## Dedekind Domains, To Contain is to Divide

### To Contain is to Divide

If H and J are ideals, H contains J iff H divides into J. The reverse implication is clear, so let's prove the forward direction.

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.

### Sum and Intersection

The intersection of two ideals G and H is the largest ideal J that is contained in both G and H, which is the largest ideal that is divisible by both G and H. In other words, J = lcm(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 larger of m and n.

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.

### Quotient Becomes a PID

Let H be a proper nonzero ideal in a dedekind domain R, and factor H into its prime powers. The ideals containing H become ideals in R/H. In fact, R/H inherits the contain/divide relationship.

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.)

### Two Generators

Every ideal in a dedekind domain can be generated using one or two generators.

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.