# Arithmetic progressions of units in a domain

Let $R$ be a domain with unity, and suppose that $R^\times$ has finite rank as an abelian group.

1. Can $R^\times$ contain infinitely long arithmetic progressions?
2. 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 $c(R)$?

This answer consists of two parts:

I: If $R$ has positive characteristic $n$, then any arithmetic progression has maximal length $n$, for in

$$a,a+b,a+2b,\ldots,a+nb$$

we must have $(a+nb)-a=nb=0$. This answers questions (1) and (2) in the negative, for $\mathrm{char}(R)>0$.

II: If $\mathrm{char}(R)=0$, suppose that $R^\times$ is finitely generated (as in the original problem statement). We claim that the Jacobson radical $J_R$ of $R$ vanishes, proven in a series of Lemmas below:

Lemma 1: $R^\times + J_R = R^\times$.

Proof: Let $u \in R^\times$ and $j \in J_R$. If $u+j \notin R^\times$, then $u+j \in \mathfrak{m}$ for some maximal ideal, so that $u \in \mathfrak{m}$ because $j \in J_R \subset \mathfrak{m}$. This contradicts that $u \in R^\times$, so that $u+j \in R^\times$. //

Lemma 2: If $R^\times$ is finitely generated, then $J_R =(0)$.

Proof: Throughout, we identify $\mathbb{Z}$ with the characteristic subring of $R$. This proof will proceed in two cases, dependent on whether or not $J_R$ contains non-zero transcendental (over $\mathbb{Q} = \mathrm{Frac}(R)$) elements. In our first case, In our first case, take $j \in J_R$ transcendental over $\mathbb{Q}$; then Lemma 1 implies that $1+j\mathbb{Z}[j] \subset R^\times$. Now, suppose that $\{\pi_i(j)\}_{i=1}^m \subset 1 + j \mathbb{Z}[j]$ is a set of irreducible polynomials. If $\{\pi_i(j)\}_{i=1}^m$ is not independent (as a multiplicative set), then there exist integers $a_i,b_i \geq 0$ such that

$$\prod \pi_i^{a_i} = \prod \pi_i^{b_i},$$

at which point the fact that $\mathbb{Z}[j]$ is a UFD implies that $a_i=b_i$ for all $i$. Thus $\{\pi_i(j)\}_{i=1}^m$ is an independent set, whence $R^\times$ has rank at least $m$. It thus suffices (for the contradiction in our first case) to evince infinitely many irreducible polynomials in $1+j\mathbb{Z}[j]$, and such an example is provided by the cyclotomic polynomials $\Phi_n(j)$ with $n>1$ (or the linear polynomials).

Our second case is just as involved. Take $j \in J_R$ algebraic; we may assume without loss of generality that $j \in J_R$ is an algebraic integer. The (multiplicative) abelian group $S \subset R^\times$ generated by $\{1+kj\}_{k \in \mathbb{Z}}$ has finite rank, by assumption on $R^\times$. The field norm $N: \mathbb{Q}(j) \to \mathbb{Q}$ restricts to a homomorphism $N:S \to \mathbb{Q}^\times$, and the image of this map has finite rank. Let

$$f(x):= N(1+xj) \in \mathbb{Z}[x].$$

Finiteness of rank implies that the set of prime divisors of $\{f(k) : k \in \mathbb{Z}\}$ is finite. Yet if $P:= \prod p_i$ represents the product of these primes, it follows that $f(1+kP) \equiv 1 \mod P$, so our hypothesis forces $f(1+kP) = \pm 1$ for all integers $k$. (This is an extension of Euclid's proof of the infinitude of the primes to a statement about the prime divisors of a polynomial.) This gives a contradiction, so that no such $j \in J_R$ exists.

Therefore $J_R=(0)$ in either case. //

Now we continue in earnest:

Suppose that $R^\times$ contains the infinite arithmetic progression

$$S:=\{a,a+b,a+2b,\ldots,a+nb,\ldots\}$$

We claim that $b$ fails to generate any given residue domain $R/\mathfrak{p}$. If not, fix $\mathfrak{p} \in \mathrm{Spec}(R)$ such that $b$ generates $R/\mathfrak{p}$. Consider the map $\varphi: S \to R/\mathfrak{p}$ given by $a+nb \mapsto (a+nb) \mod \mathfrak{p}$. This map surjects, hence there exists $a+nb \in S$ such that $a+nb \equiv 0 \mod \mathfrak{p}$. But $a+nb$ is a unit, which is a contradiction.

When $\mathfrak{p}$ is maximal, $R/\mathfrak{p} \cong \mathbb{Z}/p\mathbb{Z}$ for some rational prime $p$. Since all non-zero elements generate $\mathbb{Z}/p\mathbb{Z}$, this forces $b \in \mathfrak{p}$, so that $b \in J_R$, the Jacobson radical of $R$. But $J_R=(0)$ by Lemma 2, so that $b = 0$, a contradiction. For the case $\mathrm{char}(R)=0$, it follows that the statement of question (1) is universally false.

At this point, I do not know whether or not question (2) can be answered positively under any hypothesis. As mentioned in my original post, the answer to (2) is "no" when $R$ is a number field of degree $n$, with an explicit upper bound on length given by $\max(n,4)$. Perhaps it would be possible to look at algebraic subfields of $R$ to reach a similar (if not slightly weaker) conclusion in this more general setting.

Disclaimer: Some of the material above (Lemma 2, in particular) is sourced from my blog, Integer Miscellany. In that article I consider an ideal that is not quite the Jacobson radical (it behaves the same for these purposes, however), so I've kept this proof self-contained to prevent confusion.

• An interesting answer. But why does $\frak p$ maximal imply that $R/{\frak p}\cong\mathbb{Z}/p\mathbb{Z}$. How did you rule out the possibility of larger finite residue field? Commented May 20, 2013 at 6:56