In the post about number theory and lattices, we tried to determine when is the Euclidean distance in is actually a Euclidean norm and we were led to study the embeddings of rings such as
as lattices in
. As mentioned there, not all rings can be nicely embedded in
as lattices, and in this post we will try to find the “right” way to view them as lattices, and what other algebraic problems can be translated to this setting.
I will only assume basic knowledge about rings, and I will not assume any knowledge about algebraic number theory. I believe that the right order is to first learn about algebraic number theory and then see its connection to lattices. This being said, in my opinion the best way to show the result in this post is through lattices (and it is usually done in this way), hence it should not be problematic for those unfamiliar with the black magic of number theorists. For those who do want to do some reading first (or after), I recommend J.S. Milne great course notes.
Historically, a big part of algebraic number theory was developed in order to solve equations over the integers. Probably, the most famous is Fermat’s last conjecture (which later became a theorem), but before that, let us try to solve some more simpler equations that naturally lead to algebraic extensions of .
Consider first the problem of solving where
are integers. If there exists a solution, then clearly we must have that
is an integer, but also it must be nonnegative. One last obvious but important property is that this equation cannot have infinitely many solutions, since
(and sometimes there are no solutions).
The geometric interpretation of this problem is finding all the lattice points from which lie on the circle of radius
around the origin. Since this lattice can be also viewed as
, we can ask what does this problem mean in this setting. As usual, any sum of two squares can (and should!) be written as
, so a solution to this equation is equivalent to decomposing the number
over the ring
. This reformulation, and a little bit of knowledge about the ring
leads to a full description of the integers
which have a presentation as a sum of two squares (for
prime, one way to prove an existence of a solution is Minkowski’s theorem). A similar question can be asked about the equation
, which in a similar process leads us to consider the ring
.
To make things interesting, we can consider unbounded curves, for example the hyperbola . While it is not clear that there are only finitely many lattice points on this hyperbola, as in the circles\ellipses examples, it is still true. Indeed, if there are infinitely many solutions, then we can assume that they are positive so that
. Writing the equation as
we see that
. As these are integers, they must be equal so that
. As the image below shows, there are many integral points close to
, but there are only two points on this curve (and a similar claim is true for general
).
What happens if we consider instead the equation ? Following the examples up until now, we will start with the decomposition
so that we’re looking for solution in
. The reason this trick usually helps, is because it is easier to work with just multiplication in a ring (think the unique factorization for integers), than multiplication and addition as in the original equation. As in the case
, we can ask for which
there is a solution, but let us concentrate on the problem where
.
The geometric interpretation of are the lattice points from
contained in two hyperbolas. But now we have another way of viewing it as
, so changing our basis to
, we see that
is a lattice point from
and we’re looking for such point which satisfy . In other words, instead of working with a “normalized lattice”
and with “weird” hyperbolas, we work with a “weird” lattice and “normalized” hyperbolas.
We are thus led to consider the map defined by
which is injective and its image is exactly the right lattice above. Note that just sending
to its embedding in
will produce a dense subset, so in a sense we added one more dimension to counter this problem (actually, the fact that it is dense is exactly why the argument that
has only finitely many solutions doesn’t work here). Now that we have an embedding of the ring as a lattice, we can ask what does it mean that it has a lattice point on the hyperbolas
.
The function has the same role here as the complex conjugation
has in
. It is an automorphism of the ring, i.e. it is invertible (with itself as an inverse) and it satisfies
and
. Furthermore, if
, then
. By the definition we have above, an element
is sent to
and we ask what does it means that
.
Clearly, if , then
is invertible with inverse
. On the other hand, if
is invertible, then
for some
, and applying
we get that
also. Multiplying these two equations we obtain that
– a product of two integers equals one, so
.
We conclude that an element is sent to a lattice point on if and only if it is invertible – this is the algebraic interpretation of this geometric property. Furthermore, if
is invertible, then so is
for any
. It doesn’t yet means that we have infinitely many lattice points on
, since we might have
for some
(e.g.
), but we are quite close. Searching a little bit, we see that
is invertible (with inverse
), and moreover
. It then follows that
so we can find infinitely many points on
, or in other words, we have just used an extension of
(or equivalently of
) to show that
has infinitely many distinct solutions! Also, there was nothing special about 2 – for any
which is not a square (why?), if you can find a single nontrivial (
) solution to
, then you can find infinitely many distinct solutions (try to see if you can prove this result without using the algebraic interpretation, i.e. the multiplicative structure of the set of solutions). We are left to ask whether we can find a single nontrivial solution, or in other words, find the group of invertible elements in
which correspond to integer solutions of
. These equations for
nonsquare integer are called Pell’s equations.
Going back to our example with the ring , it can be viewed as a lattice inside
and its invertible elements are exactly the lattice points on
, and moreover there are infinitely many such points. While the hyperbolas
are “normalized” they are still not that easy to work with, and we rather work with straight lines than these hyperbolas (or algebraically speaking, we prefer addition to multiplication). Thus, we use the standard trick to move from multiplication to addition – the logarithm. Given a point
on
, we send it to
. The condition
implies that
so that these points lie on the 1-dimensional subspace
. The fact that the invertible elements form a group, imply that their images in
form an additive subgroup. Finding a nontrivial solution (=invertible element) is equivalent to show that this group is not trivial, and at least in the case
we see that this group is actually a lattice in
.
Everything mentioned up until now is not special for and is true in much greater generality. I will finish this post with the formulation of these general results, which are due to Dirichlet.
Definition: Let be a finite extension. A unital subring
is called an order if
- The ring
together with
generate
(“not too small”).
- As an abelian group we have that
where
(“not too big”).
Example: The rings inside
. Similarly we can take the rings
.
If is a finite extension of degree
, then it can be shown that there are exactly
distinct homomorphisms
. We let
be the number of such maps with
and
be the number of maps that don’t satisfy this condition (each such map comes with its complex conjugate), so that
.
In the examples we used so far, we only had two embeddings – the identity and the conjugation (both in the “complex” cases like where
, and in the “real” case like
where
). You should check that the definition and theorem below, when restricted to these example, are exactly what we proved in this post so far.
Definition: Let be a finite extension. Let
be its real embeddings and
be its nonreal embeddings. Define
We are finally ready to state the theorem for general field. Parts (2) and (3) are usually called Dirichlet’s unit theorem.
Theorem: Let be a finite extension and let
be an order. Then
is a lattice in
.
is a lattice in
.
- We have that
where
is the finite cyclic group of all the roots of unity in
.
Remark: Going back to the problem of solving the equation , the theorem above not only tells us that there are infinitely many such solutions (for
square free), in a sense they are all generated by one solution. In this case
so that
is a one dimensional space. In the case of
, every solution arises from an element of the form
for
.
Somewhere in the beginning of this post I mentioned Fermat’s last theorem\conjecture, but this was a long time ago and as this is already a long post I will leave it for a future post.