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When the catenary degree meets the tame degree in embedding dimension three numerical semigroups
Caterina Viola Cortona - September 2014
SLIDE 2 based on P . A. Garc´ ıa-S´ anchez, C. Viola, When the catenary degree meets the tame degree in embedding dimension three numerical semigroups, to appear in Involve.
. A. Garc´ ıa-S´ anchez, Z. Tripp, C. Viola,
ω-primality in embedding dimension three numerical
semigroups, preprint.
. A. Garc´ ıa-S´ anchez, J. J. Morais, GAP pakage numericalsgps
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Setup
Let S = n1, . . . , np be a p-generated numerical semigroup.
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Setup
Let S = n1, . . . , np be a p-generated numerical semigroup. A factorization of s ∈ S is an element x = (x1, . . . , xp) ∈ Np such that x1n1 + · · · + xpnp = s.
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Setup
Let S = n1, . . . , np be a p-generated numerical semigroup. A factorization of s ∈ S is an element x = (x1, . . . , xp) ∈ Np such that x1n1 + · · · + xpnp = s. The length of x is |x| = x1 + · · · + xp.
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Setup
Let S = n1, . . . , np be a p-generated numerical semigroup. A factorization of s ∈ S is an element x = (x1, . . . , xp) ∈ Np such that x1n1 + · · · + xpnp = s. The length of x is |x| = x1 + · · · + xp. Given another factorization y = (y1, . . . , yp), the distance between x and y is
d(x, y) = max{|x − gcd(x, y)|, |y − gcd(x, y)|},
where gcd(x, y) = (min{x1, y1}, . . . , min{xp, yp}).
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66 ∈ S = 6, 9, 11, c(S) = 4
The factorizations of 66 ∈ 6, 9, 11 are F(66) = {(0, 0, 6), (1, 3, 3), (2, 6, 0), (4, 1, 3), (5, 4, 0), (8, 2, 0), (11, 0, 0)} The distance between (11, 0, 0) and (0, 0, 6) is 11.
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66 ∈ S = 6, 9, 11, c(S) = 4
The factorizations of 66 ∈ 6, 9, 11 are F(66) = {(0, 0, 6), (1, 3, 3), (2, 6, 0), (4, 1, 3), (5, 4, 0), (8, 2, 0), (11, 0, 0)} The distance between (11, 0, 0) and (0, 0, 6) is 11. (11, 0, 0) (11, 0, 0) (0, 0, 6) (0, 0, 6) 11
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66 ∈ S = 6, 9, 11, c(S) = 4
The factorizations of 66 ∈ 6, 9, 11 are F(66) = {(0, 0, 6), (1, 3, 3), (2, 6, 0), (4, 1, 3), (5, 4, 0), (8, 2, 0), (11, 0, 0)} The distance between (11, 0, 0) and (0, 0, 6) is 11. (3, 0, 0) (11, 0, 0) (8, 2, 0) (0, 2, 0)|(8, 2, 0) (0, 0, 6) (0, 0, 6) 3 10
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66 ∈ S = 6, 9, 11, c(S) = 4
The factorizations of 66 ∈ 6, 9, 11 are F(66) = {(0, 0, 6), (1, 3, 3), (2, 6, 0), (4, 1, 3), (5, 4, 0), (8, 2, 0), (11, 0, 0)} The distance between (11, 0, 0) and (0, 0, 6) is 11. (3, 0, 0) (11, 0, 0) (8, 2, 0) (0, 2, 0)|(3, 0, 0) (5, 4, 0) (0, 2, 0)|(5, 4, 0) (0, 0, 6) (0, 0, 6) 3 3 9
SLIDE 11 66 ∈ S = 6, 9, 11, c(S) = 4
The factorizations of 66 ∈ 6, 9, 11 are F(66) = {(0, 0, 6), (1, 3, 3), (2, 6, 0), (4, 1, 3), (5, 4, 0), (8, 2, 0), (11, 0, 0)} The distance between (11, 0, 0) and (0, 0, 6) is 11.
(3, 0, 0) (11, 0, 0) (8, 2, 0) (0, 2, 0)|(3, 0, 0) (5, 4, 0) (0, 2, 0)|(3, 0, 0) (2, 6, 0) (0, 2, 0)|(2, 6, 0) (0, 0, 6) (0, 0, 6) 3 3 3 8
SLIDE 12 66 ∈ S = 6, 9, 11, c(S) = 4
The factorizations of 66 ∈ 6, 9, 11 are F(66) = {(0, 0, 6), (1, 3, 3), (2, 6, 0), (4, 1, 3), (5, 4, 0), (8, 2, 0), (11, 0, 0)} The distance between (11, 0, 0) and (0, 0, 6) is 11.
(3, 0, 0) (11, 0, 0) (8, 2, 0) (0, 2, 0)|(3, 0, 0) (5, 4, 0) (0, 2, 0)|(3, 0, 0) (2, 6, 0) (0, 2, 0)|(1, 3, 0) (1, 3, 3) (0, 0, 3)|(1, 3, 0) (0, 0, 3) (0, 0, 6) 3 3 3 4 4
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The catenary degree
The catenary degree of s ∈ S, c(s), is the minimum nonnegative integer N such that for any two factorizations x and y of s, there exists a sequence of factorizations x1, . . . , xt of s such that x1 = x, xt = y, for all i ∈ {1, . . . , t − 1}, d(xi, xi+1) ≤ N. The catenary degree of S, c(S), is the supremum (maximum) of the catenary degrees of the elements of S.
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The catenary degree of 77 ∈ 10, 11, 23, 35
(0, 7, 0, 0) (1, 4, 1, 0) (2, 1, 2, 0) (2, 2, 0, 1)
3 6 2 3 5 2
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The catenary degree
(0, 7, 0, 0) (1, 4, 1, 0) (2, 1, 2, 0) (2, 2, 0, 1)
3 2 3 5 2
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The catenary degree
(0, 7, 0, 0) (1, 4, 1, 0) (2, 1, 2, 0) (2, 2, 0, 1)
3 2 3 2
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The catenary degree
(0, 7, 0, 0) (1, 4, 1, 0) (2, 1, 2, 0) (2, 2, 0, 1)
2 3 2
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66 ∈ S = 6, 9, 11, t(S) = 7
The factorizations of 66 ∈ 6, 9, 11 are F(66) = {(0, 0, 6), (1, 3, 3), (2, 6, 0), (4, 1, 3), (5, 4, 0), (8, 2, 0), (11, 0, 0)} Besides, 9 divides 66 (11, 0, 0)
SLIDE 19 66 ∈ S = 6, 9, 11, t(S) = 7
The factorizations of 66 ∈ 6, 9, 11 are F(66) = {(0, 0, 6), (1, 3, 3), (2, 6, 0), (4, 1, 3), (5, 4, 0), (8, 2, 0), (11, 0, 0)} and 11 also divides 66 (8, 2, 0) (11, 0, 0)
3
SLIDE 20 66 ∈ S = 6, 9, 11, t(S) = 7
The factorizations of 66 ∈ 6, 9, 11 are F(66) = {(0, 0, 6), (1, 3, 3), (2, 6, 0), (4, 1, 3), (5, 4, 0), (8, 2, 0), (11, 0, 0)} (8, 2, 0) (11, 0, 0)
3
(4, 1, 3)
7
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The tame degree
The tame degree of S, t(S), is defined as the minimum N such that for any s ∈ S and any factorization x of s, if s − ni ∈ S for some i ∈ {1, . . . , p}, then there exists another factorization y of s such that d(x, y) ≤ N and the ith coordinate of y is nonzero (ni “occurs” in this factorization).
SLIDE 22 The catenary degree of S is less than or equal to the tame degree
c(S) ≤ t(S)
SLIDE 23 The catenary degree of S is less than or equal to the tame degree
c(S) ≤ t(S) It is known that in some cases both coincide (for instance for monoids with a generic presentation).
SLIDE 24 The catenary degree of S is less than or equal to the tame degree
c(S) ≤ t(S) It is known that in some cases both coincide (for instance for monoids with a generic presentation). We want to characterize when the equality holds if the embedding dimension of S is three.
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Embedding dimension three numerical semigroups
Let S = n1 < n2 < n3 be a numerical semigroup with embedding dimension 3. Define ci = min{k ∈ N \ {0} | kni ∈ nj, nk, {i, j, k} = {1, 2, 3}}. Then, for all {i, j, k} = {1, 2, 3}, there exist some rij, rik ∈ N such that cini = rijnj + riknk.
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Embedding dimension three numerical semigroups
We know that Betti(S) = {c1n1, c2n2, c3n3}. Hence 1 ≤ # Betti(S) ≤ 3.
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Embedding dimension three numerical semigroups
We know that Betti(S) = {c1n1, c2n2, c3n3}. Hence 1 ≤ # Betti(S) ≤ 3. Herzog proved that S is symmetric if and only if rij = 0 for some i, j ∈ {1, 2, 3}, or equivalently, # Betti(S) ∈ {1, 2}.
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Embedding dimension three numerical semigroups
We know that Betti(S) = {c1n1, c2n2, c3n3}. Hence 1 ≤ # Betti(S) ≤ 3. Herzog proved that S is symmetric if and only if rij = 0 for some i, j ∈ {1, 2, 3}, or equivalently, # Betti(S) ∈ {1, 2}. Therefore, S is nonsymmetric if and only if # Betti(S) = 3.
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The nonsymmetric case
Let S be a numerical semigroup minimally generated by {n1, n2, n3} with n1 < n2 < n3.
SLIDE 30 The nonsymmetric case
Let S be a numerical semigroup minimally generated by {n1, n2, n3} with n1 < n2 < n3.
. A. Garc´ ıa-S´ anchez, A. Geroldinger proved that c(S) = t(S) for S a nonsymmetric embedding dimension three numerical semigroup.
SLIDE 31 The nonsymmetric case
Let S be a numerical semigroup minimally generated by {n1, n2, n3} with n1 < n2 < n3.
. A. Garc´ ıa-S´ anchez, A. Geroldinger proved that c(S) = t(S) for S a nonsymmetric embedding dimension three numerical semigroup. For this reason we focus henceforth in the case S is symmetric, and thus # Betti(S) ∈ {1, 2}.
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When S has two Betti elements
When S has two Betti elements, we distinguish the three subcases: c1n1 = c2n2 c3n3; c1n1 = c3n3 c2n2; c1n1 c2n2 = c3n3;
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The case c1n1 = c2n2 c3n3
Proposition
Let S = n1, n2, n3 with n1 < n2 < n3 and c1n1 = c2n2 c3n3. Then c(S) < t(S).
Example
S = 4, 6, 7 c(S) = 3 < t(S) = 5
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The case c1n1 = c3n3 c2n2
Proposition
Let S = n1, n2, n3 with n1 < n2 < n3 and c1n1 = c3n3 c2n2. Then c(S) < t(S).
Example
S = 4, 5, 6 c(S) = 3 < t(S) = 4
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The case c1n1 c2n2 = c3n3
Proposition
Let S = n1, n2, n3 with n1 < n2 < n3 and c1n1 c2n2 = c3n3. If c2n2 ∤ c1n1, then c(S) < t(S).
Example
S = 5, 8, 12 c(S) = 4 < t(S) = 6
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The case c1n1 c2n2 = c3n3
Proposition
Let S = n1, n2, n3 with n1 < n2 < n3 and c1n1 c2n2 = c3n3. If c2n2 ∤ c1n1, then c(S) < t(S).
Example
S = 5, 8, 12 c(S) = 4 < t(S) = 6
Proposition
Let S = n1, n2, n3 with n1 < n2 < n3 and c1n1 c2n2 = c3n3. If c2n2 | c1n1, then c(S) = t(S).
Example
S = 12, 14, 21 c(S) = t(S) = 7
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When S has a single Betti element
Proposition
Let S = n1, n2, n3 with n1 < n2 < n3 and c1n1 = c2n2 = c3n3. Then c(S) = t(S).
Example
S = 6, 10, 15 c(S) = t(S) = 5
SLIDE 38 Main result
Theorem
Let S be an embedding dimension three numerical semigroup minimally generated by {n1, n2, n3}. For every {i, j, k} = {1, 2, 3}, define ci = min{k ∈ N \ {0} | kni ∈ nj, nk}. Then c(S) = t(S) if and only if either # Betti(S) 2,
- r c1n1 c2n2 = c3n3 and c2n2 divides c1n1.
SLIDE 39 ω-primality, the definition
The ω-primality function assigns to each element n ∈ S the value
ω(n) = m if m is the smallest positive integer with the property that
whenever p
i=1 aini − n ∈ S for |a| > m, there exists
b = (b1, . . . , bp) ∈ Np with b ≤ a (with the usual partial ordering on
Np) such that p
i=1 bini − n ∈ S and |b| ≤ m.
SLIDE 40 ω-primality, the definition
The ω-primality function assigns to each element n ∈ S the value
ω(n) = m if m is the smallest positive integer with the property that
whenever p
i=1 aini − n ∈ S for |a| > m, there exists
b = (b1, . . . , bp) ∈ Np with b ≤ a (with the usual partial ordering on
Np) such that p
i=1 bini − n ∈ S and |b| ≤ m.
We set
ω(S) = sup{ω(S, ni) | i ∈ {1, . . . , p}}.
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ω-primality
By definition, an element b ∈ S is a prime element if and only if
ω(S, b) = 1, and S is factorial if and only if ω(S) = 1.
Numerical semigroups other than N have no prime elements.
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ω-primality
By definition, an element b ∈ S is a prime element if and only if
ω(S, b) = 1, and S is factorial if and only if ω(S) = 1.
Numerical semigroups other than N have no prime elements. c(S) ≤ ω(S) ≤ t(S) We are interested in comparing the ω-primality with the catenary degree in embedding dimension three numerical semigroups.
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Comparing the ω-primality with the catenary degree
Theorem
Let S = n1, n2, n3 with n1 < n2 < n3 be a numerical semigroup with embedding dimension three. (a) If # Betti(S) = 3, then c(S) = ω(S) = t(S). (b) If c1n1 = c2n2 c3n3, then c(S) < ω(S). (c) If c1n1 = c3n3 c2n2, then c(S) < ω(S). (d) If c1n1 c2n2 = c3n3 and c2n2 | c1n1, then c(S) = ω(S) = t(S). (e) If c1n1 = c2n2 = c3n3, then c(S) = ω(S) = t(S).
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Comparing ω-primality with the catenary degree
In the case c1n1 c2n2 = c3n3 and c2n2 ∤ c1n1, with some examples we show that we cannot say more than we state in the theorem above.
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Comparing ω-primality with the catenary degree
In the case c1n1 c2n2 = c3n3 and c2n2 ∤ c1n1, with some examples we show that we cannot say more than we state in the theorem above.
Example
S = 19, 350, 490 in which c1n1 c2n2 = c3n3 and c2n2 ∤ c1n1. We have c(S) = ω(S) = 70.
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Comparing ω-primality with the catenary degree
In the case c1n1 c2n2 = c3n3 and c2n2 ∤ c1n1, with some examples we show that we cannot say more than we state in the theorem above.
Example
S = 19, 350, 490 in which c1n1 c2n2 = c3n3 and c2n2 ∤ c1n1. We have c(S) = ω(S) = 70. S = 62, 63, 147 in which c1n1 c2n2 = c3n3 and c2n2 ∤ c1n1. We have c(S) = 21 < ω(S) = 23.
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Questions
When ω(S) = t(S) in embedding dimension three numerical semigroup with two Betti elements?
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Questions
When ω(S) = t(S) in embedding dimension three numerical semigroup with two Betti elements? When c(S) = t(S) for embedding dimension four numerical semigroup?
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Thank you
