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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?

hxhxhx88
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2 Answers2

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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.

anon
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  • Let $R$ be a valuation domain with value group $(\mathbb{Q},+)$. Let $\mathfrak{p}$ be the unique maximal ideal. Then $\mathfrak{p} = \mathfrak{p}^n$ for all $n \geq 1$...so having a prime factorization doesn't necessarily give you a bound on lengths of factorizations. Right? – Pete L. Clark Feb 20 '14 at 05:29
  • @PeteL.Clark Right. I suppose I should make more explicit the fact we are assuming Dedekindness to arrive at the contradiction. – anon Feb 20 '14 at 05:33
  • On second thought, the reasoning only shows the ring of algebraic integers is not Dedekind, not that PIs don't admit prime factorizations. – anon Feb 20 '14 at 05:40
  • Unfortunately I agree. That the ring of all algebraic integers does not satisfy ACC on principal ideals was an exercise I assigned in the algebraic number theory course I was teaching...and it was solved in a recent problem session as you had done in your answer. Showing an explicit failure of factorization into primes seems harder; I'll think more about it and possibly assign it! – Pete L. Clark Feb 20 '14 at 05:42
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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.

Cardboard Box
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    Hi, it is a good example, but I don't think it shows that $(2)$ cannot factor into finitely many prime ideals. It is a failure attempt to factor, maybe there is some other ways that work... – hxhxhx88 Feb 20 '14 at 14:51