Consider two (primitive) elements $\pi_{i} \in \mathbb{C}$, such that $\pi_{1} = M \pi_{2}$ for $M \in \mathcal{S}_{m}$ with $$\mathcal{S}_{m}:=\Big\{\begin{pmatrix} A & B \\ 0 & D \end{pmatrix} \in \mathrm{GL}_{2}(\mathbb{Z}) : AD = m >0,\; 0 \leq B \leq D\Big\}.$$ Let $\Lambda_{i} = \mathbb{Z}\pi_{i} + \mathbb{Z}$ denote the corresponding lattice. Does this imply that: (a) the elliptic curves that correspond to these lattices are isogenous, and (b) the conductor of the order that corresponds to $\pi_{1}$ divides the conductor of the order that corresponds to $\pi_{2}$, which equals det($M$).
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1$\begingroup$ Usually $\operatorname{GL}_2(\mathbb{Z})$ are all matrices with unit determinant, so your $m$ would be $\pm 1$. $\endgroup$– Chris WuthrichCommented Oct 10, 2022 at 10:24
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$\begingroup$ But this is "almost" $SL_{2}(\mathbb{Z})$ (with the (-1) as well) and it only gives homothetic lattices (isomorphic elliptic curves), right? $\endgroup$– EAgCommented Oct 10, 2022 at 10:28
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1$\begingroup$ That is why I think you meant to use a different notation. You may take matrices in $\operatorname{GL}_2(\mathbb{Q})$ with integer entries. $\endgroup$– Chris WuthrichCommented Oct 10, 2022 at 10:32
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$\begingroup$ The set $\mathcal{S}_{m}$ I am referring to, can be found in Silverman's "Advanced Topics in the Arithmetic of Elliptic Curves", pp.143-148. My question above is my understanding of those pages but I wanted to double check that I am not "over-simplifying" the isogenous situation for elliptic curves over $\mathbb{C}$. $\endgroup$– EAgCommented Oct 10, 2022 at 10:32
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