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In the thread below a reference to a paper (published in American Journal of Math) classifying finite rings with a cyclic group of units is given.

The finite ring $R_l:=\mathbb{Z}/2^l\mathbb{Z}$ is not on this list (for $l=3$ you may check that the group of units $(R_l)^*\cong \mathbb{Z}/(2)\times \mathbb{Z}/(2)$ is non-cyclic). Has the group of units $R_l^*$ of $R_l$ been classified?

More generally if $K$ is a number field and $I \subseteq \mathcal{O}_K$ is an ideal, there is a factorization $I=\mathfrak{m}_1^{l_1}\cdots \mathfrak{m}_d^{l_d}$ into a product of distinct maximal ideals. This leads to the following question: Is the multiplicative group of units $(\mathcal{O}_K/I)^* \cong \oplus_i (\mathcal{O}_K/\mathfrak{m}_i^{l_i})^*$ known for any such ideal $I$?

Is the group of units of a finite ring cyclic?

hm2020
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2 Answers2

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Yes, see https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n#Powers_of_2, the answer is that $$ (\mathbb{Z}/2^k\mathbb{Z})^\times \cong \mathrm{C}_2 \times \mathrm{C}_{2^{k-2}} $$

Alex J Best
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  • Thanks, I altered the question - if you have references to this new question, post a response. – hm2020 Sep 13 '21 at 11:08
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    @hm2020 I'd recommend posting your new question as a new question so it gets seen, I'm not sure I know a reference, but I imagine the answer is something similar that for each maximal ideal $\mathfrak m$ there is some $n_{\mathfrak m}$ (depending on the ramification of the prime?) and all $i>n_{\mathfrak m}$ we have $(\mathcal O/\mathfrak m^i)^\cong (\mathcal O/\mathfrak m^{n_\mathfrak m})^ \times Cyclic$ – Alex J Best Sep 13 '21 at 12:26
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We have the following result:

The group of units $(\mathbf Z/2^l\mathbf Z)^{\times}\:(l\ge 2)$ is generated by the internal direct product of the subgroup of order $2$ defined by the congruence class of $-1$ and the subgroup of order $2^{l-2}$ generated by the congruence class of $5$.

If you read french, you can find a proof on Wikipedia.

Bernard
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