Timeline for Do there always exist integers $\beta_2$, $\beta_3$ such that $\{a, \beta_2 + \omega_2, \beta_3 + \omega_3 \}$ is an integral basis for the ideal $(a, \alpha )$ of a cubic field
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| Feb 8, 2012 at 7:53 | comment | added | Samuel Hambleton | 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! | |
| Feb 6, 2012 at 17:26 | comment | added | Franz Lemmermeyer | 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. | |
| Feb 6, 2012 at 2:54 | comment | added | Samuel Hambleton | 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! | |
| Feb 6, 2012 at 2:50 | vote | accept | Samuel Hambleton | ||
| Feb 4, 2012 at 19:17 | history | answered | Franz Lemmermeyer | CC BY-SA 3.0 |