Let S be an infinite compact hausdorff space and let C(S) be the ring of continuous functions from S into R1. Review the properties of this ring. In particular, M is a maximal ideal iff it consists of all functions that vanish on a point x in S.
If there is a primary decomposition then there are finitely many associate primes. What does the conductor ideal [0:f] look like? The function g is an annihilator of f iff g is zero wherever f is nonzero. Taking radicals, g is in rad([0:f]) iff some power of g is 0 wherever f is nonzero. This is actually the same constraint. The associate ideal is the set of functions that vanish wherever f is nonzero. We want to know when this associate ideal is prime.
Ignore f = 0, for the entire ring is not a prime ideal.
If f is nonzero on two points u and v, then every g in rad([0:f]) must be 0 on u and v. It is possible to build two functions g1 and g2 whose product is 0, yet g1(u) ≠ 0 and g2(v) ≠ 0. Since S is normal, a continuous function h maps u to 0 and v to 1. Rescale, so that h is 1 on u and -1 on v. Let g1 equal h when h is positive, and 0 otherwise. (Yes, this is a valid function in c(S).) Let g2 equal h when h is negative, and 0 otherwise. Now g1g2 lies in rad([0:f]), but g1 and g2 do not. This contradicts primality, hence we only need consider conductors with f(u) ≠ 0, for some point u, and g(u) = 0. These conductors are maximal ideals in c(S), and they are prime. There is one such conductor for each isolated u, so that f can be nonzero on u and 0 elsewhere.
Every point u that builds an associate ideal is isolated, and open. Since S is hausdorff, every point is also closed. The points that build the associate primes are disconnected components.
Since 0 has a primary decomposition, there are finitely many isolated points ui, leading to associate primes Pi. Let f be 0 on all isolated points ui, and 1 elsewhere. We can do this because the union over ui is closed, and its complement is also closed. Now f lies in each Pi, and some power fn lies in the intersection of Qi. In other words, fn = 0, thus f = 0. There are no points, outside of ui, where f can be 1. This contradicts the fact that S is an infinite set. Therefore 0 has no primary decomposition.
Assume S has a nonisolated point u, and let M be the maximal ideal of functions that vanish on u. Shrink M down to a minimal prime ideal P. To be an associate prime, P must vanish on some point v. If P vanishes on both u and v it cannot be prime, hence v = u. Thus P is the conductor of f, where f is nonzero only on u. The preimage of 0 is a closed set, making u an open set, which contradicts the fact that u is not an isolated point. Thus there are minimal primes that are not associate primes.