Factorization in complement-finite ideals of free monoids Nicholas - - PowerPoint PPT Presentation

Factorization in complement-finite ideals of free monoids

Nicholas R. Baeth (Joint work with Matthew Enlow) March 23, 2019

Franklin & Marshall College

Multiplication in Numerical Semigroups

Let S = n1, . . . , nt = {a1n1 + · · · + atnt : ai ∈ 0} be a numerical semigroup, a complement-finite additive subsemigroup of (0, +).

• S\{0} is a cancellative multiplicative submonoid of .
• \S < ∞
• For all s ∈ S and all n ∈ , ns ∈ S.

S\{0} is a complement-finite ideal of the free multiplicative submonoid .

1

Seemingly nice subsemigroups

Let S be a complement-finite ideal of F; that is, a (multiplicative) submonoid of a free (reduced) monoid F such that:

• F\S < ∞
• fs ∈ S s ∈ S\{1} and f ∈ F

Examples

• 1. S\{0} ⊆ where S is a numerical semigroup
• 2. S = \{p, p2, . . . , pk} ⊆ with p prime and k ≥ 1.
• 3. S = \{paqb : p, q ∈ , (a, b) ∈ A ⊆ 0 × 0}.
• 4. Generalizations of 3.

2

Examples

S = 3, 4, 5 = {3, 4, 5, . . .}

3

Examples

S = 3, 4, 5 = {3, 4, 5, . . .}

• A(S) = {p ∈ ≥3} ∪ {2p: p ∈ ≥3} ∪ {4, 8}

3

Examples

S = 3, 4, 5 = {3, 4, 5, . . .}

• A(S) = {p ∈ ≥3} ∪ {2p: p ∈ ≥3} ∪ {4, 8}
• α = 4a8b =⇒

L(α) =

• ⌈v2(α)

3 ⌉, ⌊v2(α) 2 ⌋

• .

3

Examples

S = 3, 4, 5 = {3, 4, 5, . . .}

• A(S) = {p ∈ ≥3} ∪ {2p: p ∈ ≥3} ∪ {4, 8}
• α = 4a8b =⇒

L(α) =

• ⌈v2(α)

3 ⌉, ⌊v2(α) 2 ⌋

• .
• α = p1 · · · ps(2q1) · · · (2qt)4a8b =⇒

L(α) =

• (s + t), (s + t) + ⌊v2(α)

2 ⌋

• if s >> 0.

3

Examples

S = 3, 4, 5 = {3, 4, 5, . . .}

• A(S) = {p ∈ ≥3} ∪ {2p: p ∈ ≥3} ∪ {4, 8}
• α = 4a8b =⇒

L(α) =

• ⌈v2(α)

3 ⌉, ⌊v2(α) 2 ⌋

• .
• α = p1 · · · ps(2q1) · · · (2qt)4a8b =⇒

L(α) =

• (s + t), (s + t) + ⌊v2(α)

2 ⌋

• if s >> 0.
• ρ(S) = 3/2.

3

Examples

S = \{p, p2, . . . , pk−1} ⊆ where p is prime in and k ≥ 2.

4

Examples

S = \{p, p2, . . . , pk−1} ⊆ where p is prime in and k ≥ 2.

• A(S) = {piq: q ∈ \{p}, 0 ≤ i ≤ k} ∪ {pk, . . . , p2k−1}

4

Examples

S = \{p, p2, . . . , pk−1} ⊆ where p is prime in and k ≥ 2.

• A(S) = {piq: q ∈ \{p}, 0 ≤ i ≤ k} ∪ {pk, . . . , p2k−1}
• α = pa =⇒ L(α) =

a 2k−1

, a

k

• .

4

Examples

S = \{p, p2, . . . , pk−1} ⊆ where p is prime in and k ≥ 2.

• A(S) = {piq: q ∈ \{p}, 0 ≤ i ≤ k} ∪ {pk, . . . , p2k−1}
• α = pa =⇒ L(α) =

a 2k−1

, a

k

• .
• α = q1 · · · qspa =⇒ L(α) =

s, s + a

k

• if s >> 0.

4

Examples

S = \{p, p2, . . . , pk−1} ⊆ where p is prime in and k ≥ 2.

• A(S) = {piq: q ∈ \{p}, 0 ≤ i ≤ k} ∪ {pk, . . . , p2k−1}
• α = pa =⇒ L(α) =

a 2k−1

, a

k

• .
• α = q1 · · · qspa =⇒ L(α) =

s, s + a

k

• if s >> 0.
• ρ(S) = 2k−1

k .

4

Examples

S = 6, 2k, 3k

5

Examples

S = 6, 2k, 3k

• Irreducibles that divide 62k−1 are: 6, 2k, . . . , 22k−1,

3k, . . . , 32k−1, 6 · 2i, 6 · 3i with 1 ≤ i ≤ k − 1

5

Examples

S = 6, 2k, 3k

• Irreducibles that divide 62k−1 are: 6, 2k, . . . , 22k−1,

3k, . . . , 32k−1, 6 · 2i, 6 · 3i with 1 ≤ i ≤ k − 1

• 62k−1 = (22k−1)(32k−1) and so ρ(S) ≥ ρ(62k−1) = k − 1

2.

5

Examples

S = 6, 2k, 3k

• Irreducibles that divide 62k−1 are: 6, 2k, . . . , 22k−1,

3k, . . . , 32k−1, 6 · 2i, 6 · 3i with 1 ≤ i ≤ k − 1

• 62k−1 = (22k−1)(32k−1) and so ρ(S) ≥ ρ(62k−1) = k − 1

2.

• L(62k−1) = [2, 2k − 1]

5

Irreducible Elements

Let S be a complement-finite ideal of a free (reduced) monoid F = F (P). The irreducible elements of S are those with the following forms:

• 1. p ∈ P ∩ S
• 2. px with p ∈ P ∩ S and x ∈ F\S
• 3. qr1

1 · · · qrt t with q1, . . . qt ∈ P\S and (r1, . . . , rt) almost

minimal Moreover, no irreducible element is prime in S. For each s ∈ S, the combined number of irreducibles of types (1) and (2) is independent of the factorization.

6

Factorizations

S a complement-finite ideal of F = F (P) P\S = {p1, . . . , pt} with ki = min{k: pk

i ∈ S}

• n ≥ ki =⇒ LS(pn

i ) = LT(n) where T = ki, . . . , 2ki − 1.

• ni ≥ ki i =⇒ L(pn1

1 · · · pnt t ) ⊇ t i=1 LS(pni i )

• ρ(S) ≤ M

m where

M = max{n1 + · · · + nt : pn1

1 · · · pnt t ∈ A(S)} and

m = min{n1 + · · · + nt : pn1

1 · · · pnt t ∈ A(S)}

• N ∈ t

i=1[ki, 2ki − 1] and α = pa1 1 · · · pat t ∈ A(S)

=⇒ αN = (pN

1 )a1 · · · (pN t )at and ρ(α) ≥ N a1+···+an.

7

C-monoid structure

S a complement-finite ideal of a (reduced) free monoid F.

• 1. Then S is not a Krull monoid.

[It’s not completely integrally closed.]

• 2. S is a C-monoid. Moreover, the class semigroup C∗(S, F)

has exactly two idempotent elements: {1} and S. Recall that C∗(S, F) = {[x]: x ∈ F} with [x] = [y] whenever xa ∈ S ⇔ ya ∈ S, and S is a C-monoid when C∗(S, F) < ∞.

8

A Transfer Homomorphism

Let S be an complement-finite ideal of a free monoid F and let C = C∗(S, F) = {e, c1, . . . , cn, h} denote its class semigroup, where e = {1} and h = S\{1} are the two idempotent elements. Let

B∗(S) = { cv1

1 · · · cvn n

• formal product

:

• cvi

i

• actual product

= h} ⊆ F ({c1, . . . , cn}).

The natural projection from S to B∗(S) is a transfer homomorphism.

9