# A wrong proof about Dedekind domains

I "proved" that a Dedekind domain is a PID, but as we know this is wrong (for example $\mathbb{Z}[\sqrt{-5}]$). I do not know what is wrong in my proof:

Suppose $R$ is a Dedekind domain, $I$ is any nonzero ideal of $R$, $\mathfrak p_i$ all the primes of $I$. If $r_i$ is the $\mathfrak p_i$-adic value of $I$, we can find an element $x\in R$ with its $\mathfrak p_i$-adic value $r_i$, so $I=(x)$ is principal.

Who can tell me what is wrong in my "proof"?

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If there are infinitely many primes in $R$, then are you sure you can specify the valuation of $x$ at each one? – Keenan Kidwell Sep 29 '11 at 14:26

## 1 Answer

First, your presentation seems slightly muddled. I think you are arguing the following:

Let $I$ be a nonzero ideal in a Dedekind domain $R$. Then $I$ admits a factorization into primes $I = \mathfrak{p}_1^{a_1} \cdots \mathfrak{p}_r^{a_r}$. Now you observe that (for instance by the Chinese Remainder Theorem, or Artin-Whaples approximation) there exists an element $x \in R$ such that for all $1 \leq i \leq r$, $\operatorname{ord}_{\mathfrak{p}_i}(x) = a_i$. Finally you want to claim that $I = (x)$. (I hope I am understanding correctly.)

The problem is with the very last step: it need not be the case that $I = (x)$. You have enforced that for all $1 \leq i \leq r$, $\operatorname{ord}_{\mathfrak{p}_i}(x) = \operatorname{ord}_{\mathfrak{p}_i}(I) = a_i$. But what about all the other prime ideals of $R$? The Chinese Remainder Theorem does not say that you can find such an $x$ which is divisible by the given set of primes to prescribed multiplicities and is not divisible by any other prime ideals of $R$: indeed, as you have noticed, this is necessarily false in any Dedekind domain which is not a PID.

Nevertheless the argument above does prove something. In fact it proves several things:

1) ("Moving Lemma") Given any nonzero ideal $I$ in a Dedekind domain and any finite set $S$ of nonzero prime ideals of $R$, there exists $0 \neq x$ in the fraction field $K$ such that the fractional ideal $(x) I$ is not divisible by any $\mathfrak{p} \in S$.

This implies:

2) If $R$ has only finitely many prime ideals, then it is a PID.

In fact it also implies:

3) If all but finitely many prime ideals of $R$ are principal, then all prime ideals of $R$ are principal and thus $R$ is a PID.

Fact 2) above is a standard exercise in this subject. For some reason 3) is not: it seems to have first been recorded by Luther Claborn in the 1960's: see Corollary 1.6 here. I "rediscovered it" a few years ago when teaching a course on algebraic number theory. To better appreciate the result, note that the ring $\mathbb{Z}[\sqrt{-3}]$ has infinitely many prime ideals, exactly one of which is not principal. But it is not a Dedekind domain, since it is not integrally closed in its fraction field.

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Yes,that is my meaning and you find the reason,thanks.I lose other primes,their adic values maybe positive,Chinese Remainder Theorem can not go to infinite,so we just can get that there is some idea J with IJ=(a). – Strongart Oct 1 '11 at 10:35
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