Let
n be an integer > 1. The
integers
modulo n
are
^{1}:
0,1,...,n1
We can see that there are
n of them.
The set of all integers modulo
n is denoted
by
Z_{n}.
The first problem is to try to write
n or
1 as one of the n
items in this list. More generally, if a is an
integer we want to be able to put a in the list.
The answer is as follows, given such an a we can choose a (unique)
integer
k such that a+kn is one of 0,1,...,n1.
Then we define a=a+kn. This makes sense because
the right hand side is in our list.
Let's look at an example. n=3.
and so on. More generally, we can see that

0=...,9,6,3,0,3,6,9,...

1=...,8,5,2,1,4,7,10,...

2=...,7,4,1,2,5,8,11,...
Ordinary integers can be added and multiplied
(they form a ring). How can we do the same for
the integers modulo n? Define:
a+b=a+b
a.b=ab
It turns out that these rules make Z_{n} into
a commutative ring. The identity element is 1 and the
zero element is 0.
This just means that 1.a=a and
0+a=a, for any a.
Let's look at n=3.
So we have Z_{3}={0,1,2}
and we can see that, for example,
2.2=4=1.
Thus, 2
has an inverse (itself) that is, it is a unit.
Z_{3} is a field (every nonzero element is a unit)
but this is not true in general. For example, in Z_{6}
the element 2 does not have an inverse. In fact we have
2.3=0. Actually thse two examples
are quite typical.
Proposition:

a is a unit in Z_{n} iff a
is coprime to n.

Z_{n} is a field iff n is prime.
Proof: If a
is a unit than there exists an integer b
such that a.b=1.
By definition there esists an integer
k with abkn=1. If a prime divides a
and n
then it divides 1, so (a,n)=1.
On the other hand, if (a,n)=1 then Euclid's algorithm
gives us integers r,s such that 1=ar+ns.
Thus 1=ar+ns=a.r and
a is a unit.
Thus a is not divisible by n. The second stament is clear
since a ring is a field iff every nonzero element is a unit.
Z_{n} is well adapted to studying congruences
the reason being that a=b iff a is congruent
to b modulo n, i.e. ab is divisible by n.
Results about congruences, such as Fermat's little theorem and
Wilson's theorem are rather neatly stated this way. For example, the
former says that x^{p}=x in Z_{p},
for a prime p.
Finally, we use some technology to give another identification
of the ring of integers modulo n. Define a function
f:Z>Z_{n} by
f(a)=a. The way we defined addition and multiplication
in Z_{n} make it clear that this is a
ring homomorphism. By definition it is surjective. It is clearly
not injective though. The kernel is the ideal of Z consisting
of all mutiples of n. This ideal is denoted by nZ.
By the first isomorphism theorem we can deduce that
Z_{n} is isomorphic to the quotient ring
Z/nZ.
The usual notation is to use a bar rather than an underline. Roll on
MathML!