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This is a generalization of this question.

Let $A$ be a Dedekind domain. Let $K$ be the field of fractions of $A$. Let $I$ be a non-zero ideal of $A$. Let $\alpha$ be a non-zero element of $K$ which is relatively prime to $I$. That is, $(\alpha)$ and $I$ have no common prime factors.

Are there elements $\beta$, $\gamma$ of $A$ with the following properties?

(1) $\alpha = \beta/\gamma$.

(2) $\beta$ and $\gamma$ are relatively prime to $I$.

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Makoto Kato
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  • Somebody cascade downvoted for 6(or 7) of my questions 2 hours ago. Probably this question has something to do with it because I had been inactive for several months until I posted this question(and the other relevant one). – Makoto Kato Mar 02 '13 at 21:10

2 Answers2

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If $\alpha\in A$ there is nothing to prove, and thus we can assume that $\alpha\notin A$. Writing $(\alpha)=J_1/J_2$, where $J_i$ are integral ideals relatively prime to $I$, then $J_2\subseteq (A:\alpha)$. This shows that the ideal $(A:\alpha)=\{a\in A:a\alpha\in A\}$ is relatively prime to $I.$ Now we have to find an element $\gamma\in (A:\alpha)$ which is relatively prime to $I$. This follows immediately from this result: there exists $\gamma\in (A:\alpha)$ and a non-zero ideal $M$ such that $(\gamma)=(A:\alpha)M$ and $M+I=A$. Since $(A:\alpha)$ and $M$ are relatively prime to $I$ it follows that $(\gamma)$ is also relatively prime to $I$.

Remark. After posting my answer the OP suggested to apply this result directly to $J_2$ finding $\gamma\in J_2$ such that $(\gamma)=J_2M$ with $M+I=A$ and then observe that $(\alpha)=J_1M/J_2M$.

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(Ack, you made a new question just before I started updating my answer)

Since we are interested in what happens at $I$, consider its semi-localization $A_I$ of $A$ formed by inverting every element of $A$ relatively prime to $I$.

This truly does kill every prime not dividing $I$ by the Chinese remainder theorem: for every other prime there is an $a \in A$ such that $a \equiv 0 \bmod \mathfrak{p}$ but $a \equiv 1 \bmod I$.

Asking if $\alpha = \beta / \gamma$ with $\gamma$ relatively prime to $I$ is thus equivalent to asking of $\alpha \in A_I$. This is true iff the prime factorization of $\alpha$ as a fraction over $A_I$ contains no primes to a negative power.

Given such an equation, $\alpha$ is furthermore relatively prime to $I$ if and only if $\beta$ is.

The question you actually asked can be phrased as asking if $\alpha$ is a unit in $A_I$ rather than simply being an element of $A_I$.