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Let $\alpha \in \mathbb{C}$ be an algebraic integer of degree $n$, not a unit, and let $R = \mathbb{Z}[\alpha]$. Then every element $\beta \in R$ can be written uniquely in the form

$$c_0+ c_1 \alpha+\ldots +c_{n-1} \alpha^{n-1},$$ with $c_i=c_i(\beta) \in R$. Let $S \subset R$ denote the set of $\beta \in R$ such that $c_0(\beta) \equiv 1 \mod N(\alpha)$. Equivalently, $S$ is the set of $\beta \in R$ such that $\beta \equiv 1 \mod (\alpha)$.

Let $F(S) \subset \mathrm{Frac}(R)$ denote the set of ratios of elements of $S$, i.e. the (multiplicative) abelian group generated by $S$. Here we have used the fact that $\alpha$ is not an algebraic unit to make sure that $0 \not\equiv 1 \mod (\alpha)$, so $0 \notin S$ and this construction makes sense.

Claim: As an abelian group, $F(S)$ has infinite rank.

It seems that the logical way to do this would be to show that $S$ itself contains infinitely many irreducible elements (considered irreducible as elements in $R$), but I am not sure how to proceed from here. One avenue for exploration is based on the fact that irreducible elements in $\mathcal{O}_k$ (for $k=\mathbb{Q}(\alpha)$) that happen to lie in $S$ remain irreducible over our smaller ring. Thus, it would suffice to find infinitely many irreducible elements in $\mathcal{O}_k \cap S$, which may be a better understood problem. (Perhaps with the Chebotarev density theorem?)

A second method that I have considered attempts to find infinitely many irreducible elements of a particular form in $S$, e.g. irreducibles of the form $1+m\alpha$, with $m \in \mathbb{Z}$. If the norm form polynomial $$f(x):=N(1+x\alpha) = \prod_{i=1}^n \left(1-x \alpha^{(i)}\right) \in \mathbb{Z}[t]$$ assumes infinitely many prime values, then the sequence $\{1+m\alpha\}$ contains infinitely many irreducibles (as a factorization of $1+m\alpha$ would yield a non-trivial factorization of the norm). Here, $f(x)$ satisfies the hypotheses of Bunyakovsky's conjecture, so this claim is likely true (even if this avenue is not the best for proving it). Perhaps a variant of this method could pan out, if we ask for $f(x)$ to admit infinitely many distinct prime divisors (as $x$ varies), instead of infinitely many prime values?

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up vote 1 down vote accepted

Let $p\ge 2$ be an integer dividing $\alpha$ in $R$. It exists because $\alpha$ is not a unit.

Then $F(1+p\mathbb Z)=(1+p\mathbb Z)(1+p\mathbb Z)^{-1}$ is a subgroup of $F(S)$, and it is enough to show that it is not finitely generated as group. If it was, then only finitely many prime numbers appear in the factorization of elements of $F(1+p\mathbb Z)$. But by Dirichlet there are infinitely many prime numbers in $1+p\mathbb Z$. Contradiction.

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