I’m reading through Primes of the Form
, by David Cox (link; it’s good!). Here are the high-level notes I took on the first chapter, which is about the theory of quadratic forms.
(Meta point re blog: I’m probably going to start posting more and more of these more high-level notes/sketches on this blog on topics that I’ve been just learning. Up til now I’ve been mostly only posting things that I understand well and for which I have a very polished exposition. But the perfect is the enemy of the good here; given that I’m taking these notes for my own sake, I may as well share them to help others.)
1. Overview
For example, we have the famous quadratic form

As readers are probably aware, we can say a lot about exactly which integers can be represented by
: by Fermat’s Christmas theorem, the primes
(and
) can all be written as the sum of two squares, while the primes
cannot. For convenience, let us say that:
The basic question is: what can we say about which primes/integers are properly represented by a quadratic form? In fact, we will later restrict our attention to “positive definite” forms (described later).
For example, Fermat’s Christmas theorem now rewrites as:
Theorem 3 (Fermat’s Christmas theorem for primes)
An odd prime
is (properly) represented by
if and only if
.
The proof of this is classical, see for example my olympiad handout. We also have the formulation for odd integers:
Theorem 4 (Fermat’s Christmas theorem for odd integers)
An odd integer
is properly represented by
if and only if all prime factors of
are
.
Proof: For the “if” direction, we use the fact that
is multiplicative in the sense that

For the “only if” part we use the fact that if a multiple of a prime
is properly represented by
, then so is
. This follows by noticing that if
(and
) then
. 
Tangential remark: the two ideas in the proof will grow up in the following way.
- The fact that
“multiplies nicely” will grow up to become the so-called composition of quadratic forms.
- The second fact will not generalize for an arbitrary form
. Instead, we will see that if a multiple of
is represented by a form
then some form of the same “discriminant” will represent the prime
, but this form need not be the same as
itself.
2. Equivalence of forms, and the discriminant
The first thing we should do is figure out when two forms are essentially the same: for example,
and
should clearly be considered the same. More generally, if we think of
as acting on
and
is any automorphism of
, then
should be considered the same as
. Specifically,
Definition 5
Two forms
and
said to be equivalent if there exists

such that
. We have
and so we say the equivalence is
- a proper equivalence if
, and
- an improper equivalence if
.
So we generally will only care about forms up to proper equivalence. (It will be useful to distinguish between proper/improper equivalence later.)
Naturally we seek some invariants under this operation. By far the most important is:
Definition 6
The discriminant of a quadratic form
is defined as

The discriminant is invariant under equivalence (check this). Note also that we also have
.
Observe that we have

So if
and
(thus
too) then
for all
. Such quadratic forms are called positive definite, and we will restrict our attention to these forms.
Now that we have this invariant, we may as well classify equivalence classes of quadratic forms for a fixed discriminant. It turns out this can be done explicitly.
Exercise 8
Check there only finitely many reduced forms of a fixed discriminant.
Then the big huge theorem is:
Theorem 9 (Reduced forms give a set of representatives)
Every primitive positive definite form
of discriminant is properly equivalent to a unique reduced form. We call this the reduction of
.
Proof: Omitted due to length, but completely elementary. It is a reduction argument with some number of cases. 
Thus, for any discriminant
we can consider the set

which will be the equivalence classes of positive definite of discriminant
. By abuse of notation we will also consider it as the set of equivalence classes of primitive positive definite forms of discriminant
.
We also define
; by the exercise,
. This is called the class number.
Moreover, we have
, because we can take
for
and
for
. We call this form the principal form.
3. Tables of quadratic forms
Example 11 (Examples of quadratic forms with
,
)
The following discriminants have class number
, hence having only the principal form:
-
, with form
.
-
, with form
.
-
, with form
.
-
, with form
.
-
, with form
.
-
, with form
.
-
, with form
.
-
, with form
.
This is in fact the complete list when
.
Example 12 (More examples of quadratic forms)
Here are tables for small discriminants with
. When
we have
-
, with
forms
and
.
-
, with
forms
and
.
-
, with
forms
and
.
-
, with
forms
and
.
-
, with
forms
and
.
-
, with
forms
and
.
As for
we have
-
, with
forms
and
.
-
, with
forms
and
.
-
, with
forms
and
.
-
, with
forms
,
and
.
4. The Character
We can now connect this to primes
as follows. Earlier we played with
, and observed that for odd primes
,
if and only if some multiple of
is properly represented by
.
Our generalization is as follows:
This generalizes our result for
, but note that it uses
in an essential way! That is: if
, we know
is represented by some quadratic form of discriminant
\dots but only since
do we know that this form reduces to
.
Proof: First assume WLOG that
and
. Thus
, since otherwise this would imply
. Then

hence
.
The converse direction is amusing: let
for integers
,
. Consider the quadratic form

It is primitive of discriminant
and
. Now
may not be reduced, but that’s fine: just take the reduction of
, which must also properly represent
. 
Thus to every discriminant
we can attach the Legendre character (is that the name?), which is a homomorphism

with the property that if
is a rational prime not dividing
, then
. This is abuse of notation since I should technically write
, but there is no harm done: one can check by quadratic reciprocity that if
then
. Thus our previous result becomes:
As a corollary of this, using the fact that
one can prove that
Proof: The congruence conditions are equivalent to
, and as before the only point is that the only reduced quadratic form for these
is the principal one. 
5. Genus theory
What if
? Sometimes, we can still figure out which primes go where just by taking mods.
Let
. Then it represents some residue classes of
. In that case we call the set of residue classes represented the genus of the quadratic form
.
The thing that makes this work is that each genus appears exactly once. We are not always so lucky: for example when
we have that
We now prove that:
Proof: For the first part, we aim to show
is multiplicatively closed. For
,
we use the fact that

For
, we instead appeal to another “magic” identity

and it follows from here that
is actually the set of squares in
, which is obviously a subgroup.
Now we show that other quadratic forms have genus equal to a coset of the principal genus. For
, with
we can write

and thus the desired coset is shown to be
. As for
, we have

so the desired coset is also
, since
was the set of squares. 
Thus every genus is a coset of
in
. Thus:
Definition 20
We define the quotient group

which is the set of all genuses in discriminant
. One can view this as an abelian group by coset multiplication.
Thus there is a natural map

(The map is surjective by Theorem~14.) We also remark than
is quite well-behaved:
Proposition 21 (Structure of
)
The group
is isomorphic to
for some integer
.
Proof: Observe that
contains all the squares of
: if
is the principal form then
. Thus claim each element of
has order at most
, which implies the result since
is a finite abelian group. 
In fact, one can compute the order of
exactly, but for this post I Will just state the result.
6. Composition
We have already used once the nice identity

We are going to try and generalize this for any two quadratic forms in
. Specifically,
Proposition 23 (Composition defines a group operation)
Let
. Then there is a unique
and bilinear forms
for
such that
In fact, without the latter two constraints we would instead have
and
, and each choice of signs would yield one of four (possibly different) forms. So requiring both signs to be positive makes this operation well-defined. (This is why we like proper equivalence; it gives us a well-defined group structure, whereas with improper equivalence it would be impossible to put a group structure on the forms above.)
Taking this for granted, we then have that
Theorem 24 (Form class group)
Let
,
be a discriminant. Then
becomes an abelian group under composition, where
- The identity of
is the principal form, and
- The inverse of the form
is
.
This group is called the form class group.
We then have a group homomorphism

Observe that
and
are inverses and that their
images coincide (being improperly equivalent); this is expressed in the fact that
has elements of order
. As another corollary, the number of elements of
with a given genus is always a power of two.
We now define:
Definition 25
An integer
is convenient if the following equivalent conditions hold:
- The principal form
is the only reduced form with the principal genus.
-
is injective (hence an isomorphism).
-
.
Thus we arrive at the following corollary:
Corollary 26 (Convenient numbers have nice representations)
Let
be convenient. Then
is of the form
if and only if
lies in the principal genus.
Hence the represent-ability depends only on
.
OEIS A000926 lists 65 convenient numbers. This sequence is known to be complete except for at most one more number; moreover the list is complete assuming the Grand Riemann Hypothesis.
7. Cubic and quartic reciprocity
To treat the cases where
is not convenient, the correct thing to do is develop class field theory. However, we can still make a little bit more progress if we bring higher reciprocity theorems to bear: we’ll handle the cases
and
, two examples of numbers which are not convenient.
7.1. Cubic reciprocity
First, we prove that
To do this we use cubic reciprocity, which requires working in the Eisenstein integers
where
is a cube root of unity. There are six units in
(the sixth roots of unity), hence each nonzero number has six associates (differing by a unit), and the ring is in fact a PID.
Now if we let
be a prime not dividing
, and
is coprime to
, then we can define the cubic Legendre symbol by setting

Moreover, we can define a primary prime
to be one such that
; given any prime exactly one of the six associates is primary. We then have the following reciprocity theorem:
Theorem 28 (Cubic reciprocity)
If
and
are disjoint primary primes in
then

We also have the following supplementary laws: if
, then

The first supplementary law is for the unit (analogous to
) while the second reciprocity law handles the prime divisors of
(analogous to
.)
We can tie this back into
as follows. If
is a rational prime then it is represented by
, and thus we can put
for some prime
,
. Consequently, we have a natural isomorphism
![\displaystyle \mathbb Z[\omega] / \pi \mathbb Z[\omega] \cong \mathbb Z / p \mathbb Z. \displaystyle \mathbb Z[\omega] / \pi \mathbb Z[\omega] \cong \mathbb Z / p \mathbb Z.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cmathbb+Z%5B%5Comega%5D+%2F+%5Cpi+%5Cmathbb+Z%5B%5Comega%5D+%5Ccong+%5Cmathbb+Z+%2F+p+%5Cmathbb+Z.+&bg=ffffff&fg=000000&s=0)
Therefore, we see that a given
is a cubic residue if and only if
.
In particular, we have the following corollary, which is all we will need:
Proof: By cubic reciprocity:


Now we give the proof of Theorem~27. Proof: First assume

Let
be primary, noting that
. Now clearly
, so done by corollary.
For the converse, assume
,
with
primary and
. If we set
for integers
and
, then the fact that
and
is enough to imply that
(check it!). Moreover,

as desired. 
7.2. Quartic reciprocity
This time we work in
, for which there are four units
,
. A prime is primary if
; every prime not dividing
has a unique associate which is primary. Then we can as before define

where
is primary, and
is nonzero mod
. As before
,
we have that
is a quartic residue modulo
if and only if
thanks to the isomorphism
![\displaystyle \mathbb Z[i] / \pi \mathbb Z[i] \cong \mathbb Z / p \mathbb Z. \displaystyle \mathbb Z[i] / \pi \mathbb Z[i] \cong \mathbb Z / p \mathbb Z.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5Cmathbb+Z%5Bi%5D+%2F+%5Cpi+%5Cmathbb+Z%5Bi%5D+%5Ccong+%5Cmathbb+Z+%2F+p+%5Cmathbb+Z.+&bg=ffffff&fg=000000&s=0)
Now we have
Theorem 30 (Quartic reciprocity)
If
and
are distinct primary primes in
then

We also have supplementary laws that state that if
is primary, then

Again, the first law handles units, and the second law handles the prime divisors of
. The corollary we care about this time in fact uses only the supplemental laws:
Proof: Note that
and applying the above. Therefore

Now we assumed
is primary. We claim that

Note that since
was is divisible by
, hence
divides
. Thus

since
is odd and
is even. Finally,


From here we quickly deduce