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CROWDMATH 2022: Factorizations in Additive Structures
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Solution to Problem 7.3
julmath 0
Jan 4, 2023
The answer is yes. I split the proof in a proposition, a lemma and a theorem.
Proposition: Let be a positive algebraic number such that is atomic. Suppose that is not finitely generated. Then has infinitely many Betti elements.
Proof: We have that . Let be the minimal pair of . Consider the set We know that , then is a nonempty set of positive integers. Let be its minimal element and let such that and . Consider such that . We have that is not an isolated vertex in if and only in there exists , , such that . It is not hard to see that for some . Then and . This contradictis the minimality of in . Hence is an isolated vertex in . Then is a Betti element of .
Now, for each , consider . It is clear that and , for all . Following exactly the same reasoning as before, we obtain that there exists an isolated vertex in . Therefore is a Betti element of for each . This concludes our proof.
Lemma: Let . We say that, for and , if for all and for some . Then, the set:
has only finitely many elements.
Proof: Let us proceed by induction on . If , we have that , and by the well-ordering theorem we have that has a minimal element . It follows that . Now, assume the Lemma is valid for . Let . Consider the set:
Fix , and such that . Then for every such that , we have that and . Now, let such that . Then we have that for some and some , , where
Let , such that . It is clear that for , . Applying the lemma in , we obtain that the set
has only finitely many elements.
Since we obtain that also has only finitely many elements. Our proof concludes
Theorem: Let be a positive algebraic number such that is atomic and finitely generated. Then has only finitely many Betti elements.
Proof: Let be the minimal polynomial of , . Suppose that is a Betti element of . Then, there exist such that and are not in the same connected component in . We can say that , , where . It is known that , for some . In this case we will say that produces a Betti element.
Consider the set . For and we say that if and only if for each , or .
It is not hard to see that is an equivalence relation that has finitely many equivalence classes. Let be an equivalence class of . Because of the definition of it is not hard to see that for some and , there exists a map such that for , where is the set of positive coeficients of . Now, note that if we have that for , then
It follows that there exists such that , , . Note that connects and in . Then does not produce a Betti element. Using Lemma 3 on it follows that only finitely many polynomials in produce a Betti element. Since there are only finitely many equivalence classes in , we conclude that must have only finitely many Betti elements.
Proposition: Let be a positive algebraic number such that is atomic. Suppose that is not finitely generated. Then has infinitely many Betti elements.
Proof: We have that . Let be the minimal pair of . Consider the set We know that , then is a nonempty set of positive integers. Let be its minimal element and let such that and . Consider such that . We have that is not an isolated vertex in if and only in there exists , , such that . It is not hard to see that for some . Then and . This contradictis the minimality of in . Hence is an isolated vertex in . Then is a Betti element of .
Now, for each , consider . It is clear that and , for all . Following exactly the same reasoning as before, we obtain that there exists an isolated vertex in . Therefore is a Betti element of for each . This concludes our proof.
Lemma: Let . We say that, for and , if for all and for some . Then, the set:
has only finitely many elements.
Proof: Let us proceed by induction on . If , we have that , and by the well-ordering theorem we have that has a minimal element . It follows that . Now, assume the Lemma is valid for . Let . Consider the set:
Fix , and such that . Then for every such that , we have that and . Now, let such that . Then we have that for some and some , , where
Let , such that . It is clear that for , . Applying the lemma in , we obtain that the set
has only finitely many elements.
Since we obtain that also has only finitely many elements. Our proof concludes
Theorem: Let be a positive algebraic number such that is atomic and finitely generated. Then has only finitely many Betti elements.
Proof: Let be the minimal polynomial of , . Suppose that is a Betti element of . Then, there exist such that and are not in the same connected component in . We can say that , , where . It is known that , for some . In this case we will say that produces a Betti element.
Consider the set . For and we say that if and only if for each , or .
It is not hard to see that is an equivalence relation that has finitely many equivalence classes. Let be an equivalence class of . Because of the definition of it is not hard to see that for some and , there exists a map such that for , where is the set of positive coeficients of . Now, note that if we have that for , then
It follows that there exists such that , , . Note that connects and in . Then does not produce a Betti element. Using Lemma 3 on it follows that only finitely many polynomials in produce a Betti element. Since there are only finitely many equivalence classes in , we conclude that must have only finitely many Betti elements.
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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!
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