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?

  • $\begingroup$ 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

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 $\mathfrak{a}+\mathfrak{p}=R$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.