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It is well-known that the group of units in the $2$-adic integers $\mathbb{Z}_2$ is isomorphic to the product of a finite abelian group with $\mathbb{Z}_2$. This fact generalizes to the ring of integers $\mathcal{O}$ of a finite separable extension $K$ of $\mathbb{Q}_2$: the group of units in $\mathcal{O}$ is isomorphic to the product of a finite abelian group with $\mathbb{Z}_2^{[K : \mathbb{Q}_2]}$.

Question: What is the structure of the units in the ring $R = \mathbb{Z}_2[x]/(x^2 - 4)$? Notice that this ring is not the ring of integers in its algebra of fractions: the fraction algebra of $R$ is $\mathbb{Q}_2 \times \mathbb{Q}_2$, so the ring of integers is $\mathbb{Z}_2 \times \mathbb{Z}_2$, but $R$ is evidently not equal to $\mathbb{Z}_2 \times \mathbb{Z}_2$.

What I Know: Every unit $u \in R^\times$ can be expressed as $u = r \cdot (1 + d\cdot x)$, where $r$ is a unit in $\mathbb{Z}_2$ and $d \in \mathbb{Z}_2$. The identification $R^\times \to \mathbb{Z}_2^\times \times \mathbb{Z}_2$ defined by $u \mapsto (r, d)$ is a bijection of sets. But this identification is not a group homomorphism, and I'm not sure how to determine the structure of $R^\times$ as a group.

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    If $R \rightarrow S$ is an injective map of rings with finite index, then $R^{\times}$ has finite index in $S^{\times}$ (excersise). Now apply that with your $R$ and $S = \mathbf{Z}_2 \oplus \mathbf{Z}_2$ under the obvious map, i.e. $P(x) \mapsto (P(2),P(-2))$. Hence $R^{\times}$ is a subgroup of $\mathbf{Z}^{\times}_2 \times \mathbf{Z}^{\times}_2 = \mathbf{Z}^2_2 \oplus {\pm 1} \oplus {\pm 1}$. The image of $-1$ is $(-1,-1)$ but $(1,-1)$ is not in the image because $P(2)-P(-2) \equiv 0 \mod 4$. – user760870 Apr 05 '20 at 17:11
  • @user760870: Why not turn that into an answer? – Torsten Schoeneberg Apr 06 '20 at 15:50
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    Here's a nice solution to the exercise about finite index of unit subgroup: https://math.stackexchange.com/questions/91405/subrings-of-finite-index-and-units – David Lampert Apr 08 '20 at 15:23

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