A field is a ring that is commutative and has division.
An example of a field is Q, the rational numbers.
An example of a ring that is not a field is Z, the integers.
However, Z is something called a domain, which means that if ab = 0, then a = 0 or b = 0.
For any commutative domain, one can define its fraction field. As its name suggests, it consists of all fractions whose numerator and denominator are in the domain. The fraction field of Z is, of course Q.
Another example is R[x], the ring of real polynomials in x. It is a domain, and it has a fraction field, R(x), the field of all real rational functions in x.
I'm going to assume that you understand modulo arithmetic of integers...
Let R be the the ring of integers modulo 8. This ring is not a domain because, for instance, 2 * 4 = 0. Thus, R does not have a fraction field. However, we can make some fractions. For example, we could take S = {1, 3, 5, 7} as the set of allowable denominators. This is called the localization of R at S, denoted RS. The same thing is also described in the opposite way; the localization of R at T where T = {0, 2, 4, 6}, also denoted as RT
moduli, in general, are a way of making a new ring from an old ring by specifying a set of elements that should be equal to zero, called an ideal. For example, we make Z mod 8 by declaring the set {..., -16, -8, 0, 8, 16, ...} to be equal to zero. If R is the ring, and I is the ideal, then we call the ring formed in this way the quotient ring of R over I, written R/I.
To satisfy this intuition, an ideal has special properties. I is an ideal of a ring, R, iff:
for any r in R and x in I: rx and xr are in I. (because multiplying by zero yields zero)
for any x, y in I: x - y are in I. (because zero - zero = zero)
(However, Z mod p, where p is a prime, is a field!)
An example of this is the ring S = R / (i^2 + 1)... (where R is the real numbers)
R is the ring of all real polynomials in the variable i.
i^2 + 1 is an element of R, and we use the notation (i^2 + 1) to mean the smallest ideal containing i^2 + 1.
The resulting quoteint ring consists of things that look like: {x + yi | x, y in R}. To see that this set is closed under multiplication:
(a + bi) (c + di) = ac + (ad + bc)i + (bd)i^2
= ac + (ad + bc)i + (bd)(i^2 + 1 - 1)
= ac + (ad + bc)i + (bd)(-1) (because, in S, i^2 + 1 = 0!)
= (ac - bd) + (ad + bc)i
which is, again, of the form x + yi.
In fact, you'll recognize that S is just the ring of complex numbers!
In general, if F is a field, and f is an irreducible polynomial in F[x], then it turns out that F[x]/(f), that is F[x] mod f, is always a field.