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 = � n 1 , . . . , n t � = { a 1 n 1 + · · · + a t n t : a i ∈ � 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 , p 2 , . . . , p k } ⊆ � with p prime and k ≥ 1 . 3. S = � \ { p a q b : 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 } ∪ { 2 p : p ∈ � ≥ 3 } ∪ { 4 , 8 } 3

Examples S = � 3 , 4 , 5 � = { 3 , 4 , 5 , . . . } ◦ A ( S ) = { p ∈ � ≥ 3 } ∪ { 2 p : p ∈ � ≥ 3 } ∪ { 4 , 8 } ◦ α = 4 a 8 b = ⇒ � ⌈ v 2 ( α ) 3 ⌉ , ⌊ v 2 ( α ) � L ( α ) = . 2 ⌋ 3

Examples S = � 3 , 4 , 5 � = { 3 , 4 , 5 , . . . } ◦ A ( S ) = { p ∈ � ≥ 3 } ∪ { 2 p : p ∈ � ≥ 3 } ∪ { 4 , 8 } ◦ α = 4 a 8 b = ⇒ � ⌈ v 2 ( α ) 3 ⌉ , ⌊ v 2 ( α ) � L ( α ) = . 2 ⌋ ◦ α = p 1 · · · p s ( 2 q 1 ) · · · ( 2 q t ) 4 a 8 b = ⇒ � ( s + t ) , ( s + t ) + ⌊ v 2 ( α ) � L ( α ) = if s >> 0. 2 ⌋ 3

Examples S = � 3 , 4 , 5 � = { 3 , 4 , 5 , . . . } ◦ A ( S ) = { p ∈ � ≥ 3 } ∪ { 2 p : p ∈ � ≥ 3 } ∪ { 4 , 8 } ◦ α = 4 a 8 b = ⇒ � ⌈ v 2 ( α ) 3 ⌉ , ⌊ v 2 ( α ) � L ( α ) = . 2 ⌋ ◦ α = p 1 · · · p s ( 2 q 1 ) · · · ( 2 q t ) 4 a 8 b = ⇒ � ( s + t ) , ( s + t ) + ⌊ v 2 ( α ) � L ( α ) = if s >> 0. 2 ⌋ ◦ ρ ( S ) = 3 / 2. 3

Examples S = � \ { p , p 2 , . . . , p k − 1 } ⊆ � where p is prime in � and k ≥ 2. 4

Examples S = � \ { p , p 2 , . . . , p k − 1 } ⊆ � where p is prime in � and k ≥ 2. ◦ A ( S ) = { p i q : q ∈ � \ { p } , 0 ≤ i ≤ k } ∪ { p k , . . . , p 2 k − 1 } 4

Examples S = � \ { p , p 2 , . . . , p k − 1 } ⊆ � where p is prime in � and k ≥ 2. ◦ A ( S ) = { p i q : q ∈ � \ { p } , 0 ≤ i ≤ k } ∪ { p k , . . . , p 2 k − 1 } ◦ α = p a = ⇒ L ( α ) = �� a � , � a . �� 2 k − 1 k 4

Examples S = � \ { p , p 2 , . . . , p k − 1 } ⊆ � where p is prime in � and k ≥ 2. ◦ A ( S ) = { p i q : q ∈ � \ { p } , 0 ≤ i ≤ k } ∪ { p k , . . . , p 2 k − 1 } ◦ α = p a = ⇒ L ( α ) = �� a � , � a . �� 2 k − 1 k s , s + � a ◦ α = q 1 · · · q s p a = ⇒ L ( α ) = � if s >> 0. �� k 4

Examples S = � \ { p , p 2 , . . . , p k − 1 } ⊆ � where p is prime in � and k ≥ 2. ◦ A ( S ) = { p i q : q ∈ � \ { p } , 0 ≤ i ≤ k } ∪ { p k , . . . , p 2 k − 1 } ◦ α = p a = ⇒ L ( α ) = �� a � , � a . �� 2 k − 1 k s , s + � a ◦ α = q 1 · · · q s p a = ⇒ L ( α ) = � if s >> 0. �� k ◦ ρ ( S ) = 2 k − 1 k . 4

Examples S = � 6 , 2 k , 3 k � 5

Examples S = � 6 , 2 k , 3 k � ◦ Irreducibles that divide 6 2 k − 1 are: 6, 2 k , . . . , 2 2 k − 1 , 3 k , . . . , 3 2 k − 1 , 6 · 2 i , 6 · 3 i with 1 ≤ i ≤ k − 1 5

Examples S = � 6 , 2 k , 3 k � ◦ Irreducibles that divide 6 2 k − 1 are: 6, 2 k , . . . , 2 2 k − 1 , 3 k , . . . , 3 2 k − 1 , 6 · 2 i , 6 · 3 i with 1 ≤ i ≤ k − 1 ◦ 6 2 k − 1 = ( 2 2 k − 1 )( 3 2 k − 1 ) and so ρ ( S ) ≥ ρ ( 6 2 k − 1 ) = k − 1 2 . 5

Examples S = � 6 , 2 k , 3 k � ◦ Irreducibles that divide 6 2 k − 1 are: 6, 2 k , . . . , 2 2 k − 1 , 3 k , . . . , 3 2 k − 1 , 6 · 2 i , 6 · 3 i with 1 ≤ i ≤ k − 1 ◦ 6 2 k − 1 = ( 2 2 k − 1 )( 3 2 k − 1 ) and so ρ ( S ) ≥ ρ ( 6 2 k − 1 ) = k − 1 2 . ◦ L ( 6 2 k − 1 ) = [ 2 , 2 k − 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. q r 1 1 · · · q r t t with q 1 , . . . q t ∈ P \ S and ( r 1 , . . . , r t ) 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 = { p 1 , . . . , p t } with k i = min { k : p k i ∈ S } ◦ n ≥ k i = ⇒ L S ( p n i ) = L T ( n ) where T = � k i , . . . , 2 k i − 1 � . ◦ n i ≥ k i � i = ⇒ L ( p n 1 1 · · · p n t i = 1 L S ( p n i t ) ⊇ � t i ) ◦ ρ ( S ) ≤ M m where M = max { n 1 + · · · + n t : p n 1 1 · · · p n t t ∈ A ( S ) } and 1 · · · p n t m = min { n 1 + · · · + n t : p n 1 t ∈ A ( S ) } i = 1 [ k i , 2 k i − 1 ] and α = p a 1 1 · · · p a t ◦ N ∈ � t t ∈ A ( S ) t ) a t and ρ ( α ) ≥ = ⇒ α N = ( p N 1 ) a 1 · · · ( p N N a 1 + ··· + a n . 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 , c 1 , . . . , c n , h } denote its class semigroup, where e = { 1 } and h = S \ { 1 } are the two idempotent elements. Let � c v i B ∗ ( S ) = { c v 1 1 · · · c v n = h } ⊆ F ( { c 1 , . . . , c n } ) . : n i � ����� �� ����� � ���� formal product actual product The natural projection from S to B ∗ ( S ) is a transfer homomorphism. 9