Let $K$ be a finite extension of the field of rational numbers. Let $A$ be the ring of algebraic integers in $K$. Let $I$ be a non-zero ideal of $A$. Let $\alpha$ be a non-zero element of $K$ which is relatively prime to $I$.

Are there algebraic integers $\beta$, $\gamma$ in $A$ with the following properties?

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

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

EDIT(Mar.2,2013) Here is a generalization of this question.

  • 3
    $\begingroup$ I wonder was this question downvoted because it's a poor question? Or because Makoto is the one asking it? $\endgroup$ – JSchlather Feb 28 '13 at 9:34
  • $\begingroup$ PS I interpreted "relatively prime to $I$" to mean "the prime factorizations of the invertible ideals $(\alpha)$ and $I$ don't have any primes in common". $\endgroup$ – user14972 Feb 28 '13 at 20:24
  • $\begingroup$ Can the assertion be proved without using the fact that the class number of $K$ is finite? $\endgroup$ – Makoto Kato Feb 28 '13 at 22:03
  • $\begingroup$ @Hurkyl I didn't think it's necessary to define a self-evident terminology. $\endgroup$ – Makoto Kato Feb 28 '13 at 23:06
  • $\begingroup$ The other self-evident meaning is $1 \notin I + (\alpha)$. A priori, this doesn't look equivalent to the other when $\alpha$ is an element of the fraction field. $\endgroup$ – user14972 Feb 28 '13 at 23:43

Any "obvious" algorithm works. Many/most of them could even be adapted to a proof.

The following is a quick proof that there isn't an obstacle to avoiding $I$:

Factor $(\alpha)$ as $J_1 / J_2$ where $J_i$ are integral ideals both relatively prime to $I$ (e.g. by splitting the factorization of $(\alpha)$ into the parts with positive and negative exponent).

Let $J_2^h = (g)$ for some positive integer $h$ (e.g. let $h$ be the class number of $K$).

Then we can choose $\beta = g \alpha$ and $\gamma = g$.


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