Dedekind Domains, An Introduction


A dedekind domain (biography) is a beautiful generalization of a unique factorization domain (ufd).

As you recall, a nonzero nonunit in a ufd is uniquely a product of prime elements. In the integers, 30 is 2×3×5, and that's the end of it. A dedekind domain is similar, but now we're working with ideals. Every nonzero nonunit ideal (i.e. nonzero proper ideal) is uniquely a product of prime ideals. The multiples of 30 are the multiples of 2 times the multiples of 3 times the multiples of 5, and that's the end of it.

A field has no proper nonzero ideals, hence it is dedekind by default.

Let's show that a pid is dedekind. Let R be a pid, and H a nonzero proper ideal. Since R is a pid, H = xR for some generator x. Now H is a product of ideals iff x is the product of their generators, or an associate of this product. Also, an ideal is prime iff its generator is prime. Therefore the unique factorization of x implies a unique factorization of H.

There are ufds that are not dedekind. We'll see this later on.


If P is a prime ideal, and S and T are proper ideals in R, and P*S = P*T = H, then write H as a unique product of prime ideals. This product necessarily includes P. It also includes the primes of S, or the primes of T. The primes of S are therefore equal to the primes of T, and S = T. If PS = PT, we can cancel P, like an integral domain.

If J is an ideal, and JS = JT, write J as a product of prime ideals and cancel them one at a time. This leaves S = T.

Chains of Ideals

Let S H and J be ideals, with J properly contained in H. Now SJ lies in SH, and if SJ = SH then cancel S to show J = H. Therefore SJ is properly contained in SH.

This assumes S is nonzero. Strict inequality persists even if S = R, or H = R, or J = 0.

Multiplication by S maps a chain of ideals (ascending, descending, or linearly ordered) to another chain of ideals.

Now for the converse. Let SJ lie properly in SH. If SJ = 0 then J = 0 (integral domain), which is a proper ideal in H, so set this case aside. If S = R then SJ = J and SH = H, so set this case aside.

Now J+H is an ideal that I will call H′. With SJ in SH, SH = SH′. Cancel S, and H = H′. Therefore J lies in H. Since SJ is properly contained in SH by assumption, J < H. Chains of ideals correspond under the action of S.