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For any discrete valuation ring $R$ (a local principal ideal domain other than a field), the group of units of $R$ is an infinite abelian group with at most one element of order 2 (which must be $-1$ if it exists, and it exists iff $R$ does not have characteristic 2, i.e. iff its characteristic is zero or an odd prime). In fact, there must be at most $n$ units of order dividing $n$ ($n$th roots of unity) for all positive integers $n$.

Why is the unit group of a DVR infinite? The answer is that if $p$ is a uniformizer in $R$, then for any positive integer $n$, $1+p^n \in U(R)$ (because it is clearly not divisible by $p$, so it must be a unit). This implies that $R$ has an infinite unit group.

The following two questions are about the converse:

  1. If $G$ is an infinite abelian group with no element of order 2 such that for any $n \ge 3$, there are at most $n$ elements of $G$ with order dividing $n$, must there be a discrete valuation ring of characteristic 2 whose unit group is isomorphic to $G$?
  2. If $G$ is an infinite abelian group with exactly one element of order 2 such that for any $n \ge 3$, there are at most $n$ elements of $G$ with order dividing $n$, must there be a discrete valuation ring of characteristic other than 2 whose unit group is isomorphic to $G$?

Question 1 is hard to answer in general even for the infinite cyclic group $C_\infty$ (which is isomorphic to the additive group $\mathbf{Z}$). In particular, the ring of Laurent polynomials over the field with 2 elements from this MathOverflow question is not a discrete valuation ring.

Why consider separate cases for DVRs of characteristic 2 and those of other characteristics? Because the proofs may depend on whether $G$ has an element of order 2 (so $-1 \neq 1$) or not (so $-1 = 1$).

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  • $\begingroup$ If you don't want just counter-examples, then I think it is too broad and asking for the kind of groups appearing in the DVR of subextensions of $\Bbb{Q}_p$ would be a good start. $\endgroup$ – reuns Aug 7 at 21:56
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An easy additional constraint is that if $G$ is the unit group of any DVR, then $G\otimes\mathbb{Q}$ must be infinite-dimensional. To prove this, note that if $R$ is a DVR with fraction field $K$, there is a short exact sequence $$0\to R^\times\to K^\times\to \mathbb{Z}\to 0$$ given by the valuation. If $K$ has characteristic $0$, this immediately implies $R^\times\otimes\mathbb{Q}$ must be infinite-dimensional, since $K^\times\otimes\mathbb{Q}\supseteq\mathbb{Q}^\times\otimes\mathbb{Q}$ is infinite-dimensional.

If $K$ has characteristic $p>0$, then since $K^\times$ contains an element of infinite order, $K$ cannot be algebraic over $\mathbb{F}_p$. Thus $K$ contains a copy of $\mathbb{F}_p(x)$, which again means $K^\times\otimes\mathbb{Q}$ is infinite-dimensional since there are infinitely many irreducible polynomials over $\mathbb{F}_p$.

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