# Abelian groups that arise as unit groups of discrete valuation rings

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 $$-1 = 1$$).

• 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. – reuns Aug 7 '19 at 21:56

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 $$\mathbb{F}_p$$.