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Let $K$ be a cubic extension of the rationals, let $O_K$ be the ring of integers of $K$ and let $d$ be the discriminant of $K$.

True or false:

For every ideal $A$ of $O_K$, there exists an ideal $B$ of $O_K$ belonging to the same ideal class as $A$ such that $B + (d) = (1)$.

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What possible approaches have you come up with so far? – Alex Becker Dec 30 '11 at 7:31
Possible approach: Let $p$ be a prime divisor of $d$ and let $P$ be above $p$. I would guess that there is an ideal $I$ of $O_K$ such that $P \sim I$ and $I + (d) = (1)$. – Alecia Dec 30 '11 at 8:17
up vote 2 down vote accepted

In fact, for any finite extension $K$ of rationals (not just cubic), and for any non-$0$ ideal $J$ (not just $J=(d_K)$) it is true that any ideal class contains an ideal prime to $J$. If $J=\prod P_i^{a_i}$ and $A=C\prod P_i^{b_i}$ where $C$ is prime to $J$ then take $\alpha=\beta/\gamma$, $\alpha\in K$, $\beta,\gamma\in O_K$, such that $(\gamma)=D\prod P_i^{b_i}$ ($D$ prime to $J$) and $D|(\beta)$, $J$ prime to $(\beta)$ (such $\beta,\gamma$ exist by Chinese remainder theorem; even better, use a strong approximation theorem to get $\alpha$ directly). Then $B=\alpha A\subset O_K$ is prime to $J$.

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Cool. Thank you! – Alecia Dec 30 '11 at 11:54

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