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Let $K$ be a cubic extension of the rational numbers of discriminant $D$ and $\{ 1, \omega_2, \omega_3 \}$ be an integral basis for the ring of integers $\mathcal{O}_K$ of $K$. Let $\alpha \in \mathcal{O}_K$ be primitive so that no rational prime divides $\alpha$, let the norm of $\alpha$ be equal to $a^3$, with $a \in \mathbb{Z}$, and assume that $a$ is prime to $D$.

Question: Do there always exist rational integers $\beta_2$, $\beta_3$ such that $$\{ a, \beta_2 + \omega_2, \beta_3 + \omega_3 \}$$ is an integral basis for the ideal $(a, \alpha )$ ? Even when $\gcd (disc.(\alpha ), a ) \not= 1$ ?

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  • $\begingroup$ By $\text{disc} (\alpha )$ I mean the discriminant of the minimum polynomial of $\alpha$. $\endgroup$ – Samuel Hambleton Feb 4 '12 at 9:38
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If I understand this correctly, then the answer is no. If an ideal $I$ has a basis of the desired form, then clearly any element of the ring of integers is congruent to a rational integer modulo $I$. But this implies that $I$ is a product of ideals of inertia degree $1$.

Thus if $\alpha$ is the cube of an ideal of degree $2$, such a basis cannot exist.

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  • $\begingroup$ Ah-ha! Thank you very much Dr. Lemmermeyer. I think that the rational prime $p \mid \gcd (\text{disc}(\alpha ), a)$ if and only if there is an ideal $(p, \pi )$ of inertia degree $2$ dividing $I$. Thus an ideal $I$ may have such a basis while it conjugate might not! $\endgroup$ – Samuel Hambleton Feb 6 '12 at 2:54
  • $\begingroup$ If the extension is not normal, the "conjugate" ideal is not in the same field; in normal cubic extensions there are no ideals of degree 2. If you're interested in class groups, a theorem of Kummer (generalized by Hilbert) states that every ideal class is generated by an ideal divisible only by prime ideals of degree 1. $\endgroup$ – Franz Lemmermeyer Feb 6 '12 at 17:26
  • $\begingroup$ I meant to say that the rational prime $p$ might split as $(p) = \mathfrak{p}_1 \mathfrak{p}_2$, where $\mathfrak{p}_1$ has degree $1$ and $\mathfrak{p}_2$ has degree $2$, in which case $\mathfrak{p}_1$ would have such a basis while $\mathfrak{p}_2$ does not. Your remarks are interesting anyway. I'll keep the theorem in mind. It looks like it could be handy. Thanks! $\endgroup$ – Samuel Hambleton Feb 8 '12 at 7:53

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