Proportionally modular Diophantine inequalities J. C. Rosales Porto - - PowerPoint PPT Presentation

I P B´   E P C E   A

Proportionally modular Diophantine inequalities

• J. C. Rosales

Porto 2008

I P B´   E P C E   A

• J. C. Rosales, P

. A. Garc´ ıa-S´ anchez and J. M. Urbano-Blanco, The set of solutions of a proportionally modular diophantine inequality, J. Number Theory, 128 (2008), 453-467

• M. Bullejos, J. C. Rosales, Proportionally modular diophantine

inequalities and Stern-Brocot tree, preprint

• J. C. Rosales, P

. A. Garc´ ıa-S´ anchez and J. M. Urbano-Blanco, Modular diophantine inequalities and numerical semigroups, Pacific J. Math. 218 (2005), 379-398

• J. C. Rosales, P

. A. Garc´ ıa-S´ anchez, J. I. Garc´ ıa-Garc´ ıa and

• J. M. Urbano-Blanco, Proportionally modular diophantine

inequalities, J. Number Theory 103(2003), 281-294

• J. M. Urbano-Blanco and P

. Vasco PhD theses

I P B´   E P C E   A

What we intend to study

• A proportionally modular Diophantine inequality is an

expression of the form ax mod b ≤ cx, with a, b, and c positive integers

• The set S = S(a,b,c) of solutions to such inequality is a

numerical semigroup

• A numerical semigroup is proportionally modular if it is of this

form

I P B´   E P C E   A

Some basic definitions

• If A ⊂ N, we denote by A the submonoid of (N,+) generated

by A,

A = {λ1a1 +···+λnan | n ∈ N,λ1,...,λn ∈ N,a1,...,an ∈ A}

• A is a numerical semigroup if and if gcd(A) = 1
• Every numerical semigroup S is finitely generated, and thus

there exist positive integers n1,...,np ∈ S such that S = n1,...,np. If no proper subset of {n1,...,np} generates S, we say that this set is a minimal generating set of S

• Minimal generating sets are unique and its elements are

minimal generator of the semigroup

I P B´   E P C E   A

Example S(12,32,3) = {x ∈ N | 12x mod 32 ≤ 3x} = {0,3,6,7,8,9,10,→} = 3,7,8

12×1 mod 32 = 12 > 3×1 12×7 mod 32 = 20 ≤ 3×7 12×2 mod 32 = 24 > 3×2 12×8 mod 32 = 0 ≤ 3×8 12×3 mod 32 = 4 ≤ 3×3 12×9 mod 32 = 12 ≤ 3×9 12×4 mod 32 = 16 > 3×4 12×10 mod 32 = 24 ≤ 3×10 12×5 mod 32 = 28 > 3×5

. . .

12×6 mod 32 = 8 ≤ 3×6

I P B´   E P C E   A

Some simplifications

• The inequality ax mod b ≤ cx has the same solutions as

(a mod b)x mod b ≤ cx

• If c ≥ a, then S(a,b,c) = N
• Thus we can assume that c < a < b

I P B´   E P C E   A

Lemma

If c < a < b are positive integers, S(a,b,c) = T ∩N, where T is the submonoid of (Q,+) generated by [ b

a , b a−c ]. Conversely, if

a1,a2,b1,b2 are positive integers with a1

b1 < a2 b2 , and T is the

submonoid of (Q,+) generated by [ a1

b1 , a2 b2 ], then

T ∩N = S(a2b1,a1a2,a2b1 −a1b2)

• T ∩N is the proportionally modular numerical semigroup

associated to the interval [ a1

b1 , a2 b2 ], and we denote it by

S([ a1

b1 , a2 b2 ])

I P B´   E P C E   A

Lemma

If c < a < b are positive integers, S(a,b,c) = T ∩N, where T is the submonoid of (Q,+) generated by [ b

a , b a−c ]. Conversely, if

a1,a2,b1,b2 are positive integers with a1

b1 < a2 b2 , and T is the

submonoid of (Q,+) generated by [ a1

b1 , a2 b2 ], then

T ∩N = S(a2b1,a1a2,a2b1 −a1b2)

Example

• S(12,32,3) = S([ 32

12, 32 9 ]) = S([ 8 3, 32 9 ])

• T =

k∈N[k 8 3,k 32 9 ] = {0}∪[ 8 3, 32 9 ]∪[ 16 3 , 64 9 ]∪[8, 32 3 ]∪···

• S(12,32,3) = T ∩N = {0,3,6,7,8,9,10,...}

I P B´   E P C E   A

First results

Lemma

A positive integer x belongs to S([ a1

b1 , a2 b2 ]) if and only if there exists

a positive integer y such that a1

b1 ≤ x y ≤ a2 b2

Lemma

If a2b1 −a1b2 = 1, then S([ a1

b1 , a2 b2 ]) = a1,a2

I P B´   E P C E   A

B´ ezout sequence

A sequence a1

b1 < a2 b2 < ··· < ap bp of rational numbers is a B´

ezout sequence if a1,...,ap,b1,...,bp are positive integers and ai+1bi −aibi+1 = 1 for all i ∈ {1,...,p −1} Its length is p, and a1

b1 and ap bp are its ends

Proposition

If a1

b1 < a2 b2 < ··· < ap bp is a B´

ezout sequence, S([ a1

b1 , ap bp ]) = a1,...,ap

I P B´   E P C E   A

Example

Let us solve the Diophantine inequality 50x mod 131 ≤ 3x We know that the set of solutions is S([ 131

50 , 131 47 ])

As 131 50 < 76 29 < 21 8 < 8 3 < 11 4 < 25 9 < 39 14 < 131 47 is a B´ ezout sequence,

S

• [131

50 , 131 47 ]

• = 131,76,21,8,11,25,39 = 8,11,21,25,39

I P B´   E P C E   A

Problems:

1) If a

b < c d , is there a B´

ezout sequence with ends a

b and c d ?

2) If so, find a procedure to compute such a sequence 3) Characterize proportionally modular numerical semigroups in

terms of their minimal generators

I P B´   E P C E   A

Proposition

Let a1, a2, b1 and b2 be positive integers with a1

b1 < a2 b2 and

mcd{a1,b1} = 1 = mcd{a2,b2}. Then there exists a B´

ezout sequence of length less than or equal to a2b1 −a1b2 +1 and with ends a1

b1 and a2 b2

Corollary S([ a1

b1 , a2 b2 ]) has embedding dimension less than or equal to

a2b1 −a1b2 +1

I P B´   E P C E   A

Refinement of sequences: looking for minimal generators

Proper sequences

A B´ ezout sequence a1

b1 < a2 b2 < ··· < ap bp is proper if ai+hbi −aibi+h ≥ 2

for all h ≥ 2 such that i,i +h ∈ {1,...,p}

Remark

Every B´ ezout sequence can be refined to a proper B´ ezout sequence with the same ends

Examples

• 5

3 < 12 7 < 7 4 < 9 5 is a non proper B´

ezout sequence

• 5

3 < 7 4 < 9 5 is a proper B´

ezout sequence

I P B´   E P C E   A

Proper B´ ezout sequences are not the final solution

Proposition

If a1

b1 < a2 b2 < ··· < ap bp is a proper B´

ezout sequence, then max{a1,...,ap} = max{a1,ap}

Corollary

If a1

b1 < a2 b2 < ··· < ap bp is a proper B´

ezout sequence, then there exists h ∈ {1,...,p} such that a1 ≥ a2 ≥ ··· ≥ ah ≤ ah+1 ≤ ··· ≤ ap

Example

2 1 < 3 1 < 4 1 is a proper B´

ezout sequence. However {2,3,4} is not a minimal generating system for 2,3,4

I P B´   E P C E   A

Adjacent ends and minimal generators

Adjacent fractions

Two fractions a1

b1 < a2 b2 are adjacent if a2 b2+1 < a1 b1 and b1 = 1, or a2 b2 < a1 b1−1

Proposition

If a1

b1 < a2 b2 < ··· < ap bp is a proper B´

ezout sequence with adjacent ends, then {a1,...,ap} is the minimal generating system of

S([ a1

b1 , ap bp ])

I P B´   E P C E   A

Theorem

Let S be a proportionally modular numerical semigroup. Then there exists an arrangement n1,...,np of its minimal generators and positive integers b1,...,bp such that n1

b1 < n2 b2 < ··· < np bp is a

proper B´ ezout sequence with adjacent ends

Corollary

Let S be a proportionally modular numerical semigroup with minimal generating set n1 < n2 < ··· < np con p ≥ 3. Then

n1,...,np−1 is also proportionally modular

I P B´   E P C E   A

This yields a characterization of proportionally modular numerical semigroups

Characterization

A numerical semigroup S is proportionally modular if and only if there exists a (convex) arrangement n1,...,np of its minimal generators such that

• 1. mcd{ni,ni+1} = 1 for all i ∈ {1,...,p −1}
• 2. ni−1 +ni+1 ≡ 0 mod ni for all i ∈ {2,...,p −1}

Examples

• S = 5,7,11 is not proportionally modular, since

5+11 0 mod 7 and 7+11 0 mod 5

• S = 6,8,11,13 is not proportionally modular, mcd{6,8} 1

I P B´   E P C E   A

Example

S = 8,11,21,25,39 is proportionally modular 8 11 21 8 11 21 8 11 25 21 8 11 25 39

I P B´   E P C E   A

A reformulation of the above characterization

Corollary

Let S be a proportionally modular numerical semigroup. Then there exists an arrangement n1,...,np of its minimal generators so that S = n1,n2∪n2,n3∪···∪np−1,np

• S = 7,8,9,10,12 is not proportionally modular

        

7 8 7 8 9 7 8 9 10

• S = 12,7∪7,8∪8,9∪9,10

I P B´   E P C E   A

Corollary

A numerical semigroup S is proportionally modular if and only if there exists a (convex) arrangement n1,...,np of its minimal generators fulfilling that

• 1. ni,ni+1 is a numerical semigroup for all i ∈ {1,...,p −1}
• 2. ni−1,ni,ni+1 = ni−1,ni∪ni,ni+1 for all i ∈ {2,...,p −1}

I P B´   E P C E   A

Computing some invariants for the embedding dimension three case

• Given a numerical semigroup S and n ∈ S \{0}, the Ap´

ery set

• f n in S is

Ap(S,n) = {s ∈ S | s −n S}

• Ap(S,n) = {0,w(1),...,w(n −1)} where w(i) is the least

element in S congruent with i modulo n

• #Ap(S,n) = n
• The Frobenius number of S is the largest integer not in S

F(S) = max(Ap(S,n))−n

• The set of gaps of S is H(S) = N\S

#H(S) = 1

n(w(1)+···+w(n −1))− n−1 2

I P B´   E P C E   A

If S is a proportionally modular numerical semigroup with minimal generating set {n1,n2,n3}, then we can assume that

mcd{n1,n2} = mcd{n2,n3} = 1 and that dn2 = n1 +n3 for some

d ∈ N\{0,1}

Proposition Ap(S,n2) = {0,n1,...,⌊ n3

d ⌋n1,n3,...,(n2 −⌊ n3 d ⌋−1)n3}

F(S) = max{⌊ n3

d ⌋n1 −n2,⌊ n1 d ⌋n3 −n2}

#H(S) =

n1(1+⌊

n3 d ⌋)⌊ n3 d ⌋+n3(n2−⌊ n3 d ⌋)(n2−⌊ n3 d ⌋−1)−n2(n2−1)

2n2

I P B´   E P C E   A

How to compute a B´ ezout sequence with given ends Idea

• Every B´

ezout sequence can be obtained by gluing two B´ ezout sequences, the first with decreasing numerators, and the second with increasing numerators 131 50 < 76 29 < 21 8 < 8 3 < 11 4 < 25 9 < 39 14 < 131 47

131 50 < 76 29 < 21 8 < 8 3 sequence with decreasing numerators 8 3 < 11 4 < 25 9 < 39 14 < 131 47 sequence with increasing numerators

I P B´   E P C E   A

Decreasing numerators

Theorem

Let a1 and b1 be two positive integers with a1 ≥ b1 ≥ 1 and

mcd{a1,b1} = 1

As long as bi 1, set ai+1 = b−1

i

mod ai y bi+1 = (−ai)−1 mod bi Then there exists a positive integer p such that ap = ⌈ a1

b1 ⌉, bp = 1

and a1

b1 < a2 b2 < ··· < ap bp is a B´

ezout sequence Moreover, if a1

b1 < a′

2

b′

2 < ··· <

a′

q

b′

q a B´

ezout sequence such that a1 ≥ a′

2 ≥ ··· ≥ a′ q, then q ≤ p, a′ i = ai and b′ i = bi for all i ∈ {2,...,q}

Proposition

If i ∈ {2,...,p −1}, then ai+1 = (−ai−1) mod ai and bi+1 = (−bi−1) mod bi

I P B´   E P C E   A

Example

Let us construct the maximal B´ ezout sequence with decreasing numerators and left end 131

50

a1 = 131 and b1 = 50 Then a2 = 50−1 mod 131 = 76 and b2 = (−131)−1 mod 50 = 29 By using that ai+1 = (−ai−1) mod ai and bi+1 = (−bi−1) mod bi, we get 131 50 < 76 29 < 21 8 < 8 3 < 3 1

I P B´   E P C E   A

Increasing numerators

Theorem

Let a1 and b1 be two positive integers with a1 ≥ b1 ≥ 1 and

mcd{a1,b1} = 1

As long as ai 1, set ai+1 and bi+1 as follows

1) If bi 1, ai+1 = (−bi)−1 mod ai and bi+1 = a−1

i

mod bi

2) If bi = 1, then ai+1 = ai −1 and bi+1 = 1

Then there exists a positive integer p such that ap = 1, bp = 1 and

ap bp < ··· < a2 b2 < a1 b1 is a B´

ezout sequence Moreover, if a′

s

b′

s < ··· <

a′

2

b′

2 < a1

b1 is a B´

ezout sequence such that

a′

s

b′

s ≥ 1 and a′

s ≤ ··· ≤ a′ 2 ≤ a1, then s ≤ p, a′ i = ai and b′ i = bi for all

i ∈ {2,...,s}

Proposition

If i ≥ 2 and bi 1, then ai+1 = (−ai−1) mod ai and bi+1 = (−bi−1) mod bi

I P B´   E P C E   A

Example

Let us construct the maximal B´ ezout sequence with fractions greater than or equal to one, with increasing numerators and with right end 131

47

a1 = 131 and b1 = 47 Then a2 = (−47)−1 mod 131 = 39 and b2 = 131−1 mod 47 = 14 By using that if bi 1, then ai+1 = (−ai−1) mod ai and bi+1 = (−bi−1) mod bi, we obtain 131 47 > 39 14 > 25 9 > 11 4 > 8 3 > 5 2 > 2 1 > 1 1

I P B´   E P C E   A

Example

Let us construct the B´ ezout sequence with ends 131

50 and 131 47

1) We compute the maximal B´

ezout sequence with decreasing numerators and left end 131

50

131 50 < 76 29 < 21 8 < 8 3 < 3 1

2) Then we keep constructing the maximal B´

ezout sequence with increasing denominators and right end 131

47 until we find an

element in the above sequence 131 47 > 39 14 > 25 9 > 11 4 > 8 3

3) The desired sequence is

131 50 < 76 29 < 21 8 < 8 3 < 11 4 < 25 9 < 39 14 < 131 47

I P B´   E P C E   A

Theorem

Let a, b, c and d be positive integers such that

mcd{a,b} = mcd{c,d} = 1 and a

b < c d

Then there exists a unique proper B´ ezout sequence with ends a

b

and c

d

Lemma

Let a1, b1, a2 and b2 be positive integers Then a1

b1 < a2 b2 is a B´

ezout sequence if and only if a1+b1

b1

< a2+b2

b2

is a B´ ezout sequence

I P B´   E P C E   A

Example

Let us construct the unique proper B´ ezout sequence with ends

8 25

and 20

7

1) We construct the proper B´

ezout sequence with ends 8+25

25 = 33 25

and 20+7

7

= 27

7

33 25 < 4 3 < 3 2 < 2 1 < 3 1 < 7 2 < 11 3 < 15 4 < 19 5 < 23 6 < 27 7

2) Subtracting 1 to all the terms, we obtain the desired sequence

8 25 < 1 3 < 1 2 < 1 1 < 2 1 < 5 2 < 8 3 < 11 4 < 14 5 < 17 6 < 20 7