Abstrct We give closed form for the ctenry degree of ny element in numericl monoid generted by generlized rithmetic sequence in embedding dimension three. While it is known in generl tht the lrgest nd smllest nonzero ctenry degrees re ttined t Betti elements, the current literture contins no informtion bout the other relizble ctenry degrees. By clssifying ech element in terms of its Betti element divisors, we identify ll the ctenry degrees chieved nd where they occur. In ddition, our reserch provides the dissonnce number nd the period vlue, even though previous works hve shown only tht the ctenry degree is periodic using non-existentil proof. Problem Let S be numericl semigroup generted by generlized rithmetic sequence of embedding dimension 3. Chrcterize the ctenry degree of ech elements in S. Motivtions & Definitions Let S be numericl semigroup generted by generlized rithmetic sequence, sy S =, h + d, h + d,..., h + wd, where, h, d, w N nd nd d re reltively prime (ssures the cofiniteness of S). For the set of genertors to be miniml, it sufficies to require tht 1 w 1. Since = forces S to hve embedding dimension, which hs been studied in detil, it will be ssumed tht 3. Let s S. Define the set of fctoriztions of s s { } w Z(s) := (α 0, α 1,..., α w ) s = α 0 + α i (h + id), α i N 0. We sy tht the length of z = (α 0, α 1,..., α w ) Z(s) is i=1 z := w α i. i=0 1
Let z = (α 0, α 1,..., α w ) nd z = (β 0, β 1,..., β w ) be fctoriztions of s. The gretest common divisor of z nd z is defined s gcd(z, z ) := (min{α 0, β 0 },..., min{α w, β w }) nd the distnce between z nd z s d(z, z ) := mx{ z gcd(z, z ), z gcd(z, z ) }. The distnce stisfies some desired properties. For instnce, if k S then the distnce is norm on Z(k) stisfying tht d(z + z 1, z + z ) = d(z 1, z ) for ll z 1, z Z(k) nd z fctoriztion of n rbitrry element of S. For ech nonzero s S, consider the grph s with vertex set Z(s) in which two vertices z, z Z(s) shre n edge if gcd(z, z ) 0. If s is not connected, then s is clled Betti element of S. We write Betti(S) = {s S : s is disconnected} for the set of Betti elements of S. A sequence of fctoriztions in Z(s) z = z 0, z 1,..., z n 1, z n = z is clled n N-chin if d(z i, z i+1 ) N for ech 1 i n 1. The previous definitions help us to define one of the key invrints studied in fctoriztion theory, the ctenry degree. The ctenry degree of n element s, denoted c(s), is the miniml N N 0 such tht for ny two fctoriztions there is n N-chin between them. The ctenry degree of the numericl semigroup S, denoted c(s), is defined s c(s) := sup s S {c(s)}. It is known tht c(s) = /x h + d from Theorem 3.8 [3]. When h = 1, we sy tht the genertors form n rithmetic sequence. For this cse, the ctenry degree of ech element hs been completely chrcterized in Theorem 3.1 from [5]. So it will be ssumed tht h. The gol of this reserch is to chrcterize the ctenry degree of ech element in numericl semigroup generted by generlized rithmetic sequence of embedding dimension 3.
Closed Form For The Ctenry Degree Theorem 1. Let S =, h+d, h+d nd x S. Define y 1 = x b 1, y = x b, nd y 3 = x b 3 where b 1, b, nd b 3 re the Betti elements of S. If we cn write x = q + id with q, i N 0 nd i { 0,..., 1 }, then 0 if y j / S, j c(x) = h + 1 if only y 1 S + d + i (h ) + (mod) i if y S or y 3 S Proof. From Theorem 13, we know tht if y j S, j, the element hs ctenry degree zero. From Theorem 41, we know tht if only y 1 S, then the element hs ctenry degree h + 1. These proofs stisfy the first two cses of our closed form. We need to chrcterize the remining ctenry degrees, which occur when y S or y 3 S. Recll, from Theorem CITATION OF PERIOD- ICITY, tht the element (c(s) + x, 0, 0) hs the sme ctenry degree s (c(s), 0, 0), specificlly the ctenry degree equls c(s). In Theorem 19, we know tht multiples of the first genertor fter β 3 hve ctenry degree c(s) nd since they re multiples of, re equivlent to zero modulo. From Proposition 40, we know tht if x is divisible by β or β 3 then the ctenry degree of x is either +d+k(h ) for some k {1,..., } or c(s). From Theorem 19 nd Lemm 0, we know tht if x hs ctenry degree c(s), x is either multiple of or hs form (m, 1, ) with even. Thus we wnt our third cse to output ctenry degree c(s) only when these situtions occur. Now, to clssify the remining elements nd ctenry degrees, note tht x cn lwys be written s x = q + id with 0 i 1 becuse x = α 0 () + α 1 (h + d) + α (h + d). Thus we cn lwys group x in terms of multiples of nd d. We wnt to write the ctenry degree of x in terms of this i vlue, outputting c(s) if i = 0 nd + d + k(h ) for i 0. This still covers our extr even c(s) cses by Lemm 0 nd Remrk 7. 3
Now if x is divisible by β or β 3 : write x = q + id where 0 i 1 to ensure i. Then we know tht since q 0 mod. x mod q + id mod id mod Since we wnt to be ble to find the + d + k(h ) ctenry degrees, for i 0 this implies x = (m, α 1, k) for k {1,..., }, α1 {0, 1}, m N. x = (m, α 1, k) = m + α 1 (h + d) + ( k)(h + d) = (m + α 1 h + h kh + d)() + (α 1 k)d. Since (, d) = 1, x (α 1 k)d mod. Therefore, we know x (α 1 k)d mod id mod. Since (, d) = 1, we cn remove the d term by dding the inverse to both sides of the modulr equivlence. Therefore we hve x i mod α 1 k mod. Since k {1,..., } nd α1 {0, 1}, 0 > α 1 k. We know 0 i < thus i = α 1 k in order for i α 1 k mod. i = α 1 k k = i + α 1 The vlue of k must lwys be n integer, nd specificlly the sme integer i + α1 for α 1 = 0 or α 1 = 1. Thus k = which elimintes the need for the α 1 term s the constnt is the sme regrdless. i Thus for x = q + id, c(x) = + d + (h ) when i 0. We cn combine the x 0 mod cse nd the x = q + id cses. We know tht when is even, if x 0, then x = (m, 0, ) nd (m + (c(s), 0, 0) re equl since ( )(h + d) = ( h + d).thus if we wnt to write ll ctenry degrees, including c(s) in terms of the + d + i (h ) ctenry eqution, this eqution must remin the sme for ll even vlues. 4
If is odd, the eqution must remin the sme for ll vlues of x not congruent to 0 mod. This mens tht when i 0, c(x) = + d + i (h ). However, when i = 0, we wnt to obtin ctenry degree c(s). Currently, in the eqution + d + i (h ), if i = 0, we get ctenry degree + d + (h ) = + d + h 1 = c(s) 1. We need to dd 1 to the eqution only when is odd nd i = 0. The following eqution stisfies these conditions: i i c(x) = + d + (h ) + (mod) When is even, mod = 0 so no dditionl term will be dded. Additionlly, ctenry degree c(s) = + d + (h ) which will be included in this cse for i = 0. When is odd, mod = 1. When i 0, i = 0 becuse i < since i 1. Thus no dditionl term will be dded when i 0. When is odd nd i = 0, i = = 1 nd dds the dditionl term needed to obtin ctenry degree c(s) for i = 0. Therefore the eqution c(x) = +d+ i (h )+( mod ) i gives ctenry degree c(s) nd + d + k(h ) t the correct times. Since we hve covered ech type of divisibility by Betti elements, nd thus found the ctenry degree for ny x S, we hve the closed form of ctenry degree for numericl semigroup generted by generlized rithmetic sequence. 5
Contents 1 Betti Set 6 Ctenry Degree 0 1 3 Ctenry Degree h + 1 5 4 Ctenry Degree c(s) 5 5 Intermedite Vlues For The Ctenry Degree Set 7 6 Periodicity & Dissonnce 4 7 Conclusion 54 1 Betti Set Min Result: Betti(S) = { (h + d), C(S), /(h + d) } Proposition. The element (h + d) hs only two fctoriztions: (0,, 0) nd (h, 0, 1). Also, the distnce between the two fctoriztion is h + 1. Proof. Let z 1 = (0,, 0) nd z = (h, 0, 1). It is esy to check tht both z 1 nd z re fctoriztions of (h + d). If z = (α 0, α 1, α ) Z((h + d)) with α i N 0, then (h + d) = α 0 + α 1 (h + d) + α (h + d). Hence, α 1. If α 1 = we hve tht nd thus, z = z 1. If α 1 = 1 then (h + d) = α 0 + (h + d) + α (h + d). (h + d) = α 0 + 1(h + d) + α (h + d) h + d = α 0 + α (h + d). 6
which is contrdiction since this will force α = 0 but does not divide h + d. Finlly, if α 1 = 0 then nd so α 1. If α = 1, then nd z = z. If α = 0 then (h + d) = α 0 + α (h + d). h = α 0 (h + d) = α 0 d = (α 0 h) which is contrdiction since gcd(, d) = 1 nd 3. Then, if z Z((h+ d)) then z = z 1 or z = z, nd it is esy to check tht z 1, z Z((h + d)). Also, notice tht gcd(z 1, z ) = (0, 0, 0) nd so d(z, z ) = h + 1. Theorem 3. The element (h + d) is Betti element with ctenry degree h + 1. Proof. This is n immedite result from Proposition. Proposition 4. If (α 0,..., α w ) Z(c(S)) nd α 0 0, then α 0 = c(s) nd α i = 0, for ll i {1,..., w}. Proof. It is esy to see tht (c(s), 0,..., 0) is fctoriztion of c(s). Let z = (α 0,..., α w ) Z(c(S)). Notice tht fter removing the first genertor, we obtin fctoriztion of n element in the rithmetic sequence generted by the genertors: h + d,..., h + wd. In the utiliztion of the Moving Boxes theorem on rithmetic sequences, notice z cn be trnsformed into ny of following fctoriztions: Cse 1. (α 0, 0,..., α k, 0,..., 0), α k 0, or Cse. (α 0, 0,..., α i, α i+1, 0,..., 0), α i, α i+1 0. We will show tht in either cse, we obtin contrdiciton. Cse 1. Notice tht ( ) c(s) = h + d w 7
= (α 0 ) + α k (h + kd) = α 0 + α k h + α k kd. Then, we hve tht α k k. Hence, there exists t N, t > 0, such tht α k k = t nd so α k = t/k. If α k > or k > 1, then d < α k kd. Also, w h α0 + α k h = th/k + α 0 since w /k. This is contrdiction since ( ) h + d α 0 + α k (h + kd). w Cse. We hve ( ) h + d w = α 0 + α i (h + id) + α i+1 (h + (i + 1)d) = α 0 + α i h + α i+1 h + (α i i + α i+1 (i + 1))d. Thus (α i i + α i+1 (i + 1)) nd there exists t N, t > 0, such tht t = α i i + α i+1 (i + 1). Since α i i + α i+1 (i + 1), d α i i + α i+1 (i + 1). Notice tht iα i + iα i+1 α i+1. Therefore, < iα i + iα i+1 or < α i+1 becuse (α i i + α i+1 (i + 1)). If < iα i + iα i+1, then w h i h < (α i+1 + α i )h but this is contrdiction. Now if < α i+1, then w h < αi+1 h which is contrdiction. Thus, α 0 > 0 implies tht α 0 = c(s). Notice tht the element c(s) hs t lest two fctoriztions (c(s), 0,..., 0) Z(c(S)) nd nother with first coordinte 0. The fct tht fctoriztion with first coordinte 0 exists comes from direct impliction of Omidli s pper [3]. Theorem 5. The element c(s) is Betti number with ctenry degree c(s). In prticulr, it is the first multiple of with nonzero ctenry degree. Proof. In Proposition 4 we sw tht c(s) hs just one fctoriztion with nonzero first coordinte. Consequently, such fctoriztion is n isolted point in c(s), mking the grph disconnected nd c(s) Betti element. By Omidli s pper [3], the first multiple of tht hs more thn one fctoriztion is exctly c(s). Let us show tht this element hs ctenry degree c(s). Tke the fctoriztions z = (c(s), 0,..., 0) nd z = (α 0, α 1,..., α w ) z. By Lemm 4, we hve tht α 0 = 0 nd by the Moving Boxes theorem 8
for rithmetic sequences CITATION CITATION CITATION, we hve tht z = (0, 0,..., α i, α i+1, 0,..., 0) where α i nd α i+1 re not both zero. Notice tht gcd(z, z ) = (0,..., 0). If α i + α i+1 > c(s), then when we compre fctoriztions z nd z we see tht: c(s) < (α i + α i+1 )(h + id) α i (h + id) + α i+1 (h + (i + 1)d) nd this is contrdiction becuse these two fctoriztions should be equl to the sme element. Thus, α i +α i+1 c(s) nd so d(z, z ) = mx{c(s), α i + α i+1 } = c(s). Theorem 6. The element ( )(h + d) is Betti element. Proof. We wnt to show tht is the smllest multiple of the third genertor tht hs fctoriztion bsed in the monoid M =, h + d. In order to do this, we will use the fct tht Γ symmetric iff s Γ iff F(Γ) s Γ. The Frobenius number of M =, h + d is (h + d) h d = h + d h d. We wnt to show tht ny integer less thn ( ) multiplied by (h + d) nd subtrcted from the Frobenius number is in the monoid generted by M. We will denote ll cses less thn ( ) by ( k) where k {1,..., } for odd nd k {1,..., 1}. ( ) k (h + d) = (h + d) (h + d) k(h + d) = h h (h) (d) kh d F(M ) ( ) k (h + d) = h + d h d (h h h + = d h d + d + kh + kd If is even: d h d + h + d + kh + kd. 9 (h) (d) kh d)
If is odd: d h d + ( 1 1 h) + d + kh + kd. We will first evlute the cse where is even: ( ) F(M ) k (h + d) = d h d + h + d + kh + kd ( ) h = () kh 1 + (h + d)(k 1) k cnnot exceed 1 or the originl eqution becomes negtive. When k = h, we hve ()( h 1) becomes negtive. Since we hve found fctoriztion for ll positive integer multiples of (h + d), ( ) must be the smllest multiple of the first two genertors. We will now evlute the cse where is odd: ( ) F(M ) k (h + d) ( ) 1 = d h d + h + 1 d + kh + kd (( ) ) 1 = () h kh 1 + h + (h + d)(k ) When k= 1, ( 1 1 1 )h kh 1 + h = ( )h h 1 + h = h 1. h > 1 nd thus h 1 is positive. If k ws greter, the initil ( k) term would be zero. Since we hve found fctoriztion for ll positive integer multiples of (h + d), ( ) must be the smllest multiple of the first two genertors. This is defining chrcteristic of Betti elements in embedding dimension 3. Therefore, ( )(h + d) is Betti element for embedding dimension 3. Theorem 7. The betti elements of M =, h + d, h + d re: 10
Tble 1: Betti Elements Element Ctenry Degree Fctoriztions β 1 = (h + d) h + 1 (0,, 0), (h, 0, 1) β = ( )(h + d) + d + (h ) (0, 0, ), (,, 0) β 3 = c(s) c(s) = h + d (c(s), 0, 0), (0,, ) Note tht if is n even number, the nd nd 3rd betti elements re equl: ( ) (c(s)) = h + d = h + d = (h) + d ( ) = = ( (h) + d ) (h + d). Also notice tht if is odd, then c(s) > + d + (h ): h > 1 h + d > h + d + 1 h + h + d > h h + d + 1 h + h + d > + d + h h + 1 h + h + d > + d + h h + ( ) ( ) + 1 1 h + d > + d + (h ) h + d > + d + (h ) c(s) > + d + (h ). 11
Ctenry Degree 0 The first ctenry degree tht we will chrcterize is ctenry degree zero. A zero ctenry degree is chieved when n element only hs one unique fctoriztion. To prove ll cses when n element hs ctenry degree zero, we will find it s unique fctoriztion nd then prove tht no other exists. To bound this region, we will show thn ny lrger element hs t lest two fctoriztions, nd thus positive ctenry degree. Proposition 8. The element m for m {1,..., c(s) 1} hs ctenry degree zero. Proof. This is direct consequence from Lemm 3.7 in [3]. Proposition 9. Consider the element n = h+d+m(h+d) S where m { 0,..., / }. Then, Z(n) = { (0, 1, m) }, i.e., n hs just one fctoriztion. Proof. Suppose (α 0, α 1, α ) Z(n). Note tht α 0 < m + 1 in ny fctoriztion nd α < m + 1 in ny fctoriztion. Cse 1: (α 0, 0, 0) α 0 () = h + d + mh + md ( = h + mh + d + md ) For this to be fctoriztion, d+md must be divisible by. Since cnnot divide d, must divide 1 + m. 1 + m 1 + ( 1) = 1 + = 1. So 1 + m never reches multiple of, nor is ever zero. Thus the fctoriztion cnnot exist. Cse : (0, α 1, 0) α 1 (h + d) = h + d + mh + md = h(1 + m) + (1 + m) This is only fctoriztion when m = 0. If m = 0, (0, 1, α ) = (0, 1, 0). There this is exctly the originl fctoriztion. 1
Cse 3: (0, 0, α ) α (h + d) = h + d + mh + md ( ) 1 = h(1 + m) + d + m 1 + m 1 + m for ny m. Hence, there re no fctoriztions of this form. Cse 4: (α 0, α 1, 0) α 0 () + α 1 (h + d) = h + d + mh + md α 0 () = h + d + mh + md α 1 (h) α 1 (d) ( = h + mh α 1 h + d + md α ) 1d In order for this fctoriztion to exist, must divide d + md α 1 d. Since cnnot divide d, must divide 1+m α 1. We hve shown tht 1+m <. So this eqution must equl negtive multiple of. α 1 < m + 1 which mens 1 + m α 1 > 0 but less thn. Thus 1 + m α 1 nd there does not exist fctoriztion of this form. Cse 5: (α 0, 0, α ) α 0 () + α (h + d) = h + d + mh + md α 0 () = h + d + mh + md α h α d ( α 0 () = h + mh α h + d + md α ) d In order for this fctoriztion to exist, must divide d + md α d = d(1 + m α ). Since cnnot divide d, must divide 1 + m α. We know 1 + m < so 1 + m α must equl negtive multiple of. α < m + 1 so 1 + m α > 1 + m (m + 1) = 1 + m m = 1. So 1 + m α cnnot rech negtive multiple of since 3. If 1+m α = 0, then α will not equl n integer. Therefore, fctoriztions 13
of this form cnnot exist. Cse 6: (0, α 1, α ) We know (0, 1, m) is fctoriztion lredy. So we wnt to find fctoriztion with α 1 > 1. We will let α 1 = 1 + α 1 to cover ll cses with α 1 1. (1 + α 1)(h + d) + α (h + d) = h + d + mh + md α (h + d) = mh + md α 1(h) α 1(d) = h(m α 1) + d(m α 1/) m α 1 = m α 1/ only when α 1 = 0 which is the originl fctoriztion. Cse 7: (α 1, α 1, α ) h + d + mh + md = α 0 () + α 1 (h + d) + α (h + d) α 0 () = h + d + mh + md α 1 h α 1 d α h α d ( α 0 () = h + mh α 1 h α h + d + md α ) 1d α d In order for this to be fctoriztion, must divide d + md α 1 d α d = d(1 + m α 1 α ). cnnot divide d so must divide 1 + m α 1 α. 1+m < nd α 1 < m+1 nd α < m+1 so 1+m α 1 α must equl negtive multiple of in order to be divisible by. Putting our inequlities together we get 1 + m α 1 α > m. The lrgest m cn be is 1 so 1 + m α1 α > m. So 1 + m α1 α never reches multiple of nd thus cnnot be divisible by. If 1 + m α 1 α = 0, then α 1 = 1 + m α nd α 0 = h + mh α 1 h α h = h + mh h mh + α h α h = mh + α h. But α < m + 1 so mh + α h is negtive or zero. If it is zero, the fctoriztion is (0, 1, m) which is the originl fctoriztoin. Therefore, we only hve one fctoriztion of the element h + d + mh + md, m {0,..., 1} which is (0, 1, m). Since there is only one fctoriztion, the ctenry degree of this element must be zero. 14
Proposition 10. The element (α 0, 1, 0) for α 0 {0,..., c(s) 1} for even nd α 0 {0,..., c(s) h 1} for odd hs ctenry degree zero. Proof. For even: c(s) 1 = h + d 1 = h + d 1. (c(s) 1, 1, 0) = h + d + h + d. For odd: c(s) h 1 = h + d h 1 = ( +1 )h + d h 1 = h + 1h + d h 1 To void confusion in proof, we will let the α 0 term from our originl fctoriztion be equl to the constnt m. Cse 1: (α 0, 0, 0): α 0 () = m + h + d ( = m + h + d ) But (, d) = 1 nd cnnot divide d. Thus there re no fctoriztions of this form for either even or odd. Cse : (0, α 1, 0): α 1 (h + d) = m + h + d ( = (h + d) 1 + m ) h + d A multiple of cn only be fctor of h + d when m = c(h + d) since the monoid is miniml. But m = h + d > h + d 1 > h + d h 1 nd this fctoriztion produces n element lrger thn (α 0, 1, 0). Thus the only fctoriztion of this form is m = 0. This gives fctoriztion (0, 1, 0) = (α 0, 1, 0) which is the originl. Cse 3: (0, 0, α ) α (h + d) = m + h + d = (h)(1) + d ( ) 1 + m In order for this to be fctoriztion, we need m = x(h + d) + d where the extr d term will equte 1 nd 1 from the bove (h)(1) + d ( 1 ). Then m = xh + (x+1)d. Since cnnot divide d, must divide (x + 1). Thus we need x + 1 = n for some n Z. n 0 since x is n integer. The next 15
integer n could be is 1, requiring 1 + x =. Cn x = 1? If this is true, then: m = d + c(h + d) ( ) 1 + c m = ch + d ( ) + 1 = h + d > h + d 1 > h + d h 1 So m is too lrge for both the even nd odd cses. Hving x be lrger integer merely increses the vlue of the lredy too lrge m. Thus this is contrdiction nd there re no fctoriztions of this form. Cse 4: (α 0, α 1, 0) Obviously (m, 1, 0) is fctoriztion. We wnt to see if there re ny others. To do this, we will hve α 1 = (1 + α 1), where α 1 Z. α 0 () + α 1 (h + d) = m + h + d α 0 () + (1 + α 1)(h + d) = m + h + d α 0 () + α 1(h + d) = m α 0 () = m α 1(h) α 1(d) ( ) = m α 1h α 1d (, d) = 1 so must divide α 1. The smllest α 1 could be is. But then α 0 = m h d which is negtive since the lrgest m cn be is h + d 1. Thus ny lrger α 1 would only mke α 0 even more negtive. Thus the only solution is when α 1 equls zero, which is exctly the originl fctoriztion. Cse 5: (α 0, 0, α ) α 0 () + α (h + d) = m + h + d α 0 () = m + h + d α (h) α (d) ( = m + h α h + d α ) d 16
So must divide d α d = d(1 α ). Since (, d) = 1, must divide 1 α. α > 0 so 1 α = x for some integer x. x cnnot be zero or α would not be n integer. 1 α = x implies tht α = 1+x. If is even, x is even. Thus 1 + x is odd nd not n integer when divided by two. This cnnot be fctoriztion for even. If is odd, x must lso be odd to hve 1+x Z. If x = 1, we get α = nd α0 = m + h ( )h d. We wnt to ensure tht α 0 is nonnegtive. The lrgest m cn be is h + d h 1. So α 0 h + d h 1 + h h d = 1. So α0 is negtive nd this is contrdiction. If x is greter odd integer, α 0 will become even more negtive for we re lwys subtrcting the α nd xd terms. Thus there re no fctoriztions of this form. Cse 6: (0, α 1, α ) α 1 (h + d) + α (h + d) = m + h + d α 1 (h + d) = m + h + d α (h) α (d) = (h + d)(1 α ) + m α d We need to know whether m α d cn equl x(h + d). This gives us m = xh+d( x+α ). If x+α = n for some n N 0, then m = nh α h+nd. If n = 0, then m = α h < 0 which is contrdiction. If n = 1, then m = h α h + d. Substituting this bck into our bove eqution, we hve α 1 (h + d) = (h + d)(1 α ) + h α h + d α d = (h + d)( α ) which is lwys negtive or zero, contrdiction. If n, m h α h + d h + d which is outside the rnge of vlues m cn tke. Thus there re no fctoriztions of this form. Cse 7: (α 0, α 1, α ) m + h + d = α 0 + α 1 (h + d) + α (h + d) α 0 = m + h + d α 1 h α 1 d α h α d ( = m + h α 1 h α h + d α ) 1d α d In order for this to be fctoriztion, must divide d α 1 d α d = d(1 α 1 α ). Since (, d) = 1, must divide 1 α 1 α. So we must hve 1 α α = c for some c Z. Since α 1, α N, c 0. We know tht to keep the fctoriztion from getting too big, α 1 < + 1 nd α <. Combining these inequlities we cn see tht 1 α 1 α > 3. So we only need to check when c = 1 nd c =. 17
If c = 1, then 1 α 1 α =. This implies α 1 = 1 α +. Substituting these vlues bck into α 0 we get α 0 = m + h α h (1 α + )h d = m + h α h h + α h h d = m + α h h d. The smllest m cn be is zero. If m = 0, then we require α h > h + d. But α < so this is impossible. If is odd, the lrgest m cn be is h + d h 1, giving α 0 = h+d h 1+α h h d = h h 1+α h. In order for α 0 to be positive, t the very lest, α = + 1. However, if we substitute this bck into our eqution for α 1, α 1 1 ( + 1) + = 1 ( 1) + = 1 +1 + = 0. This is contrdiction. If is even, the lrgest m cn be is h+d 1, giving α0 = h+d 1+α h h d = h 1+α h. So t the very lest, α >. But if α >, α1 < 1 + = 1 + = 1. But we hve ssumed α 1 > 0 so this is contrdiction. We now must check if c =. This gives 1 α 1 α = α 1 = 1 α +. Substituting this into our α 0 eqution we get α 0 = m + h (1 α + )h α h d = m + α h h d. For odd, the mximum m vlue is h + d h 1. So α0 h + d h 1 + α h h d = h h d h 1 + α h. Obviously α > for this fctoriztion to not be negtive. However, we originlly restricted α < so this is contrdiction. If is even, the mximum m vlue is h + d 1. So α 0 h + d 1 + α h h d = h d h 1 + α h Obviously α must be greter thn for this fctoriztion to not be negtive. However, we originlly restricted α less thn. So there re no fctoriztions of this form. Thus the only fctoriztions of the element m + h + d re (m, 1, 0). Thus the element hs ctenry degree zero. Proposition 11. An element with fctoriztion (α 0, α 1, α ) with α 0 {0,..., h 1}, α 1 {0, 1}, α {0,..., 1} hs only one fctoriztion nd thus ctenry degree zero. The element (α 0, 0, ) for α0 {0,..., h 1} with odd lso hs only one fctoriztion nd thus ctenry degree zero. Proof. We will cll our fctoriztion (α 0, α 1, α ) s (x, y, z) to void confusion in the proof. We will prove this sttement for (α 0, α 1, α ) with α 0 {0,..., h 1}, α 1 {0, 1}, α {0,..., 1} nd then mention the different odd cse with (α 0, 0, ) for α0 {0,..., h 1} in prethesis only when the proof differs. We cll it the extr cse. Proofs will be done by ssuming we cn fctor (x, y, z) s fctoriztion of different form, nd then finding contrdiction. First we will estblish size restrictions on the vlues of fctoriztion (α 0, α 1, α ). The mximum element 18
in our originl set hs x = h 1, y = 1, z = 1. This gives element (h 1)+h+d+( 1)(h+d) = h+ h+ d d < h+d. Since ll elements re less thn h+d, we cn restrict α 0 < h+d, α 1 <, nd α <. For the dditionl cse (α 0, 0, ), odd, note tht < / nd < for odd. We will only check the cses where α1 = 0 for this element. Cse 1: (α 0, 0, 0) α 0 = x + yh + yd + zh + zd ( ) yd + zd = x + yh + zh + In order for this to be fctoriztion, we need to divide yd+zd = d(y+z). Since (, d) = 1, we require to divide y + z. If y = 0, z = c. The lrgest z cn be is 1. (The extr cse hs z = ). So z <. This mens the only fctoriztion is when z = 0. The resulting fctoriztion is (x, 0, 0) for y = z = 0. This is identicl to the originl fctoriztion. If y = 1, 1 + z = c. The lrgest z cn be is 1. So 1 + z = 1 + ( 1) = 1 + = 1 <. So this is not divisible for ny z. There re no other fctoriztions of this form. Cse : (0, α 1, 0) α 1 (h + d) = x + yh + yd + zh + zd = (h + d)(y + z) + x + zd In order for this to be fctoriztion, x+zd = c(h+d) x = ch+cd zd x = ch + cd zd. So in order for this to work, cd zd = d(c z) must be divisible by. So (c z) must be divisible by. If c = 0 in the first plce, x + zd = 0(h + d) only if x = z = 0. This is the originl fctoriztion cse. If c = z, then c z = 0. This mkes x = ch + (c z)d = zh. Since x < h, the gin only works if z = x = 0. Since z <, c z never cn rech negtive multiple of. Any lrger c vlue continues to mke x too lrge s it is greter multiple of h. Thus there re no fctoriztions of this form. 19
Cse 3: (0, 0, α ) α (h + d) = x + yh + yd + zh + zd = (h + d)(z) + yh + yd + x. We need to be ble to fctor n (h + d) term out from yh + yd + x. If y = 0, this requires fctoring h + d from x. Since x < h, this is not possible. If y = 1, we need to fctor from x + h + d. x + h + d = c(h + d) x = ch + cd h d x = ch h + cd d. This requires dividing cd d = d(c 1). So must divide c 1 since (, d) = 1. Notice c 1 cnnot equl zero or c will not be n integer. Then 3 so c 1 cnnot equl negtive multiple of. Assume c 1 =. Then x = ( +1 )h h + d > h which is outside the rnge of cceptble x vlues. Any greter c 1 vlue gives n even lrger x vlue. Thus no fctoriztion of this type cn exist. Cse 4: (α 0, α 1, 0) α 0 + α 1 (h + d) = x + yh + yd + zh + zd α 0 = x + yh + yd + zh + zd α 1 h α 1 d ( = x + yh + zh α 1 h + yd + zd α ) 1d So must divide yd + zd α 1 d = d(y + z α 1 ). Since (, d) = 1, must divide y + z α 1. If y = 0, z α 1 = c. The lrgest z vlue is 1 (or in the dditionl cse). So z α 1 = α1 < (likewise, z α 1 α1 < ). So c = 0 or c < 0. The smllest z vlue is zero nd the lrgest α 1 <. So z α 1 never reches negtive multiple of. If z α 1 = 0 then α 1 = z. Substituting this bck into α 0 gives α 0 = x + (0)h + zh (z)h + 0 = x zh. This fctoriztion is negtive unless z = 0. This then gives fctoriztion (x, 0, 0) for y = z = 0 which is exctly the originl. If y = 1, 1 + z α 1 = c. The lrgest z cn be is 1. So 1 + z α1 1 α1 <. So 1 + z α 1 = c. The smllest z = 0 nd lrgest α 1 <. so 1 + (0) α 1 > 1 >. So c cnnot be negtive number. If c = 0, then 1 + z = α 1. Substituting this vlue into α 0 gives α 0 = x + h + zh (1 + z)h + 0 = x zh which is negtive for z > 0. So this gives one fctoriztion when z = 0 which is (x, 1, 0) which is exctly the originl fctoriztion for y = 1, z = 0. 0
Cse 5: (α 0, 0, α ) α 0 + α (h + d) = x + yh + yd + zh + zd α 0 = x + yh + yd + zh + zd α h α d ( = x + yh + zh α h + yd + zd α ) d For this to be fctoriztion, we need yd + zd α d divisible by. Since (, d) = 1, must divide y + z α. If y = 0, z α = c. The lrgest z = 1 (or for odd) so z <. Thus we need negtive multiple of or c = 0. If z α = 0 then α = z. Plugging this into α 0 gives α 0 = x + yh + zh zh = x + yh = x nd gives fctoriztion (x, 0, z) which is the originl fctoriztion for y = 0. The smllest z vlue is zero. Thus α < implies α >. So we need only check one more cse when c, the negtive multiple is equl to 1. z α = implies α = z +. This requires to be even. Substituting this vlue into α 0 = ( ) x + yh + zh α h + yd+zd α d gives α0 = x + zh (z + )h d = x h d. Since x < h nd 3, α 0 < 0 which is contrdiction. If y = 1, 1+z α = c. The lrgest z = 1. So 1+z α 1 <. So c must be zero or negtive. If 1+z α = 0 then α = 1 +z. But α must be n integer so this is contrdiction. The smllest z = 0 so the smllest vlue 1+z α cn be is greter thn since α <. So we only need to check one cse where c = 1. If 1+z α = then α = z + 1+. Substituting this bck into α 0 = ( ) x + yh + zh α h + yd+zd α d gives α 0 = x + (1)h + zh α h + yd+zd α d = x + h + zh ( +1 + z)h d = x + h + 1 h d. Since x < h this vlue for α 0 is lwys negtive. This is contrdiction nd there re no fctoriztions of this form. Cse 6: (0, α 1, α ) α 1 (h + d) + α (h + d) = x + yh + yd + zh + zd α 1 (h + d) = x + yh + yd + zh + zd α h α d = (h + d)(y + z α ) + x + zd α d In order for this to be fctoriztion, we need x + zd α d = c(h + d). This implies x + zd ch cd = α d which provides the eqution α = z c + x ch. Since (, d) = 1 so d must divide x ch. If c = 0, α d = x + z. d 1
Substituting this into α 1 gives α 1 = y + z x z = y x. Since y {0, 1}, d d α 1 < 0 unless x = 0. Thus the only fctoriztion is (0, y, z) which is merely the originl when x = 0. If c is some other integer, then α = (x ch) + z c. Then x ch < 0 since d x < h so (x ch). Thus α d + z c. Since z {0,..., 1} (or z = for odd) nd 1 <, we get α < 0 which is contrdiction. Thus there re no fctoriztions of this form. Cse 7: When (α 0, α 1, α ) we hve Then, α 0 + α 1 (h + d) + α (h + d) = x + yh + yd + zh + zd. α 0 = x + yh + yd + zh + zd α 1 h α 1 d α h α d ( α 0 = x + yh + zh α 1 h α h + yd + zd α ) 1d α d. In order for this to be fctoriztion, we need to divide yd + zd α 1 d α d = d(y +z α 1 α ). Since (, d) = 1, must divide y +z α 1 α. We wnt to find y + z α 1 α = c. The lrgest y nd lrgest z vlues give y + z = 1 + = 1 < (nd if z =, we get y + z = 0 + < so the proof still holds). So c must be negtive or zero. If y + z α 1 α = 0 implies y + z α = α 1. Substituting this bck into α 0 gives α 0 = x + yh + zh (y + z α )h α h + 0 = x zh + α h = x + (α z)h. If y = 0, z α = α 1. So α < z to keep α 1 > 0. Thus x + (α z)h = x nh for some n > 0. Since x < h, α 0 < 0 which is contrdiction. If y = 1, we hve 1 z α > 0 which implies α z. So α 0 = x + (α z)h x + (z z)h. So α 0 = x or is x nh for n > 0 which cnnot exist by the previous rgument. The only fctoriztion is (x, 1, z) which is exctly the originl. Cn y + z α 1 α = c? Using the fct tht α 1 < nd α <, we get the equivlence α 1 α > 3. So we only need check two cses: c = 1 nd c =. If c = 1, we hve y + z α 1 α = implies α 1 = y + z α +. Substituting this into our eqution for α 0 we get α 0 = x + yh + zh α 1 h
α h d = x+yh+zh (y +z α +)h α h d = x zh+α h h d. We hve α < to keep the fctoriztion from becoming too lrge. Thus α h < h implies α h h < 0. Specificlly, (α )h = mh for some m N. We now hve α 0 = x zh mh d. Since x < h, this fctoriztion is lwys negtive nd thus we hve contrdiction. If c =, we hve y + z α 1 α = implies α 1 = y + z α +. Substituting this vlue bck into α 0 we get α 0 = x+yh+zh α 1 h α h d = x + yh + zh (y + z α + )h α h d = x zh + α h h d. By the sme rgument s the c = 1 cse, α 0 < 0 nd thus cnnot be fctoriztion. Therefore, for ny possible cse of fctoriztion, the only possible is the originl fctoriztion (x, y, z). Since ll of these elements hve only one fctoriztion, their ctenry degree must be zero. Proposition 1. The element (0, 0, α ), α {1,..., 1} hs ctenry degree zero. Proof. Lter on, we show n element (0, 0, k) hs the following fctoriztions: z = (0, 0, k) nd z = ((α + k)h + d, (α + k), α ). Notice tht if k >, the second coefficient of z is negtive, regrdless of the α vlue. Thus, there is only one fctoriztion of this element: (0, 0, k) nd thus (0, 0, α ) for α {1,..., 1} hs ctenry degree zero. Combining ll the cses we hve proved in the bove proposition, we rrive t the following theorem. Theorem 13. The following elements hve only one fctoriztion nd thus hve ctenry degree zero. If is n even number: (α 0, α 1, 0), α 0 {0,..., c(s) 1}, α 1 {0, 1} (α 0, α 1, α ), α 0 {0,..., h 1}, α 1 {0, 1}, α {0,..., 1} If is n odd number: (α 0, α 1, 0), α 0 {0,..., c(s) 1 α 1 h}, α 1 {0, 1} (α 0, α 1, α ), α 0 {0,..., h 1}, α 1 {0, 1}, α {0,..., α1 } 3
Proposition 14. Any element not in the rnges from Theorem 13 hs nonzero ctenry degree. Proof. The proofs of this sttement involve using the three essentil moves found in the fctoriztions of the betti elements in fctoriztion (α 0, α 1, α ). If α 0 c(s), we cn use the move from (α 0 c(s), α 1 +, α + ) to get new fctoriztion. If α 1 > 1, we cn use unboxing nd get new fctoriztion through (α 0 + h, α 1, α + 1). If α for even, we cn use the move (α 0 + c(s), α 1 + 0, α ). If odd nd α, we cn use the move (α 1 + h + d, α1 +, α ) to crete new fctoriztion. If even nd α 0 c(s), α 1 = 1, we cn use the move (α 1 c(s), α 1 +, α + ). If odd nd α0 c(s) h, α 1 = 1, we cn use the move (α 0 h + d, α1 1, α + ) If α0 h nd α > 0 we cn use the boxing move (α 0 h, α 1 +, α 1). These cover ll elements outside the rnges given bove. Thus ny element not in these rnges hs more thn one fctoriztion nd thus does not hve ctenry degree 0. Remrk 15. Note tht these rnges for ctenry degree zero re equivilent to the elements tht stisfy x β 1 / S, x β / S, x β 3 / S. The fctoriztions of β 1 re β 1 = (h, 0, 1) = (0,, 0). Thus if n element with fctoriztion (α 0, α 1, α ) hs α 1, there exists nother fctoriztion with (α 0 + h, α 1, α + 1). Similrly, if the α 0 coordinte is t lest h while the α coordinte is positive, there exists nother fctoriztion of form (α 0 h, α 1 +, α 1). The fctoriztions of β re β = (0, 0, ) = ( h+d,, 0). Thus ny fctoriztion with α will hve nother fctoriztion equl to (α 0 + h + d, α1 + ( ), α + ). Any fctoriztion with first coordinte α 0 + h + d nd α1 = 0 (if even) or α 1 = 1 (if odd) hve corresponding fctoriztion (α 0 ( h+d), α1 ( ), α + ). The fctoriztions of β 3 re (c(s), 0, 0) = (0,, ) Thus ny fctoriztion with α 1 c(s) hs nother fctoriztion equl to (α 0 c(s), α 1 +, α + ). Also, ny fctoriztion with α nd α1 = 1 if odd or α 1 = 0 if even hs corresponding fctoriztion (α 0 + c(s), α 1 ( ), α ). 4
If we do not llow n element to be divisible by ny Betti element, we see tht we get the exct equtions for ctenry degree zero s described in theorem 13. 3 Ctenry Degree h + 1 The first nonzero ctenry degree is h + 1. We know tht this ctenry degree is chieved t the Betti element (h + d). Through use of Betti element divisors, we will chrcterize which elements hve ctenry degree h + 1 nd which elements hve ctenry degree strictly greter thn h + 1. Proposition 16. Given tht the Betti elements for S =, h + d, h + d re {(h + d), (h + d), c(s)}, for n S such tht (h + d) divides n, but (h + d) nd c(s) do not divide n, c(n) = h + 1. Proof. Tke n S such tht c(n) 0, s described in Theorem [ctenry degree 0 theorem]. Suppose (h+d) divides n, but (h+d) nd c(s) do not divide n. Since these re Betti elements by Corollry 3.7, [9] Thus, c(n) = c((h + d)) = h + 1. Proposition 17. moved to end Theorem 18. moved to end c((h + d)) c(n) c((h + d)). 4 Ctenry Degree c(s) Ctenry degree c(s) = h + d is the lrgest ctenry of the monoid. We know this fct from [3] s well s [9] which stted tht the lrgest ctenry degree must be obtined t Betti element. Our nlysis defines which elements give this lrgest ctenry degree nd shows tht there re n infinite number of them in the monoid. Theorem 19. Ech multiple of greter thn or equl to c(s) hs ctenry degree c(s). 5
Proof. Let us show by induction in k tht d((k, 0, 0), (α 0, α 1, α )) c(s) for ll (α 0, α 1, α ) Z(k). For embedding dimension n, we hve shown in Lemm 4 tht c(s) hs exctly one fctoriztion with nonzero first coordinte, nmely (c(s), 0, 0). Additionlly, we hve lso shown tht ny other fctoriztion hs the quntity zero in the first coordinte. In such cses, it is esy to verify tht the distnce between those two fctoriztions is greter or equl thn c(s). For k > c(s), ssume d((k, 0, 0), (α 0, α 1, α )) c(s) where (α 0, α 1, α ) Z(k) nd (k, 0, 0) (α 0, α 1, α ). Consider the element (k + 1) nd define z 1 = (k + 1, 0, 0) nd z = (α 0, α 1, α ) where z 1, z Z((k + 1)), z 1 z. If (α 0 1, α 1, α ) Z(k), then by the property of the distnce function on pge we hve: d((k + 1, 0, 0), (α 0, α 1, α )) = d((k, 0, 0), (α 0 1, α 1, α )) c(s) If α 0 = 0, it is esy esily verifible tht d(z 1, z ) c(s) nd our induction finlized. Hence, given k with k c(s), we hve tht ny N-chin between (k, 0, 0) nd ny other fctoriztion must hve N c(s). Moreover, c(s) c(k) c(s). Thus, c(k) = c(s). Lemm 0. Suppose is even. Then, c(c(s) + h + d) = c(s). Proof. We will show tht the only fctoriztion of c(s) + h + d tht hs nonzero first coordinte is (c(s), 1, 0). Notice tht by looking t the Betti elements, we cn see tht (0, 1, /) Z(c(S)+h+d); thus, c(s)+h+d hs, t lest, two fctoriztions. Define z 1 = (c(s), 1, 0) nd z = (α 0, α 1, α ), z 1 z. If α 0 c(s) then z 1 = z contrdicting our ssumptions. Thus, α 0 < c(s). If α 0 = 0, then d(z 1, z ) c(s). Suppose, by wy of contrdiction, tht 0 < α 0 < c(s). If α 1 = 0, then α 1 nd c(s) + h + d = α 0 + α (h + d) (c(s) α 0 α h + h) = (1 + (α 1))d. Since gcd(, d) = 1, 1 + (α 1) but this is contrdiction since is even. Now if α 1 1, then c(s) + h + d = α 0 + α 1 (h + d) + α (h + d) c(s) = α 0 + (α 1 1)(h + d) + α (h + d). 6
Since c(s) hs exctly one fctoriztion with nonzero first coordinte, α 0 = c(s), α 1 = 1, nd α = 0. However, this contrdicts our ssumption tht z 1 z. Therefore, the only fctoriztion of c(s) + h + d tht hs nonzero first coordinte is (c(s), 1, 0). Then, d(z 1, z ) c(s), implying tht c(c(s)+ h + d) = c(s). 5 Intermedite Vlues For The Ctenry Degree Set This section will chrcterize the ctenry degree vlues tht re found between the 1st nd nd Betti elements. We will show tht the ctenry degree + d + k(h ) is relizble t the element ( k)(h + d) nd h + d + ( k)(h + d) for k {1,..., }. Theorem 1. The element ( k)(h + d) hs the following fctoriztions. z = (0, 0, k) nd z = ((α +k)h+d, (α +k), α ) for k {1,..., } fixed nd α {0,..., k} Proof. Let ( k)(h + d) be n element in the monoid, h + d, h + d. Fix k <. We will find ll the fctoriztions of the following seven cses: 1. α 0 = α 1 = 0, α > 0. α 1 = α = 0, α 0 > 0 3. α 0 = α = 0, α 1 > 0 4. α 0 = 0, α 1, α > 0 5. α 1 = 0, α 0, α > 0 6. α = 0, α 0, α 1 > 0 7. α 0, α 1, α > 0 Cse 1: When α 0 = α 1 = 0, α > 0 we hve α (h + d) = ( k)(h + d) 7
which implies α = k. Thus (0, 0, k) is fctoriztion. Cse : When α 1 = α = 0, α 0 > 0, then α 0 () = ( k)(h + d) This requires ( k)(h + d) to be divisible by. α 0 () = h + d kh kd α 0 () = ()(h + d kh kd ) Since (, d) = 1, it follows tht. Thus, must divide, k, or k. Also, k or k < 0 so k {0,..., }. If k =, then k nd the fctoriztion would be (kh + d kh d, 0, 0) = (0, 0, 0) which is not nonzero fctoriztion nd thus contrdiction. Hence, k nd so k <. Since must divide k nd k <, the only time this cn hppen is when is n even number nd k =. This gives us the fctoriztion ( h + d, 0, 0) if k = kd nd even. Otherwise, is not divisible by nd there is no solution. Cse 3: When α 0 = α = 0, α 1 > 0 we hve α 1 (h + d) = ( k)(h + d) α 1 (h) + α 1 (d) = h kh + d kd α 1 (h) + α 1 (d) = h( k) + d( k). This hs fctoriztion when k is equl to k which only occurs when k = 0. Thus, there is no fctoriztion of this type. Cse 4: When α 0 = 0, α 1, α > 0 we hve α 1 (h + d) + α (h + d) = ( k)(h + d) α 1 (h + d) + α (h + d) = h kh + d kd. Obviously α = ( k) is solution but then α 1 = 0. Thus α < k. h(α 1 + α ) + d(α 1 + α ) = h kh + d kd h(α 1 + α ) + d(α 1 + α ) = h( k) + d( k). This gives us the two liner equtions α 1 +α = k nd α 1 +α = k. From the first eqution, α 1 = k α. We wnt to show tht α 1 + α = 8
k. If ( k α ) +α = k +α is equl to k then α = k. But substituting α = k bck into the eqution forces α 1 = 0 which is contrdiction. Cse 5: When α 1 = 0, α 0, α > 0 we hve α 0 () + α (h + d) = ( k)(h + d) α 0 () + α (h + d) = h kh + d kd. If (kd) then must be even nd k = which we lredy showed in previous cse. Thus, the kd term must come from α. So let α = (m k) where k < m <. Then we hve, α (h + d) = mh + md kh kd. We ( now need α 0 () + mh ) + md = h + d which implies tht α 0 = ( m)h + ( m)d. Cn (( m)d)? If so, ( m)d h md = = h md. When does md? Let c = m. Since (, d) = 1, if cd then c. This implies tht m. Since m <, the only time this hppens is if m = with even. If m = then k < nd α = (m k) = ( k). Then simplified, α 0 = h + d. Therefore, if k < with even then there exists the fctoriztion ( h + d, 0, k). This fctoriztion is equl to ( + d, 0, α ) with α = ( k) which is prt of the fctoriztion ((α + k)h + d, (α + k), α ). Cse 6: When α = 0, α 0, α 1 > 0 we hve α 0 () + α 1 (h + d) = ( k)(h + d) α 0 () + α 1 (h) + α 1 (d) = h kh + d kd. We hve lredy shown the cse when kd so we will ssume this term comes from the α 1 term. Then, α 0 () + (m k)(h + d) = h kh + d kd with k < m to crete 9
the kd term. Next, α 0 () + mh + md kh kd = h kh + d kd α 0 () = h + kh + d mh md ( α 0 () = () h + kh + d mh md ). Now the question is: when does md? We know tht m > k nd in order to ensure tht this remins fctoriztion of the given element, m <. Thus, the only time md is if m =. If m =, then ( α 0 = h + kh + d mh md ) = h + kh + d h d = kh + d nd α 1 = (m k) = ( k). Thus (kh + d, k, 0) is fctoriztion. Note tht this fctoriztion is equl to ((α + k)h + d, (α + k), α ) for α = 0. Cse 7: Finlly, when α 0, α 1, α > 0 we proceed by letting α 1, nd α be rbitrry integers nd then solve for α 0. In doing so, we shll find reltionship between α 1 nd α nd find ll solutions to the cse. Overll, we wnt to show tht α 0 () + α 1 (h + d) + α (h + d) = ( k)(h + d). Let α = z for some z N nd let α 1 = y for some y N ( k)(h + d) = α 0 () + α 1 (h + d) + z(h + d) α 0 () + α 1 (h) + α 1 (d) = h kh + d kd zh zd α 0 () + yh + yd = h kh + d kd zh zd α 0 () = h kh + d kd zh zd yh yd ( α 0 () = () h kh + d zh yh kd + zd + yd Since α 0 is n integer, we now need to know when cn divide kd + zd + yd. Recll, we know tht k < nd z < k nd k < y < k. 30 ).
First solve we will solve for when kd + zd + yd equls. Let z = z. Then y = (z + k). Substituting these solutions bck into our eqution for α 0 () gives α 0 = (k + z)h + d. Now if kd + zd + yd = c for c N nd c, then we hve z = z, y = c (z + k). However, when we substitute these into our eqution for α 0, we get α 0 = (k + z + (1 c))h + d( c). Since k <, nd z < k, z + k < nd with c, this solution occurs when α 0 0. Thus the c solutions cnnot exist in this cse nd we hve found ll of the fctoriztions. In conclusion, the fctoriztions of ( k)(h + d) re s follows: (0, 0, k) if even nd k = /, ((/)h+d, 0, 0) which is equl to (kh+d, 0, 0) = ((α + k)h + d, (α + k), α ) for α = 0 if even nd k < /, ((/)h + d, 0, (/) k) which is equl to ((α + k)h + d, (α + k), x 1) for α = / k if k, (kh + d, k, 0) which is equl to ((α + k)h + d, (α + k), α ) for α = 0 ((k +z)h+d, (z +k), z) which is equl to ((α +k)h+d, (α + k), α ) with α = z. Bsiclly, ll elements ( k)(h + d) hve fctoriztion (0, 0, k) nd ((α + k)h + d, (α + k), α ) for k {1,..., } fixed nd α {0,..., k}. Note tht if k >, the second fctoriztion hs negtive term in the second genertor nd cnnot exist. Therefore there is only one fctoriztion of ( k)(h + d) with ctenry degree 0. Proposition. The fctoriztion of the element ( k)(h + d) with the lrgest first coordinte is ( h + d, ( ), k ). Proof. From 1 we know the element ( k)(h + d) hs the following fctoriztions. z = (0, 0, k) nd z = ((k + α )h + d, (k + α ), α ) for k {1,..., } fixed nd α {0,..., k}. The element with the lrgest first coordinte will be member of z since the first coordinte of z is 0. The first coordinte of z = (k + α )h + d. Since k fixed, to mximize 31
this vlue, we will choose α to be s lrge s possible. The lrgest vlue of α is k. Inputting this vlue for α into z we get. = (( k + z = ((k + α )h + d, (k + α ), α ) ) ( ) k h + d, k + k, ( ) = h + d,, k ) k Proposition 3. Fctoriztions (0, 0, k) = ((α +k)h+d, (α +k), α ) for fixed k {1,..., } nd α {0,..., + k}. Proof. ((α + k)h + d, (α + k), α ) = ((α + k)h + d)() + ( (α + k))(h + d) + (α )(h + d) = α h + kh + d + h α h kh + d α d kd + (α )(h + d) = α h kh + d + h α d kd + (α )(h + d) = α h kh + d + h α d kd + α h + α d = h kh + d kd = ( k)(h + d). Proposition 4. The distnce between z = (0, 0, k) nd z = ((α +k)h+ d, (α + k), α ) is d(z, z ) = + d + (α + k)(h ) + d + (k)(h ). Proof. Let z = (0, 0, k) nd z = ((α + k)h + d, (α + k), α ). Then, gcd(z, z ) = (0, 0, min{ k, α }) = (0, 0, α ), since α {0,..., } but k {,..., 1}. Then, z gcd(z, z ) = (0, 0, k α ), nd z gcd(z, z ) = ((α + k)h + d, (α + k), 0). 3
Hence, d(z, z ) = mx{ z gcd(z, z ), z gcd(z, z ) } = mx{ k α, + d + (α + k)(h )} = + d + (α + k)(h ), since the nd term hs positive α, k nd constnt terms. Note tht the smllest distnce occurs when α = 0 giving distnce + d + k(h ). Proposition 5. Let z = ((α + k)h + d, (α + k), α ) nd z l = z with α = l. The distnce between z i nd z j is d(z i, z j ) = (i j)h + i j h + 1, for i > j wlog. Proof. Let z i = ((i+k)h+d, (i+k), i) nd z j = ((j+k)h+d, (j+k), j). Assume i > j without loss of generlity. We cn see tht gcd(z i, z j ) = ((j + k)h + d, (i + k), j). z i gcd(z i, z j ) = ((i + k)h + d, (i + k), i) ((j + k)h + d, (i + k), j) = ((i j)h, 0, i j) z j gcd(z i, z j ) = ((j + k)h + d, (j + k), j) ((j + k)h + d, (i + k), j) = (0, (i j), 0) The distnce between z i nd z j is: d(z i, z j ) = mx{ z gcd(z i, z j ), z gcd(z i, z j ) } = mx{(i j)h + i j, (i j)} = (i j)h + i j since h > 0. Note tht when i j = 1, d(z i, z j ) = h + 1. Theorem 6. The element ( k)(h+d) hs fctoriztions z = (0, 0, k) nd z = ((α +k)h+d, (α +k), α ) nd ctenry degree +d+k(h ). Proof. Fix k {1,..., }. Then α {0,..., k}. d(z, z ) = +d+(α +k)(h ). The lrgest α cn be is α = k = N. The smllest α cn be is α = 0. d(z, z 0) = + d + (k + 0)(h ) = + d + k(h ). d(z, z N ) = ( + d + ( k + k)(h ) = + d + (h ). 33
d(z i, z j) = (i j)h+i j for i j is greter thn the distnce for i j = 1 which gives distnce d(z i, z i+1) = h + 1. +d+ (h ) +d+k(h ) > h+1. (i j)h+i j > h+1 when i j 1. d(z, z j) > h + 1 for ny x 1. Next, we rrnge our grphicl representtion of vertices nd edges in the following wy: Strt with the fctoriztion z. Moving clockwise, list z 1, z,..., z N until bck to z. Then, between ech consecutive z fctoriztion, there is n h + 1 distnce between ny two djcent fctoriztions, with the exxception of z. The interior distnces between ny two elements re greter thn h+1. This mens s we remove distnces till disconnecting, the interiors will lwys be removed before h + 1 which connects the outside of the grph. Now the distnces from z to ll other fctoriztions follow the pttern + d + (α + k)(h ). The smllest of these is distnce + d + k(h ) which is greter thn h + 1. Thus, ll distnces from z to z j will be removed before + d + k(h ). The next lrgest distnce is + d + k(h ). However, removing this distnce will disconnect the fctoriztion z from the rest of the grph. Thus, + d + k(h ) is the ctenry degree of this element. Note, if h =, the ctenry degree is + d. Remrk 7. Note tht if is even nd k =, then + d + k(h ) = + d + (h ) = h + d = h + d = c(s). Theorem 8. The element h + d + ( k)(h + d) hs the following fctoriztions: z = (0, 1, k) nd z = ((k + α )h + d, + 1 (k + α ), α ) for k {1,..., } fixed nd α {0,..., +1 k} Proof. Let h+d+( k)(h+d) be n element in the monoid, h+d, h+ d. Fix k {1,..., }. Ensuingly, we wish to find other fctoriztions in terms of (α 0, α 1, α ). In order for the element α 0 ()+α (h+d)+α (h+d) to not exceed h + d + ( k)(h + d), we require α 0 < h + d, α 1 <, nd α <. We will find ll the fctoriztions of the following seven cses: 1. α 0 = α 1 = 0, α > 0. α 1 = α = 0, α 0 > 0 3. α 0 = α = 0, α 1 > 0 4. α 0 = 0, α 1, α > 0 5. α 1 = 0, α 0, α > 0 34
6. α = 0, α 0, α 1 > 0 7. α 0, α 1, α > 0 Cse 1: (α 0, 0, 0) α 0 () = h + d + h kh + d kd ( = h + h kh + d + d kd ) Here we require d kd divisible by. Since cnnot divide d, we need 1 k = c for c Z. By our originl restrictions of α 1, > 1 k >. So we need only check when c = 0 nd c = 1. If c = 0, then 1 k = 0 implies k = 1/ which is contrdiction since k is n integer. If c = 1, 1 k = implies k = +1. Thus, must be odd. Substituting this vlue bck into α 0 gives α 0 = h + h kh + d d = h + h +1 h + d. The fctoriztion is 1 then ( h + d, 0, 0) for k = nd when is odd. Cse : (0, α 1, 0) α 1 (h + d) = h + d + h kh + d kd = (h + d)(1 + k) + d kd There is no wy to pull n h + d term out of d kd nd be left with n integer. Thus, there re no fctoriztions of this form. Cse 3: (0, 0, α ) α (h + d) = h + d + h kh + d kd = (h + d)( k) + h + d But h + d cnnot be fctored out of h + d. Thus there is no fctoriztion of this form. Cse 4: (α 0, α 1, 0) α 0 () + α 1 (h + d) = h + d + h kh + d kd = h + d + h kh + d kd α 1 h α 1 d ( = h + h kd + d α 1 h + d kd α ) 1d 35
So we require d kd α 1 d divisible by. Since cnnot divide d we need 1 k α 1 divisible by. So 1 k α 1 = c for c Z. This implies α 1 = 1 k c. And from our originl quntittive restrictions of α 1, we hve 1 k α 1 > 4. So we need only check c = 0, 1,, 3. If c = 0, then 1 k α 1 = 0 which implies α 1 = 1 k < 0 which is contrdiction. If c = 1, then 1 k α 1 = which implies α 1 = k + 1. This mkes α 0 = h + h kh + d ( k + 1)h = kh + d. So we hve the fctoriztion (kh + d, + 1 k, 0). If c, then α 1 1 k + nd α 0 h + h kh + d α 1 d = h + kh. Since k <, h + kh < 0. So α 0 < 0 which is contrdiction. Cse 5: (α 0, 0, α ) α 0 () + α (h + d) = h kh + d kd + h + d = h kh + d kd + h + d α (h) α (d) ( = h kh + d + h α h + d kd α ) d So we require d kd α d to be divisible by. cnnot divide d so we need 1 k α to be divisible by. So 1 k α = c for c Z. By the originl restrictions on α, 1 k α > 4. Thus we need only check for c = 0, 1,, 3. If c = 0 then α = 1/ k which is not n integer; this is contrdiction. If c = 1, then 1 k α = α = +1 k. Substituting bck in, we hve α 0 = h +1 h + d + h. This gives the fctoriztion (h + h +1 +1 h + d, 0, k) for when odd. If c =, then 1 k α = α = 1/ k + which is not n integer; this is contrdiction. If c = 3, then 1 k α = 3 α = 1/ k + 3. Substituting this for α 0 gives α 0 = h d + h ( 1+3 )h. Now 3 = h + h so h d + h 1h h h < 0 nd thus α 0 < 0. So there cn not be fctoriztion of this subcse. Cse 6: (0, α 1, α ) 36
α 1 (h + d) + α (h + d) = h + d + h kh + d kd α 1 (h + d) = h + d + h kh + d kd α h α d We know tht (0, 1, k) is fctoriztion. We wnt to see if α 1 cn be greter thn 1. (1 + α 1)(h + d) + α (h + d) = h + d + h kh + d kd α (h + d) = h kh + d kd α 1(h) α 1(d) = (h + d)( k) α 1h α 1d The α 1 terms only fctor when α 1 = c(h + d) but α 1 < nd h > 1 so this cnnot be fctoriztion, unless α 1 = 0. Cse 7: (α 0, α 1, α ) α 0 () + α 1 (h + d) + α (h + d) = h + d + h kh + d kd α 0 () = h + d + h kh + d kd α 1 (h) α 1 (d) α (h) α (d) ( α 0 () = h kh + d α 1 h α h + h + d kd α ) 1d α d So we require d kd α 1 d α d divisible by. And since nd dd re reltively prime, cnnot divide d; ensuingly, 1 k α 1 α = c for some c Z. From the originl quntittive restrictions on α 1, α, 1 k α 1 α > 6. So we only need to check when c = { 5,..., 1, 0}. So α 1 = 1 c k α nd α 0 = h kh+d cd α 1 h α h+h. If c 3, then α 0 h kh+d 3d α 1 h α h+h = h kh d (1 3 k α )h α h = h + kh + α h d < 0 implies α 0 < 0; this is contrdiction. Ensuingly, we only need to look t c = 0, 1,. If c = 0, then α 1 = 1 k α < 0 ; this is contrdiction. If c = 1, we hve α 0 = h kh+d h h+kh+α h α h+h d = kh + d + α h. This gives fctoriztion (kh + d + α h, + 1 (k + α ), α ). If c =, we hve α 0 = h + kh + α h nd α 1 = 1 k α. We require ll α terms > 0. The α 1 eqution requires α > k. Using this equivlence, we get α 1 < 1 which implies α 1 = 0. The fctoriztion we hve found is ( h + kh + α h, 0, α ). However, this is not equl to the fctoriztion (0, 1, k): 37