## Dedekind Domains, An Introduction

### 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.

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.