## Dedekind Domains

Numbers can be factored: 21 = 3*7. So can ideals: in Z, (21) = (3)(7); in Z[i], the SQRT(-1) extension of Z, we can factor 5 as (1+2i)(1-2i), and so on.

Just like elements of a ring can be prime and irreducible, so can ideals be prime and maximal. A prime ideal is one that satisfies the condition that if the product ab is in the ideal, then so is either a or b; it can be proven that it’s equivalent to saying that if the ideal product AB is contained in the ideal, then so is either A or B. An ideal is maximal if there are no ideals between it and the entire ring.

For example: in Z, the ideal (5) is both prime and maximal. If ab is in (5), then 5 divides ab, which means 5 divides a or b, i.e. a or b is in (5). Now, if there is an ideal M containing (5) and another element c, then c is not divisible by 5, so that the greatest common divisor of c and 5 is 1. Therefore the ideal contains 1, which means it’s the whole ring.

Note that just like primality and irreducibility don’t apply to units, so do primality and maximality apply only to proper ideals, i.e. ideals smaller than the whole ring.

The language I’m using here suggests that there’s a close connection between prime elements and prime ideals. In fact, an element a is prime if and only if (a) is a prime ideal. Similarly, an element a is irreducible if and only if (a) is maximal among principal ideals; while (a) doesn’t have to be a maximal ideal, any ideal that properly contains (a) and is properly contained in the ring must be generated by at least two elements.

Every maximal ideal is prime: if M is a maximal ideal, and a and b are not in M, then (M, a) and (M, b) are both equal to the entire ring. Therefore (M, a)(M, b) is equal to the entire ring (note that this requires the ring to have 1; in the ring of all even integers, it won’t work). But if ab is in M, we get that the ideal product (M^2, aM, bM, ab) is contained in M, which means M is equal to the entire ring. So we have a contradiction, and ab is not in M, so M is prime.

Note that here, the arrow of causation is reversed: prime implies irreducible but not necessarily vice versa, and maximal implies prime but not necessarily vice versa.

In a principal ideal domain, every maximal ideal is maximal principal, of course, so every irreducible is prime, and every prime ideal is maximal. But there is a bigger class of rings in which every prime ideal is maximal, which don’t have to have unique factorization; this class includes all rings formed by attaching roots to Z, and many others.

To be precise, in algebraic number theory, we study Dedekind domains, which are characterized by the following properties:

1. Every prime ideal is maximal

2. The ascending chain condition holds for all ideals (we say the ring is Noetherian)

3. Every element of the fraction field of the ring that satisfies a monic polynomial with coefficients in the ring is itself in the ring (we say it’s integrally closed)

In condition 3, the fraction field is a construction involving fractions of the rings, which we can add, multiply, and express in lower terms just like ordinary fractions. For example, the fraction field of Z is simply Q, the field of rational numbers, and the fraction field of Z[i] is Q[i] = {a+ib: a and b are in Q}.

Generally, condition 2 is fairly trivial; it’s easy to come up with a non-Noetherian ring, but it’s not interesting in number theory. Condition 3 is complicated, but important, not least because every unique factorization domain satisfies it. For a good example of a ring that doesn’t, consider Z[SQRT(-3)]: the element (1+SQRT(-3))/2 satisfies the polynomial x^2 – x + 1 = 0, which is monic, and can be constructed from fractions in the ring, but is not in the ring itself.

The most important property of Dedekind domains is that every Dedekind domain with unique factorization is a principal ideal domain. There are two ways to prove that – an annoyingly difficult proof from first principles, and a proof that follows easily from unique factorization into ideals.