# Multiplicative set of positive algebraic integers

Let $$S$$ be a set of algebraic integers such that:

• $$\mathbb{N}_{\ge 1} \subseteq S \subset \mathbb{R}_{\ge 1}$$,
• $$\alpha, \beta \in S \Rightarrow \alpha \beta \in S$$,
• $$\alpha, \beta \in S \Rightarrow \alpha, \beta \in \mathcal{O}_{\mathbb{Q}(\alpha \beta)}$$ the ring of integers of $$\mathbb{Q}(\alpha \beta)$$.

Observations:

• Obviously, the set $$S=\mathbb{N}_{\ge 1}$$ works,
• Consider $$\alpha = 2(3-\sqrt{5})$$ and $$\beta = \frac{3+\sqrt{5}}{2}$$, both are algebraic integers greater than $$1$$. Note that $$\alpha \beta = 4$$. By the third assumption above, if $$\alpha, \beta \in S$$ then $$\alpha, \beta \in \mathcal{O}_{\mathbb{Q}(\alpha \beta)} = \mathbb{Z}$$, contradiction. So $$\alpha$$ or $$\beta \not \in S$$,
• Let $$p,q,n \in \mathbb{N}$$ such that $$n^{p/q} \not \in \mathbb{N}$$. Consider $$\alpha = n^{p/q}$$ and $$\beta = n^{p(q-1)/q}$$. We can deduce as above that $$\alpha$$ or $$\beta \not \in S$$, and by induction, we deduce in fact that $$n^{p/q} \not \in S$$. In particular, $$S=\mathbb{Z}[n^{p/q}]_{\ge 1}$$ does not work.

Question 1: What more can we say about such a set? Can we determine/classify all the algebraic integers ($$\ge 1$$) not in $$S$$? Are there some references where such sets are studied?

The motivation comes from subfactor theory (see What are all the possible indices for the finite depth subfactors?), where we can add the following assumptions:

• $$S$$ is included in the set of cyclotomic integers,
• $$4 \cos(\pi/n)^2 = (\zeta_n + \overline{\zeta_n})^2 \in S$$ for all $$n\ge 3$$.

Note that $$4 \cos(\pi/5)^2 = \frac{3+\sqrt{5}}{2}$$, so by the second additional assumption, the second observation above shows that $$2(3-\sqrt{5}) \not \in S$$, and more generally $$n(3-\sqrt{5}) \not \in S$$ for all $$n \in \mathbb{N}$$.

Question 2: What more can we say in this case? Can we determine/classify all the cyclotomic integers ($$\ge 1$$) not in $$S$$?

There is some structure, but I don't think it's reasonable to say there is a classification, without more information.

It's equivalent to ask for a set $$S'$$ of algebraic integers (each $$\geq 1$$) such that $$\alpha_1,\dots, \alpha_n \in S' \implies \alpha_1,\dots ,\alpha_n, \in \mathbb Q\left(\prod_{i=1}^n \alpha_i\right)$$ and then define $$S$$ to be the set generated under multiplication by $$S'$$ and $$\mathbb N^{ \geq 1}$$.

For this construction, removing any element of $$S'$$ gives another valid $$S'$$.

If $$S'$$ is finite, we can also add "generic" elements to $$S'$$ and preserve the $$S'$$ property. Given a finite $$S'$$, choose any number field $$K$$ (other than $$\mathbb Q$$), and choose a prime $$p$$ that splits completely in $$K$$ and for which all valuations lying over $$p$$ are trivial on all numbers in $$S'$$ (possible because there are infinitely many split primes, finitely many numbers in $$S'$$, and each has finitely many nontrivial valuations). Then take any algebraic integer $$\alpha \in \mathcal O_K$$ which has positive $$\mathfrak p$$-adic valuation for some prime $$\mathfrak p$$ lying over $$p$$ and zero $$\mathfrak p'$$-adic valuation for all other primes $$\mathfrak p'$$ lying over $$p$$.

If we multiply any positive power of $$\alpha$$ by any product $$\beta$$ of integers in $$S'$$, then $$\alpha^n \beta$$ still have positive the same positive $$\mathfrak p$$-adic valuation / zero $$\mathfrak p'$$-adic valuation property, hence every element of the Galois group that fixes $$\alpha^n \beta$$ fixes $$\mathfrak p$$, hence the field $$\alpha^n \beta$$ generates contains $$K$$. Because the field $$\alpha^n \beta$$ generates contains $$K$$, which contains $$\alpha$$, it also contains $$\alpha^ n \beta/\alpha^n=\beta$$, and hence (by the assumption that $$S'$$ is valid) contains all the elements of $$S'$$ that multiply to $$\beta$$.

If we add the condtion $$4 \cos (\pi/n)^2 = (\zeta_{2n} + \overline{\zeta}_{2n})^2 \in S'$$, this construction won't quite work. It's not too hard to see that the norm of this integer is $$p$$ if $$n$$ is $$2$$ times a power of a prime $$p$$ and $$1$$ otherwise, so for each prime $$p$$, there are some integers with a nontrivial $$\mathfrak p$$-adic valuation for some $$\mathfrak p$$ lying over $$p$$. However, when $$n = 2 p^r$$, the prime $$p$$ spits into only one prime in the field $$\mathbb Q( \zeta_{2n} + \overline{\zeta}_{2n})$$ (because the inertia group of $$p$$ inside the Galois group of $$\mathbb Q(\zeta_{2n})$$, which is $$(\mathbb Z/4)^\times \times (\mathbb Z/p^r)^\times$$, is $$(\mathbb Z/p^r)^\times$$, and complex conjugation is $$-1$$, and these generate the whole group). Thus, a slightly modified construction works where we instead choose a prime $$p$$ such that each element of $$S'$$ has the same valuation at all primes of the field it generates that lie above $$p$$.

In general, the key tool to understand such sets will be $$p$$-adic valuations. Every finitely generated subgroup of a number field is (modulo some roots of unity that can be ignored here) a lattice $$\mathbb Z^r$$, and subfields define sublattices. So you're looking for a subset of a lattice such that any nonnegative integer combination doesn't intersect a particular sublattice, except if the elements you're taking nonnegative combinations of are already in the sublattice. The obstruction to this happening will be a linear form, vanishing on the sublattice, that takes positive values on all elements lying in that sublattice. Such a form can be written as a linear combination of $$p$$-adic valuations and logs of absolute values at $$\infty$$ (corresponding to all the real embeddings of the field, not just one). Using this picture, we can define many such sets $$S$$ by choosing appropriate linear combinations.

To narrow down what $$S$$ is, I suspect you'll want to get more arithmetic information, specifically, on the valuations of its elements.