This is a post in the series Purposeful Mathematics, where we try to mathematically answer the question “What is this used for?” in particular situations.
In today’s post, we explain the very basic idea of algebraic number theory, that is, to extend the usual tools and notions of the integers, such as divisors, primes, , to some more general structures, and use this generalities as new perspectives to solve problems in number theory. For those who want to dive in this fascinating topic, our recommendation is the book Algebraic Number Theory by J. S. Milne. There are numerous examples of this process, we choose one of the typical ones, which is a result of Pierre de Fermat: the only positive integer solution of the equation
is
.
Suppose that is a pair of positive integers satisfying the above equation. We have the following equality, which seems to be irrelevant to this problem.
If we replace the numbers and
by coprime positive integers
, then from the fact that
is a cube, we can conclude that
and
are cubes, and then we may keep on solving using this new piece of information. However,
and
are not integers. But in this particular case, it turns out that we still can have the same argument as above, because the set
is a Euclidean domain. Concretely, there exists a function
such that for each
,
, there exists
, such that
, and either
or
; in the case
, we say that
is divisible by
, denoted by
. This generalizes the notion of Euclidean division of integers. In particular, the function
here is given by
A simple calculation shows that , for all
. From this, we deduce that if
and
for some
, then
is divisible by
. Also, if
, then
for some . Choose integers
such that
, and let
. Then
. If
is nonzero, then
.
Now, consider the set equipped with a new structure, the map
, called the norm map. An element
is called a unit if there exists
such that
. It is not hard to check that if
is a unit, then
, which implies that
and
, or
. A nonzero element
is called a prime if it is non a unit, and there does not exists
such that both
and
are not units, and
. We show that, if
is a prime in
, and
for some nonzero
, then
or
. Suppose that
is not divisible by
, then
for some
, and
. Divide
by
, we get
, for some
, and either
or
. Keep on doing this process (recall the Euclidean algorithm for computing greatest common divisors), and note that in this process, the value of the norm map at the reminder decreases after each iteration, until it eventually reach some element
, such that the next iteration reduces
to
. Then look at the iteration process, by going backwards, it is not hard to see that
,
, and there exists
such that
. Because
, either
is a unit, or
, where
is a unit. If
, then
, but
(as
), this cannot happen. Hence,
is a unit, and then
is divisible by
, because
is divisible by
.
The above result actually shows that nonzero, nonunit elements in can be uniquely (up to multiplication by units) factorized into a product of primes in
. This allows us to apply the methods on integers described above to attack our problem. First, if
is the greatest common divisor of
and
, in the sense that
is the largest in the set of all
, where each
is a common divisor of
and
, then
divides
. However, if
, then it is not hard to see that
must be zero; otherwise we have
, or
and
have no nonunit common divisor. Exploiting the fact that
, we have
, for some integers
. Then
, and
. Thus
, and
. From this, we have that
, and
. We conclude that
is the unique solution of the given equation.
In the next post, we will be diving a little bit deeper into the interconnection of number theory, algebra, and complex analysis, via the properties of elliptic curves. See you later!