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I am trying to show that the quotient of a Dedekind domain $A$ by an ideal $\mathfrak{a}$ is a PIR (principal ideal ring). Now by using the Chinese Remainder Theorem and the fact that a direct product of PIRs is a PIR it will suffice to prove the following:

Let $\mathfrak{p}$ be a prime ideal of a Dedekind domain $A$. Then for any $n \in \Bbb{N}$ we have that $A/\mathfrak{p}^n$ is a PIR.

Now it is proven in Proposition 9.2 of Atiyah - Macdonald that a Noetherian local domain of dimension $1$ is a PID. Applying this proposition mutadis mutandis I think it will suffice to prove that $A/\mathfrak{p}^n$ is isomorphic to some kind of localization. Now we know that $$(A/\mathfrak{p}^n)_{\mathfrak{p}} \cong A_\mathfrak{p} \otimes_A A/\mathfrak{p}^n.$$

If $A/\mathfrak{p}^n$ has the structure of an $A_\mathfrak{p}$ -module then the right hand side is in fact a tensor product over the localization and so $$A_\mathfrak{p} \otimes_A A/\mathfrak{p}^n \cong A_\mathfrak{p} \otimes_{A_\mathfrak{p} } A/\mathfrak{p}^n \cong A/\mathfrak{p}^n.$$

My Question is: Can we make $A/\mathfrak{p}^n$ into an $A_\mathfrak{p}$ - module?

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3 Answers 3

up vote 2 down vote accepted

To answer the question asked, one method is to mimic the proof that $m$ is invertible in $\mathbb{Z} / \mathbb{n} \mathbb{Z}$ if and only if $m$ and $n$ are relatively prime.

But since we're merely in a Dedekind domain rather than a Euclidean domain, the key is to use ideal arithmetic.

If $s \notin \mathfrak{p}$, then the ideals $(s)$ and $\mathfrak{p}^n$ are relatively prime, and so $(s) + \mathfrak{p}^n = (1)$. Therefore, there exists some $t$ such that $$ s t \equiv 1 \pmod{\mathfrak{p}^n}$$

Among the ways to finish the proof off are to observe that this means

$$ A[s^{-1}] / \mathfrak{p}^n \cong (A / \mathfrak{p}^n)[s^{-1}] \cong A / \mathfrak{p}^n $$

(I assume the notation is understood) which leads to an isomorphism

$$ A_\mathfrak{p} / \mathfrak{p}^n \cong A / \mathfrak{p}^n $$

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Yes that was the intuition I had earlier, see my comment to Alex Youcis' answer above. –  user38268 Dec 11 '12 at 3:44
    
In your isomorphism in the last line shouldn't it be $\mathfrak{p}^n_{\mathfrak{p}}$ on the left? –  user38268 Dec 11 '12 at 3:48
    
@BenjaLim: That falls under the heading of "assuming notation is understood". Technically I had in mind $\mathfrak{p}^n A_\mathfrak{p}$, although that is equal to $\mathfrak{p}_\mathfrak{p}^n$. Also, in the previous line, that should be $\mathfrak{p}^n A[s^{-1}]$, or maybe something like $\mathfrak{p}^n[s^{-1}]$. –  Hurkyl Dec 11 '12 at 16:34

EDIT: I'm sorry, that wasn't very polite of me. Yes, there is an $A_\mathfrak{p}$-module structure--you can check below to see why.

I'm confused. I think their is an easier way to do this.

Lemma: Let $A$ be a Dedekind domain, and $\mathfrak{p}$ a prime ideal. Then, $A_\mathfrak{p}$ is a PID.

Proof: You know that $A_\mathfrak{p}$ is a Dedekind domain, and since it has only finitely many prime ideals (it's local), it is a PID. $\blacksquare$

So, now let $\mathfrak{a}=\mathfrak{p}_1^{e_1}\cdots\mathfrak{p_g}^{e_g}$. Then, as you've said, we get that $A/\mathfrak{a}\cong A/\mathfrak{p}_1^{e_1}\times\cdots\times A/\mathfrak{p_g}^{e_g}$ and so it suffices to prove that $A/\mathfrak{p}^e$ is a PIR. But, note that $A/\mathfrak{p}^e$ is local (since the only prime dividing $\mathfrak{p}^n$ is $\mathfrak{p}$) with prime $\mathfrak{p}$. Thus, we know that $A/\mathfrak{p}^e$ is isomorphic to its localization. But, since localization commutes with quotients, the localization of $A/\mathfrak{p}^e$ is $A_\mathfrak{p}/\mathfrak{p}^eA_\mathfrak{p}$. But, by the lemma $A_\mathfrak{p}$ is a PID and so any quotient of it is a PIR.

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On how to make $A/\mathfrak{p}^n$ a $A_\mathfrak{p}$ - module I think the idea is the following. Consider $A = \Bbb{Z}$ and take $\mathfrak{p} = 2$ and $n = 3$. Then to make $\Bbb{Z}/(8)$ into a $\Bbb{Z}_{2}$ - module is equivalent to asking if the non-multiples of $2$ are invertible in $\Bbb{Z}/8$, which is true. –  user38268 Dec 11 '12 at 2:48
    
Also, I think I'm doing something stupid above in the question. By the correspondence theorem the prime ideals of $A/\mathfrak{p}^n$ are in bijective correspondence with those that contain $\mathfrak{p}^n$. But now to contain is the same as to divide in a Dedekind domain and so those ideals that contain $\mathfrak{p}^n$ are $\mathfrak{p}^i$ for $0 \leq i\leq n$. Of these the only ones prime are when $ i = 0$ and so we conclude that the quotient is in fact a local ring. –  user38268 Dec 11 '12 at 2:51
    
I hope you don't mind that I have accepted Hurkyl's answer above, as it answered the question that I had originally. Thanks though for your answer. –  user38268 Dec 11 '12 at 3:46

This is a proof of Alex Youcis' lemma above.

We prove that a Dedekind domain $A$ that is local (with maximal ideal $\mathfrak{m}$) is a PID. Suppose we are given an ideal $I \subseteq A$. Then it is clear that $I \subseteq \mathfrak{m}$. Now $\sqrt{I} = \mathfrak{m}$ for there is only one maximal ideal and prime ideals are maximal in a Dedekind domain.

Since $I$ is also Noetherian we conclude by Corollary 7.16 of Atiyah - Macdonald that there is some $n > 0$ such that $\mathfrak{m}^n \subseteq I$. However in a Dedekind domain, to contain is to divide and so we conclude that $I|\mathfrak{m}^n$ and so $I$ itself is a power of $\mathfrak{m}$. Now because $A$ is a Dedekind domain we see that $\mathfrak{m} \neq \mathfrak{m}^2$ (because of unique factorization). Hence we can choose an $x \in \mathfrak{m}$ such that $x \notin \mathfrak{m}^2$. Now consider $(x)$; by assumption there is a $k > 0$ such that $(x) = \mathfrak{m}^k$. Because $x \notin \mathfrak{m}^2$ we conclude that $ k = 1$ and so $\mathfrak{m} = (x)$. Since $(x)^k = (x^k)$ we conclude that any ideal $I$ in a Dedekind local domain is principal.

$\hspace{6.5in} \square$

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