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Let $K$ be a imaginary quadratic field. Let $\mathcal O$ be its ring of integers. Suppose that $2$ divides the discriminant of $K$.

What is the structure of the multiplicative group $(\mathcal O /2^k \mathcal O)^\times$?

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  • $\begingroup$ This is almost a duplicate of this question, asked at almost the same time: math.stackexchange.com/questions/3180079/… $\endgroup$
    – Furlo Roth
    Apr 10, 2019 at 0:27
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    $\begingroup$ And just as in that case, there is an isomorphism $\mathcal{O}^{\times}_2 \simeq \mu_{K_2} \oplus (\mathbf{Z}_2)^2$, where $\mu_{K_2}$ denotes the roots of unity in $K_2$, which (under your hypothesis) is either cyclic of order $2$ or of order $4$. Once again, the answer is best understood by learning about local fields. $\endgroup$
    – Furlo Roth
    Apr 10, 2019 at 0:32
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    $\begingroup$ Have you tried looking at a specific ring, like, say, $\mathbb Z[\sqrt{14}]$, and seeing if it gives you any general insights? $\endgroup$
    – Robert Soupe
    Apr 10, 2019 at 4:07
  • $\begingroup$ See math.stackexchange.com/a/3186089/300700 $\endgroup$
    – nguyen quang do
    Apr 13, 2019 at 12:31

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