The conductor of submonoids of ℕ^{n}.

Les Reid and Leslie G. Roberts

Let ℕ = {0, 1, 2, 3, … }. The integers that can be written as the sum of multiples of 3 and 5 are 3 5 6=3+3 8=3+5 9=3+3+3 10=5+5 and every larger integer (since we now have three consecutive integers). However 7 cannot be so expressed).

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More generally, given relatively prime a, b ∈ ℕ, then every integer greater than or equal to c := (a − 1)(b − 1) = ab − a − b + 1 can be expressed as a sum of multiples of a and b, but ab − a − b cannot. Let S be the set of all integers of the form Aa + Bb (A, B ∈ ℕ). Then C is closed under addition and contains 0, i.e. is a submonoid of ℕ. Let C = {x ∈ ℕ|x ≥ c}. Then C = {x ∈ ℕ|x + ℕ ⊆ S}, and C (or abusively c) is called the conductor of S in ℕ.

Problem: Generalize the above to submonoids of ℕ^{n}.

Definition: If S ⊆ \overline{S} are submonoids of ℕ^{n} then the conductor C of S in \overline{S} is defined by

C = \{x ∈ \overline{S}|x + \overline{S} ⊆ S\}.

Note that 0 ∈ \overline{S} so if x ∈ C then x + 0 ∈ S i.e. C ⊆ S.

Hypotheses: S is generated as a submonoid of ℕ^{n} by n + 1 elements {v_{1}, v_{2}, …, v_{n+1}}, any n of which are linearly independent. Furthermore we assume that {v_{1}, v_{2}, …, v_{n+1}} generates ℤ^{n} as an abelian group and that \overline{S} = ℤ^{n} ∩ σ where σ is the cone spanned by \{v_{1}, v_{2}, …, v_{n+1}\} (so that \overline{S} is the normalization of S in ℕ^{n}). Note that

\overline{S} = \{x ∈ ℕ^{n}|dx ∈ S for some d ∈ ℕ\0\}.

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Theorem: Under the above hypotheses the conductor C of S in \overline{S} is non-empty and

C = λ + (σ^{∘} ∩ \overline{S}).

where λ = (∑_{i=1}^{n+1} d_{i}v_{i})/2 − ∑_{i=1}^{n+1} v_{i}, σ^{∘} is the interior of σ and d_{i} is the order of the subgroup of ℤ^{n} generated by {v_{1}, …, v_{i−1}, v_{i+1}, …, v_{n}}.

Example: If n = 1 and S is the submonoid of ℕ generated by relatively prime integers v_{1} = a and v_{2} = b then d_{1} = b, d_{2} = a, λ = (ba + ba)/2 − (a + b) = ab − a − b, \overline{S} = ℕ and σ^{∘} ∩ ℕ = ℤ_{+}, yielding the same conductor as before.

Example: Let n = 2, v_{1} = [ \begin{matrix}2 \\ 1 \end{matrix} ], v_{2} = [ \begin{matrix}0 \\ 2 \end{matrix} ],

v_{3} = [ \begin{matrix}3 \\ 6 \end{matrix} ]. Then d_{1} = |det ( \begin{matrix}0 & 3 \\ 2 & 6 \end{matrix} )| = 6,

d_{2} = 9, d_{3} = 4. Thus

λ = (d_{1}v_{1} + d_{2}v_{2} + d_{3}v_{3})/2 − v_{1} − v_{2} − v_{3} = [ \begin{matrix}7 \\ 15 \end{matrix} ]

If v_{1} = [ \begin{matrix}3 \\ 1 \end{matrix} ], v_{2} = [ \begin{matrix}1 \\ 3 \end{matrix} ], v_{3} = [ \begin{matrix}1 \\ 2 \end{matrix} ] then

λ = [ \begin{matrix}3 \\ 10 \end{matrix} ].

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Interesting Observations:

1. The translation to higher dimensions is not what one first might think of. The closed interval x ≥ c becomes the interior of a cone. Thus the closed interval x ≥ c should thus be thought of as the interior of the interval x ≥ c − 1.

2. The conductor in dimension greater than one need not be principal (as a \overline{C} ideal).

3. λ is symmetric in the v_{i} even though the v_{i} play a different geometric role.

4. In dimensions greater than two the cone σ need not be “simplicial”.

Motivation: Subrings of ℤ[t_{1}, …, t_{n}] generated by “terms” x_{M}t^{M}, x_{M} ∈ ℤ, M = (m_{1}, …, m_{n}) (where t^{M} = t^{m_{1}}t^{m_{2}} ⋯ t^{m_{N}}) arose naturally in our study of weakly subintegral ring extensions. The set of all M such that x_{M} ≠ 0 is a submonoid of ℕ^{n} and an attempt to describe geometrically all submonoids T of ℕ^{n} with the property that if x ∈ ℕ^{n} and dn ∈ T for d >> 0 then n ∈ T led to a discussion of the conductor of monoids. Reference: Les Reid and L.Roberts, Monomial Subrings in Arbitrary Dimension, To appear in the Journal of Algebra. (downloadable from my web page)

Our Theorem on conductors and the first example for n = 2 are respectively Theorems 5.2 and Example 5.3 of this paper.