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Let $A$ be a Dedekind domain, and $a\in A-\{0\}$. I have to prove that every ideal of $A/(a)$ is principal.

This is a particular case of the exercise 9.7 in Atiyah's Introduction to Commutative Algebra, and therefore there are lots of solutions in the internet. What's the problem then? That I am not supposed to use the fact that any nonzero ideal of $A$ is a product of prime ideals or even that any ideal of $A$ admits a primary decomposition. Why? Because (if I am not wrong) I have to prove those results from the problem I am trying to solve.

The only thing I have done (and I am not even sure that it is right) is to show that $\dim A/(a)=0$, because a chain of prime ideals in $A/(a)$ $$ P_{1}\subsetneq P_{2} $$ would give, because of the correspondence theorem, a chain of prime ideals in $A$ $$ \{0\}\subsetneq Q_{1}\subsetneq Q_{2}, $$ which contradicts that $\dim A=1$. How could I continue from here?

$\textbf{Remark:}$ Could it help to know that $A/(a)$ is isomorphic as an $A$-module to $\oplus_{i=1}^{n}A/\mathfrak{p}_{i}^{r_{i}}$ for some maximal ideals $\mathfrak{p_{i}}$ of $A$?

user26857
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H. Jackson
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1 Answers1

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Let's assume that you know that $A/(a)$ is isomorphic as an $A$-module to $\oplus_{i=1}^{n}A/\mathfrak{p}_{i}^{r_{i}}$ for some maximal ideals $\mathfrak{p_{i}}$ of $A$.

Show that $A/\mathfrak p^r$ is a PIR by using that this is isomorphic to $A_{\mathfrak p}/\mathfrak p^rA_{\mathfrak p}$, and $A_{\mathfrak p}$ is a DVR.

Now you are done.

user26857
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  • Thank you for your help. I have another question. If I am not wrong, it is also true that given an ideal $I\neq (0), A$ of $A$, $A/I\simeq \oplus_{i=1}^{n}A/\mathfrak{p}{i}^{r{i}}$. According to this, the same argument is valid for arbitrary proper and no trivial ideals, isn't it? – H. Jackson Oct 24 '15 at 16:56
  • @H.Jackson Right. (Asked here.) – user26857 Oct 24 '15 at 16:58
  • By the way, I am having problems trying to prove that $A/\mathfrak{p}^{k}\simeq A_{\mathfrak{p}}/\mathfrak{p}^{k}A_{\mathfrak{p}}$. I guess the isomorphism is $$ A/\mathfrak{p}^{k}\rightarrow A_{\mathfrak{p}}/\mathfrak{p}^{k}A_{\mathfrak{p}}, a+\mathfrak{p}^{k}\mapsto \frac{a}{1}+\mathfrak{p}^{k}A_{\mathfrak{p}}, $$ but I am having problems trying to prove that the map is surjective. – H. Jackson Oct 26 '15 at 14:17
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    @H.Jackson If $R$ is a local ring with maximal ideal $m$, then $R\simeq R_m$ since every element from $R-m$ is invertible. – user26857 Oct 26 '15 at 14:21