Let $R$ be a domain with unity, and suppose that $R^\times$ has finite rank as an abelian group.
- Can $R^\times$ contain infinitely long arithmetic progressions?
- Can $R^\times$ contain arithmetic progressions of arbitrarily long (unbounded) length?
I suspect that the answer to each of these questions is no, but I have been unable to prove either.
A brief note by Morris Newman (PDF) discusses this question for $R$ a number field (for which our question applies by the Dirichlet unit theorem). For $R$ a number field with degree $n \geq 4$ the lengths of such progressions are bounded by $n$ (and this bound is sharp). For $n=2,3$ our progressions are instead bounded of length $4$. Thus each of our numbered claims hold for number fields.
Edit: This edit reflects some recent thoughts I've had on this problem. Suppose that $R^\times$ is free on $\alpha_1,\ldots, \alpha_m$ (so we've simplified by ignoring torsion elements). Then an arithmetic progression $$ \beta_1:=\prod^m \alpha_i^{k_i(1)},\beta_2:=\prod^m \alpha_i^{k_i(2)},\beta_3:=\ldots$$ encodes a number of polynomial relations in the $\alpha_i$, given by $$p_i(\alpha_1,\ldots, \alpha_m):=\beta_{i+1}-2\beta_i+\beta_{i-1}=0.$$ If the zero sets of these polynomials have no common component, then the fact that $(\alpha_1,\ldots,\alpha_m)$ lies in the intersection of the zero sets should imply a bound on the number of relations that may hold (perhaps $m+1$ such, e.g.). Because of this tentative connection, I have added the tag "algebraic geometry" to this question.
Edit 2: I've posted a solution below which addresses all aspects of my original questions, save the following (which remains open):
Question: Let $R$ be a domain with characteristic $0$. If $R^\times$ has finite rank, does there exist a constant $c(R)$ such that any arithmetic progression in $R^\times$ has length at most $c(R)$?