A simple example of a prime ideal is the ideal (0) in the ring Z: if ab = 0, then of course a = 0 or b = 0. In fact, the same argument shows that in any domain the zero ideal is prime.

The ideal (6) of multiples of 6 in Z is not a prime ideal: 2 * 3 = 6 and neither 2 nor 3 is a multiple of 6.

For every prime number p the ideal (p) of multiples of p in the ring Z is maximal: if an ideal J strictly contains p then it contains an integer m which is not a multiple of p. But then p and m are relatively prime and, by the extended Euclidean algorithm, there is a relation am + bp = 1. But this implies that 1 is contained in the ideal J and that J = Z. Hence each ideal that strictly contains (p) coincides with Z, so (p) is maximal.

The ideal (0) in Z is prime but not maximal: for example, the ideal (2) is proper and contains (0).

The ideal (m) in Z/nZ is prime if and only if gcd(m,n) is a prime number.

If n is prime, then, as Z/nZ is a field, the only maximal ideal of Z/nZ is (0). If p is a proper prime divisor of n, then (p) is a maximal ideal of Z/nZ.

In Z[X] the ideal (X) is prime: if the product of two polynomials is divisible by X, then at least one of them is already divisible by X.

In the ring R[X], the ideal (X) is maximal: if the ideal J strictly contains (X), then it contains a polynomial f with a nonzero constant term a. But then it follows that the ideal J contains a itself and so also the element 1. We conclude that J = R.

This case is very similar to modular arithmetic. For example, (X² + 1) is prime in Q[X]/(X⁴ - 1), but not in C[X]/(X⁴ - 1).

The same ideal is maximal in Q[X]/(X⁴ - 1), and not in C[X]/(X⁴ - 1). In the latter case, (X - i) and (X + i) are two maximal ideals containing (X² + 1).

In the ring R = Z + Zi, the ideal (i - 2) is prime and maximal. Of course this requires an argument.

The ideal (2) is not prime: (1 - i)(1 + i) = 2.