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Let $R$ be a Dedekind domain, let $\mathfrak{p}$ be a nonzero prime ideal, and let $a\in\mathfrak{p}^r\setminus\mathfrak{p}^{r+1}$. Then $(a)\subseteq \mathfrak{p}^r$ and so $\mathfrak{p}^r\mid (a)$. Since this is in a Dedekind domain, this is equivalent to the existence of an ideal $\mathfrak{a}\vartriangleleft R$ such that $(a)=\mathfrak{p}^r\mathfrak{a}$.

I want to show that $\mathfrak{p}$ and $\mathfrak{a}$ are comaximal, i.e. $\mathfrak{a}+\mathfrak{p}=R$.

By multiplying by the ideal inverse, it suffices to show that $\mathfrak{p}^r\mathfrak{a}+\mathfrak{p}^{r+1}=\mathfrak{p}^r$, although I suspect there's a more direct way. The key idea should be that $\mathfrak{a}+\mathfrak{p}$ is the smallest ideal containing both $\mathfrak{a}$ and $\mathfrak{p}$, but I don't see how to use that $a\not\in\mathfrak{p}^{r+1}$. Can anyone spell this out?

Edit: I now want to show that $\mathfrak{a}\mathfrak{p}^r\cap\mathfrak{p}^{r+1}=\mathfrak{a}\mathfrak{p}^{r+1}$. Here's what I tried.

Given the result that Eric showed, we have $v_\mathfrak{p}(\mathfrak{a}+\mathfrak{p})=0$. Then $$v_\mathfrak{p}(\mathfrak{a}\mathfrak{p}^r\cap\mathfrak{p}^{r+1})=\max(v_{\mathfrak{p}}(\mathfrak{a}\mathfrak{p}^r), v_\mathfrak{p}(\mathfrak{p}^{r+1}))=v_\mathfrak{p}(\mathfrak{p}^{r+1})=r+1$$ Then since the intersection is the largest ideal contained in both, which must have $r+1$ prime factors, the intersection must be $\mathfrak{a}\mathfrak{p}^{r+1}$. Is there anything wrong here?

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  • $\begingroup$ The line $v_\mathfrak{p}(\mathfrak{a})=-v_\mathfrak{p}(\mathfrak{p})=-1$ doesn't make sense since valuations are nonnegative for integral ideals. You only have $v_\mathfrak{p}(\mathfrak{a} + \mathfrak{p}) \geq \min\{v_\mathfrak{p}(\mathfrak{a}), v_\mathfrak{p}(\mathfrak{p})\}$, and since $v_\mathfrak{p}(\mathfrak{a}) \neq v_\mathfrak{p}(\mathfrak{p})$ we have actually have equality: $v_\mathfrak{p}(\mathfrak{a} + \mathfrak{p}) = \min\{v_\mathfrak{p}(\mathfrak{a}), v_\mathfrak{p}(\mathfrak{p})\} = \min\{0,1\} = 0$. $\endgroup$ – André 3000 Oct 17 '17 at 23:02
  • $\begingroup$ As I said, it's not true that $v_\mathfrak{p}(\mathfrak{a}) = -1$. I tried to indicate that the "property" you seem to be using, namely additivity of the valuation, is not true, but maybe that wasn't clear. $v_\mathfrak{p}(\mathfrak{a}) = 0$ since $\mathfrak{a}$ and $\mathfrak{p}$ are relatively prime. Or what I should have written above: $0 = v_\mathfrak{p}(R) = v_\mathfrak{p}(\mathfrak{a} + \mathfrak{p}) \geq \min\{v_\mathfrak{p}(\mathfrak{a}), v_\mathfrak{p}(\mathfrak{p})\} = \min\{v_\mathfrak{p}(\mathfrak{a}),1\} = v_\mathfrak{p}(\mathfrak{a})$. $\endgroup$ – André 3000 Oct 17 '17 at 23:38
  • $\begingroup$ Basically you are doing the same when you take a natural number and write it as $2^em$. If you pick $e$ maximal, $m$ will be odd. $\endgroup$ – MooS Oct 18 '17 at 13:00
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Since $\mathfrak{p}$ is a maximal ideal of $R$, $\mathfrak{a}+\mathfrak{p}$ can only be equal to either $\mathfrak{p}$ or $R$. If it is equal to $\mathfrak{p}$, then $\mathfrak{a}\subseteq\mathfrak{p}$, so $a\in\mathfrak{p}^r\mathfrak{a}\subseteq\mathfrak{p}^{r+1}$. This is a contradiction, so we must have $\mathfrak{a}+\mathfrak{p}=R$.

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  • $\begingroup$ Yes, you can just multiply by $\mathfrak{p}^r$. $\endgroup$ – Eric Wofsey Oct 17 '17 at 21:16

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