MIT PRIMES/Art of Problem Solving
CROWDMATH 2022: Factorizations in Additive Structures
Additive Factorizations
Polymath project forum
Polymath project forum
3
M
G
BBookmark
VNew Topic
kLocked
Additive Factorizations
Polymath project forum
Polymath project forum
3
M
G
BBookmark
VNew Topic
kLocked
No tags match your search
MBetti elements
Exercise
resource1
algebra
polynomial
Resource 1
atomicity
catenary_degree
Resource 2
Betti elements
elasticity
factoriality
number theory
prime numbers
Problem 7
Resource 4
set of lengths
algebraic number
almost arithmetic progression
bifurcus monoid
calculus
catenary degree
confusing
crowdmath
CrowdMath Project
hilbert monoid
modular arithmetic
modular congruences
monoids
notation
polyomial
Problem 1
Problem 4
Problem 5
Problem 6
problem-2
Resource 3
sets of lengths
welcome
No tags match your search
MG
Topic
First Poster
Last Poster
Problem 7.2 solution
gourmet_salad 0
Jan 2, 2023
This can be viewed as an extension of existing results in the paper "Factorization invariants of Puiseux monoids generated by geometric sequences" (arxiv: https://arxiv.org/abs/1904.00219). In that paper, three results are shown about , for rational :
Lemma 1.
- is atomic iff is not the reciprocal of any integer
- If is atomic, the Betti elements of are - If is atomic, the set of lengths of any element is an arithmetic progression with difference
Lemma 2. Let be a positive rational, and let such that is an irreducible n-th root of . Then the polynomial is irreducible in .
Proof: This is seen in the textbook Algebra by S. Lang, in page 297.
Corollary. Let be a positive rational, and let be a positive irreducible n-th root of . Then are linearly independent in .
Proof: The proof follows from Lemma 2; If these powers of were not linearly independent, the polynomial (whose unique positive real root is ) would be reducible in .
Claim. Let be a positive rational that is not the reciprocal of any integer, and let be a positive irreducible n-th root of . Let be the map defined by . Then the following statements hold:
is an isomorphism.
If , then for all . ( is the isomorphism mapping a collection of atoms to the element they form.)
Proof. It's easy to check that is additive and maps the identity to itself, and thus it is a homomorphism. To show that it is an isomorphism, we will show that every factorization in corresponds to a unique -tuple of factorizations in , showing both surjectivity and injectivity. Since is generated by , its set of atoms must be a subset of this generating set. Moreover, as , any atom can be written in the form , where , , and . Thus, any factorization in can be written as For some coefficients . This can be rewritten further as
Lemma 1 shows that the set of atoms of is its generating set. Combining this with the previous equation, it's clear to see that the unique -tuple of factorizations in that maps to is . Thus, is an isomorphism.
implies that , and the linear independence shown in Corollary 1 shows that for all .
Corollary 1. Let be a positive rational that is not the reciprocal of any integer, and let be a positive irreducible n-th root of . The set of atoms of is its generating set, .
Proof: Any element that is generated by at least two generating elements can not be an atom, so . On the other hand, by the isomorphism , the factorizations of (where ) are isomorphic to the factorizations of the element in , where is at the -th index. As shown in Lemma 1, is an atom of , thus must be an atom, hence .
Corollary 2.Let be a positive rational that is not the reciprocal of any integer, and let be a positive irreducible n-th root of . The set of Betti elements of is Proof: For a fixed , we identify all Betti elements of the form , where . By the mapping , the factorizations of are isomorphic to the factorizations of the element in , where is at the -th index. [Previous result] shows that the only Betti elements of are . Thus, by isomorphism, the only Betti elements of this form are
Lastly, we prove that all elements of that involve at least two different (linearly independent) powers of can not be Betti elements. Let be such an element, where , and at least two of the 's are nonzero. By the mapping , the factorizations of are isomorphic to the factorizations of in . Let and be two arbitrary factorizations of , and let and be the -tuples of factorizations corresponding to them by . , and so by property in Claim 1, for all . As a result, is a sequence consisting of factorizations of . Moreover, since at least two of these factorizations are non-zero, each two consecutive factorizations have non-zero greatest common divisor, and so this is a path within the factorization graph . Therefore, is not a Betti element, and by isomorphism, is not a Betti element.
Corollary 3. Let be a positive rational that is not the reciprocal of any integer, and let be a positive irreducible n-th root of . For any , is an arithmetic progression with difference .
Proof: Let , where for all . By the mapping , the factorizations of are isomorphic to the factorizations of in . The length of a certain factorization of is equal to . Let . From Lemma 1, the length of the factorization of can take any value in . Moreover, since the terms of the -tuple are independent, . Thus, can take any value in
.
Hence, is an arithmetic progression with difference , and by isomorphism, the same applies to .
Please let me know if anything is unclear!
Lemma 1.
- is atomic iff is not the reciprocal of any integer
- If is atomic, the Betti elements of are - If is atomic, the set of lengths of any element is an arithmetic progression with difference
Lemma 2. Let be a positive rational, and let such that is an irreducible n-th root of . Then the polynomial is irreducible in .
Proof: This is seen in the textbook Algebra by S. Lang, in page 297.
Corollary. Let be a positive rational, and let be a positive irreducible n-th root of . Then are linearly independent in .
Proof: The proof follows from Lemma 2; If these powers of were not linearly independent, the polynomial (whose unique positive real root is ) would be reducible in .
Claim. Let be a positive rational that is not the reciprocal of any integer, and let be a positive irreducible n-th root of . Let be the map defined by . Then the following statements hold:
is an isomorphism.
If , then for all . ( is the isomorphism mapping a collection of atoms to the element they form.)
Proof. It's easy to check that is additive and maps the identity to itself, and thus it is a homomorphism. To show that it is an isomorphism, we will show that every factorization in corresponds to a unique -tuple of factorizations in , showing both surjectivity and injectivity. Since is generated by , its set of atoms must be a subset of this generating set. Moreover, as , any atom can be written in the form , where , , and . Thus, any factorization in can be written as For some coefficients . This can be rewritten further as
Lemma 1 shows that the set of atoms of is its generating set. Combining this with the previous equation, it's clear to see that the unique -tuple of factorizations in that maps to is . Thus, is an isomorphism.
implies that , and the linear independence shown in Corollary 1 shows that for all .
Corollary 1. Let be a positive rational that is not the reciprocal of any integer, and let be a positive irreducible n-th root of . The set of atoms of is its generating set, .
Proof: Any element that is generated by at least two generating elements can not be an atom, so . On the other hand, by the isomorphism , the factorizations of (where ) are isomorphic to the factorizations of the element in , where is at the -th index. As shown in Lemma 1, is an atom of , thus must be an atom, hence .
Corollary 2.Let be a positive rational that is not the reciprocal of any integer, and let be a positive irreducible n-th root of . The set of Betti elements of is Proof: For a fixed , we identify all Betti elements of the form , where . By the mapping , the factorizations of are isomorphic to the factorizations of the element in , where is at the -th index. [Previous result] shows that the only Betti elements of are . Thus, by isomorphism, the only Betti elements of this form are
Lastly, we prove that all elements of that involve at least two different (linearly independent) powers of can not be Betti elements. Let be such an element, where , and at least two of the 's are nonzero. By the mapping , the factorizations of are isomorphic to the factorizations of in . Let and be two arbitrary factorizations of , and let and be the -tuples of factorizations corresponding to them by . , and so by property in Claim 1, for all . As a result, is a sequence consisting of factorizations of . Moreover, since at least two of these factorizations are non-zero, each two consecutive factorizations have non-zero greatest common divisor, and so this is a path within the factorization graph . Therefore, is not a Betti element, and by isomorphism, is not a Betti element.
Corollary 3. Let be a positive rational that is not the reciprocal of any integer, and let be a positive irreducible n-th root of . For any , is an arithmetic progression with difference .
Proof: Let , where for all . By the mapping , the factorizations of are isomorphic to the factorizations of in . The length of a certain factorization of is equal to . Let . From Lemma 1, the length of the factorization of can take any value in . Moreover, since the terms of the -tuple are independent, . Thus, can take any value in
.
Hence, is an arithmetic progression with difference , and by isomorphism, the same applies to .
Please let me know if anything is unclear!
0 replies
The last problem of CrowdMath 2022 has been posted!
felixgotti 0
Jan 1, 2023
Hi everyone! Check out the tab of problems. The last problem of CrowdMath 2022 (Problems 7), which has several parts, has just been posted. The problem is aimed to better understand the sets of lengths, catenary degrees, and the sets of Betti elements of the monoids we are studying in our research project. There is no resource corresponding to this last problem, but I have stated the new and needed definitions as part of the problem statement. Feel free of course to post any question you may have. Enjoy this last problem, and Happy New Year!!!
0 replies
No more topics!