This may be an easy question for the right people, but I did not find an answer anywhere. I am trying to figure out what one can say about the ideal $a\mathbb{Z}[\lambda] \cap b \mathbb{Z}[\lambda]$ in $\mathbb{Z}[\lambda]$, where $\lambda$ is algebraic and $a, b \in \mathbb{Z}[\lambda]$. Does it have to be principal and if not, what would be an example of this? Does the ideal have to be generated by two elements? Would this change if one replaced $\mathbb{Z}[\lambda]$ by the ring of integers in $\mathbb{Q}(\lambda)$?
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6$\begingroup$ This is equivalent to asking whether $a,b$ have a lcm in $\mathbb{Z}[\lambda]$. This is not always true, for example in $\mathbb{Z}[i\sqrt{5}]$, one can show that the elements $a=6$ and $b=2(1+i\sqrt{5})$ have no gcd, hence they have no lcm. $\endgroup$– François BrunaultCommented Mar 23, 2023 at 16:00
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3$\begingroup$ If $\mathbb{Z}[\lambda]$ is the ring of integers of $\mathbb{Q}(\lambda)$, then every ideal in $\mathbb{Z}[\lambda]$ is generated by two elements. $\endgroup$– David LoefflerCommented Mar 23, 2023 at 16:01
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1$\begingroup$ Maybe the previous remark should be elucidated a little bit: letting $R=\mathbb{Z}[\lambda]$, the intersection of the ideals $aR$ and $bR$ equals $abI^{-1}$, where $I$ is the gcd of $aR$ and $bR$, which is invertible if $R$ is the full ring of integers. It is easy to see that $I = (a,b)$. Since every integral ideal $I$ is of this form (and since every ideal class is represented by an integral ideal), it follows by combining these observations that, when we let $a,b$ range over all possible values of $R$, the class of $aR \cap bR$ takes every value in the ideal class group of $R$. $\endgroup$– R.P.Commented Mar 23, 2023 at 21:19
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