Ideals in a non-dedekind domain that cannot be factored into product of primes

Question

For a domain $R$ to be a Dedekind domain it need to satisfy 3 conditions: one-dimensional, Noetherian, integrally closed.

I have got three domains satisfying all but one of those three:

1) $\mathbb{C}[x,y]$: not one-dimensional

2) Ring of all algebraic integers: not Noetherian

3) $\mathbb{Z}[\sqrt{-3}]$: not integrally closed

I know an equivalent definition of Dedekind domain is that all ideals of it can be factored into product of prime ideals.

Can anyone show me an ideal in 2) and 3) that cannot be factored into primes?

Answer 1

For (2), let $A/\mathbb Q$ be the ring of all algebraic integers. The failure of unique factorization of ideals in $A$ comes because you can always keep factoring things. For example, $$ 2 A = (\sqrt{2})^2 A = (\sqrt[4]{2})^4 A = \cdots . $$ Since no $\sqrt[2^n]{2}$ is a unit, this breaks factorization into (finitely many) prime ideals.

Answer 2

3) The source of the failure here: $\Bbb Z[\sqrt{D}]$ is "missing" $\frac{1+\sqrt{D}}{2}$ (which ${\cal O}_{\Bbb Q(\sqrt{D})}$ has) if $D\equiv1~(4)$.

Observe $(2,1+\sqrt{D})=(2,1-\sqrt{D})$ because $1+\sqrt{D}\equiv1-\sqrt{D}$ mod $2$. Hence

$$\begin{array}{ll} (2,1+\sqrt{D})^2 & =(2,1+\sqrt{D})(2,1-\sqrt{D}) \\ & =(4,2(1+\sqrt{D}),2(1-\sqrt{D}),1-D) \\ & =(2)(2,1+\sqrt{D}).\end{array}$$

The hypothesis $D\equiv1~(4)$ is used simplifying $(4,1-D)=(4)$. If $\Bbb Z[\sqrt{D}]$ were Dedekind we could cancel out and obtain $(2,1+\sqrt{D})=(2)~\Rightarrow~2\mid(1+\sqrt{D})\Rightarrow \frac{1+\sqrt{D}}{2}\in\Bbb Z[\sqrt{D}]$, a contradiction.