -98 cycle length 16
4. -2 211 98 cycle length 16
5. 7 211 -28 cycle length 18
6. -7 211 28 cycle length 18
7. 14 211 -14 cycle length 20
8. -14 211 14 cycle length 20
9. 5 205 -164 cycle length 20
10. -5 205 164 cycle length 20
11. 10 205 -82 cycle length 24
12. -10 205 82 cycle length 24
13. 13 195 -140 cycle length 22
14. -13 195 140 cycle length 22
15. 26 195 -70 cycle length 26
16. -26 195 70 cycle length 26
form class number is 16
jagy@phobeusjunior:

My software tends to report, for each cycle, a form with particularly large $${B.}$$ Now, as you can see, $${\langle 1,211,-196 \rangle}$$ and $${\langle -1,211,196 \rangle}$$ are distinct $${SL_2 \mathbb Z}$$ classes, distinct cycles.

When this happens, in slang I made up $${1 \neq -1,}$$ we can prove that $$\langle A,B,-C \rangle$$ and $$\langle -A,B,C \rangle$$ are ALWAYS DISTINCT $${SL_2 \mathbb Z}$$ classes. Because, you see,$$\langle -1,B,AC \rangle \circ \langle A,B,-C \rangle = \langle -A,B,C \rangle.$$ So, the mapping to ideals says IDENTIFY $${ \langle A,B,-C \rangle }$$ and $${ \langle -A,B,C \rangle ,}$$ send them to the same ideal. Why not? We already have an equality of ideals in $$\left[ A, \frac{B + \sqrt \Delta}{2} \right] = \left[ -A, \frac{B + \sqrt \Delta}{2} \right].$$ So, we just take the form from each pair with positive $${A,}$$ which cuts the form class number by half, down from 16 to 8, and the ideals are $$\left[1 , \frac{211 + \sqrt{45305}}{2} \right],$$ $$\left[ 2, \frac{211 + \sqrt{45305}}{2} \right],$$ $$\left[ 7, \frac{211 + \sqrt{45305}}{2} \right],$$ $$\left[ 14, \frac{211 + \sqrt{45305}}{2} \right],$$ $$\left[5 , \frac{205 + \sqrt{45305}}{2} \right],$$ $$\left[10 , \frac{205 + \sqrt{45305}}{2} \right],$$ $$\left[13 , \frac{195 + \sqrt{45305}}{2} \right],$$ $$\left[26 , \frac{195 + \sqrt{45305}}{2} \right].$$

One other valuable idea that I can impart here is that of genus. If you check, with either the forms or the ideals, you will find that each one squares to the identity. We say that each genus has only one class. It follows automatically the=at the ideal class group is not cyclic, it is $${(\mathbb Z/ 2 \mathbb Z)^3. }$$ I worked a fair amount to get the group table, then I thought, why not just use the Legendre symbols? And that works perfectly.

$$\begin{array}{rrrrr} & 5 & 13 & 17 & 41 \\ 1 & + & + & + & + \\ 2 & – & – & + & + \\ 7 & – & – & – & – \\ 14 & + & + & – & – \\ 5 \rightarrow 46 & + & – & – & + \\ 10 \rightarrow 23 & – & + & – & + \\ 13 \rightarrow 38 & – & + & + & – \\ 26 \rightarrow 19 & + & – & + & – \end{array}$$ As you can see, if the first coefficient has a common divisor with the discriminant, we switch to some other number represented by the quadratic form. This table, because there is just one class per genus, immediately tells the multiplication. Just multiply the Legendre symbols for each factor $${5,13,17,41.}$$

The Cohen-Lenstra heuristics say that the vast majority of class groups are cyclic. I did an example where that works, despite having more than one genus. Here, most forms do not square to the identity.

jagy@phobeusjunior: ./indefCycle_All_Reduced 1345
1345 factored 5 * 269
1. 1 35 -30 cycle length 6
2. -1 35 30 cycle length 6
3. 2 35 -15 cycle length 8
4. -2 35 15 cycle length 8
5. 3 35 -10 cycle length 8
6. -3 35 10 cycle length 8
7. 4 33 -16 cycle length 10
8. -4 33 16 cycle length 10
9. 8 33 -8 cycle length 10
10. -8 33 8 cycle length 10
11. 16 33 -4 cycle length 10
12. -16 33 4 cycle length 10
form class number is 12
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus\$

Now, as forms, $$\langle 8,33,-8 \rangle^2 = \langle 64, 33,-1 \rangle$$ is minus the identity. However, as ideals this is identified with the principal class. I worked it all out, the multiplication table is

$$\begin{array}{rrrrrrr} \bullet & 1 & 4 & 16 & 8 & 2 & 3 \\ 1 & 1 & 4 & 16 & 8 & 2 & 3 \\ 4 & 4 & 16 & 1 & 2 & 3 & 8 \\ 16 & 16 & 1 & 4 & 3 & 8 & 2 \\ 8 & 8 & 2 & 3 & 1 & 4 & 16 \\ 2 & 2 & 3 & 8 & 4 & 16 & 1 \\ 3 & 3 & 8 & 2 & 16 & 1 & 4 \end{array}$$ The two genera should be evident; $${1,4,16}$$ are quadratic residues $${\pmod 5}$$ and $${\pmod {269}.}$$ Then $${8,2,3}$$ are nonresidues. The group is cyclic, either $${2}$$ or $${3}$$ is a generator.

To give one straightforward example,(non-reduced) $${ \langle 1, 0,-3 \rangle }$$ and $${ \langle -1, 0,3 \rangle }$$ are sent to the same ideal. In this case, because the discriminant is divisible by at least one prime $${q \equiv 3 \pmod 4,}$$ the two forms are in different genera and it is easy to say which (positive) primes are represented. $${x^2 – 3 y^2}$$ represents all primes $${p \equiv 1 \pmod {12},}$$ while $${3 x^2 – y^2}$$ represents $${2,3}$$ and all primes $${p \equiv 11 \pmod {12}.}$$

Finally, something quite tricky. The ideal viewpoint identifies the quadratic forms $${ \langle 1, 13,-13 \rangle }$$ and $${ \langle -1, 13,13 \rangle }$$ of discriminant $${221 = 13 \cdot 17.}$$ However, the traditional concern is to find out the (positive) primes integrally represented by a form. This then moves to what is called class field theory; the primes represented by $${ \langle 1, 13,-13 \rangle }$$ are those (answer by Noam Elkies) for which $$x^8 + 34 x^6 + 83 x^4 + 34 x^2 + 1 \pmod p$$ factors into all linear factors. Put another way, $${ \langle 1, 13,-13 \rangle }$$ represents $${17}$$ and all primes $${(p|13) = (p|17) = 1}$$ such that $$x^4 + x^3 + x^2 + 2 x + 4 \pmod p$$ factors into all linear factors. Go Figure. In comparison, the primes represented by $${ \langle 5, 11,-5 \rangle }$$ are simply all those that are quadratic nonresidues $${\pmod {13}}$$ and $${\pmod {17},}$$ becuase the other class in its (form) genus is $${ \langle -5, 11,5 \rangle }$$ which represents exactly the same numbers. Oh, and this is a Markov form: the nonzero number represented of smallest absolute value is $${5,}$$ which is called the “minimum” and might be written $${m.}$$ Then, $${\Delta / m^2 = 221 / 25 =8.84 < 9,}$$ which means this must be a Markov form, parametrised by the Markov Numbers, which include $${5.}$$ There is considerable rigidity when $${9m^2 > \Delta,}$$ i.e. extremely large minimum.