If a product of various prime powers contains H then any one of those prime powers contains H. Focus on Pn containing H.
Localize about P, creating a dvr. If S is the complement of P, the new ring is denoted RP, or R/S.
The proper ideals in R/S correspond to the powers of P. Inject Pn into R/S, then contract back to R, and get Pn. This doesn't hold for all rings, but it holds true for a dedekind domain and its localization.
In ject H into R/S, giving an ideal that I will call H/S. Since there are no zero divisors, product and extension commute, hence H/S is the product of the images of the prime powers that build H. Most of these extensions are the entire ring R/S. The only extension that survives is Pn/S, which contains H/S. Therefore H/S = PJ/S for some J ≥ n.
If G contains H, localize about every prime in G, and apply the above. The exponents on the prime powers in the factorization of H are at least as large as the exponents on the corresponding prime powers in the factorization of G. Subtract exponents to find the quotient G/H. In other words, G divides into H. To contain is to divide.
As a corollary, finitely many ideals contain a given ideal H. These are the factors of H. The number of factors equals the product of the successors of the exponents on the primes of H.
The sum of two ideals G and H is the smallest ideal J that contains both G and H, which is the smallest ideal that divides both G and H. In other words, J = gcd(G,H). To find Pe in J, find Pm in the factorization of G and Pn in the factorization of H, and set e to the smaller of m and n.
This can be generalized to the sum or intersection of a finite set of ideals.
Suppose all the powers of an ideal H contain an ideal J. The powers of H divide J, and if H is divisible by P, arbitrarily high powers of P divide J. Yet the factorization of J presents a specific power of P. This is a contradiction, hence the powers of H collectively contain only 0.
The sum of any two distinct prime powers is their greatest common divisor, which is R. Apply the chinese remainder theorem, and R/H is isomorphic to a finite direct product of quotient rings, where each ring is R/Pe, where Pe is a factor of H.
Select x in R/H so that x becomes 1 in almost every projection R/Pe. However, for one prime, which I will call Q, x is an element in Q, but not in Q2, when projected into the ring R/Qf.
Factor the ideal generated by x in R/H. If P divides [x] then reduce mod P and x goes to 0. However, x is 1 mod P. Therefore [x] is a power of Q.
Since Q contains x, but Q2 does not, [x] has to be Q. Therefore Q is principal, generated by x.
Run this construction for every prime in R/H. Every prime is principal, and R/H is a pid. (Technically a pid is suppose to be an integral domain, but you know what I mean; every ideal is principal.)
Given any ideal H, take a nonzero element x in H, and consider the image of H in R/[x]. The image of H is principal, generated by y. In other words, y generates all the cosets of [x] in H. Therefore x and y generate H.