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May 29, 2014 at 11:12 comment added Albertas Maybe it's worth to add that such a $\delta$ needs not exist among the integers of the field $\mathbb{Q}(\alpha, \beta)$.
Feb 17, 2011 at 20:20 history edited Esteban Crespi CC BY-SA 2.5
Substituted $H$ (wrong) with ${\mathcal O}_H$
Feb 17, 2011 at 20:16 vote accept Esteban Crespi
Feb 17, 2011 at 16:41 comment added Franz Lemmermeyer I don't see what your problem with Hecke's proof is; the ideal generated by $\alpha$ and $\beta$ in the extension is $(A)$, and intersecting it with the base field shows that the ideal downstairs consists of all multiples of $A$ lying in $K$.
Feb 17, 2011 at 16:39 answer added Franz Lemmermeyer timeline score: 6
Feb 16, 2011 at 23:32 answer added Pete L. Clark timeline score: 20
Feb 16, 2011 at 23:25 history asked Esteban Crespi CC BY-SA 2.5