Proof
Observe that for two
elements a, b 
R the following holds:
I if and only if
(a + I) * (b + I) = I.There are four assertions to be verified.
I prime implies R/I is a domain
R/I is a domain implies I prime
I maximal implies R/I is a field
R/I is a field implies I maximal
Suppose that I is a prime ideal. We need to show that
the quotient ring has no zero divisors.
Suppose that a + I and b + I are elements whose
product is the zero element: (a + I) * (b + I) = I.
This comes down to ab belonging to I. As I is a prime
ideal a or b belongs to I. In other words:
a + I = I or b + I = I. This shows
that R/I is a domain.
Suppose that R/I is a domain. If ab
I, then it follows that a + I = I or
b + I = I. But this means a I
or b I.
I, then it follows that a + I = I or
b + I = I. But this means a I
or b I.
Suppose that I is a maximal ideal. Let a + I be a
nonzero element of R/I; that is, a does not
belong to I. Then the ideal (a) + I is strictly
bigger than I. Maximality of I implies
(a) + I = R. In particular, there exist
b R and c
I with
ab + c = 1. Thus, ab + I = 1 + I,
from which we derive (a + I)(b + I) = 1 + I.
Hence a + I is invertible in R/I.
This establishes that R/I is a field.
R and c
I with
ab + c = 1. Thus, ab + I = 1 + I,
from which we derive (a + I)(b + I) = 1 + I.
Hence a + I is invertible in R/I.
This establishes that R/I is a field.
Conversely, suppose that
R/I is a field. Let J be an ideal
of R
strictly containing I.
Then there is a J\I, so
a + I 
I. Thus being nonzero,
a + I has a
multiplicative inverse: for some b R we have
(a + I)(b + I) = 1 + I. But then
ab - 1 I, that is, 1 (a) + I, so
1 J, whence J = R.
This establishes that I is a maximal ideal.
J\I, so
a + I
I. Thus being nonzero,
a + I has a
multiplicative inverse: for some b R we have
(a + I)(b + I) = 1 + I. But then
ab - 1 I, that is, 1 (a) + I, so
1 J, whence J = R.
This establishes that I is a maximal ideal.