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$\mathbb{Z}[\alpha]$ is the quadratic integer ring associated to the squarefree integer $d$.

Let $m\in M \cap\mathbb{Z}$ and $\beta=a+b\alpha\in M$. If $\delta=x+y\alpha$, write $y=qb+r$, where $0\le r <b$, then $\delta-q\beta=x-qa+r\alpha$ by the Euclidean division in $\mathbb{Z}$. Moreover, $x-qa=q'm+s$ where $0\le s <m$.

Then we have $\delta=(q\beta+q'm) +(s+r\alpha$) where $q\beta+q'm\in M$.

How can we conclude that $M=m\mathbb{Z}\oplus(a+b\alpha) \mathbb{Z}$?

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Your result won't hold for any $m\in M\cap\mathbb Z$ and $\beta=a+b\alpha\in M$ (e.g. they could be $0$). You have to choose them carefully.

Let $m$ be such that $M\cap\mathbb Z=m\mathbb Z.$ Then, $M/m\mathbb Z$ is a subgroup of $\mathbb Z[\alpha]/\mathbb Z\simeq\Bbb Z$ hence it is generated by some $(a+b\alpha)+m\Bbb Z\in M/m\mathbb Z,$ which ends the proof.

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  • $\begingroup$ Sorry, I didn't understand your argument. Is there a way I can use that rewriting of delta? $\endgroup$ Commented Jan 7, 2023 at 10:24
  • $\begingroup$ This is not a "rewriting of [your?] delta". Which step precisely don't you understand? $\endgroup$ Commented Jan 7, 2023 at 10:46

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