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!
