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CROWDMATH 2023: Arithmetic of Power Monoids
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MPower Monoid
Exercise 3
Puiseux monoid
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Exercise 2
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atomicity
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induction
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Exercise 1.2
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exercise 1.5
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exercise 2.4
Exercise 2.6
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floor function
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Are all numerical monoids finitely generated?
Number_Basher 1
N
Nov 22, 2023
by Number_Basher
Are all numerical monoids finitely generated? My intuition agrees with the statement, but I cannot formulate a proof.
1 reply
minor progress regarding open problem 1b
smileapple 21
N
Nov 19, 2023
by jujumumu
Here are a few ideas that could potentially be useful for Open Problem 1b (unimodality of ). However, most of them probably still need lots of extra work and are very incomplete.
Part I, notation
For no particular reason, let us call some singular if it is noninvertible and not an atom. Because is just , it follows that is singular iff we can write it as where the both have at least a nonzero element.
Because , we must have Hence, if , then implying that Similarly It follows that
Part II, classification of singular sets (small cases)
What I did next was to find possible "structures" of singular for cases where was small.
So if , then because and thus contains . But is invertible, so is neither an atom nor is singular.
If , then However, if such that is a nonzero element of , then so this implies that as so that But this is impossible, so can't be singular if
Now consider , so Then we can write where the represents some unknown element. Then all the elements must also be in , as established previously. Hence WLOG we let be in the first component set : Then, let : Then, so we must have in order to satisfy Thus one singular form would be
Next, consider , in which We use the same - setup as before: Here we have two cases.
Case 4a
If , then WLOG let as Then and clearly , , , and must all be distinct. Thus and since Thus another singular form here is
Case 4b
If , then is in both , so clearly Thus, WLOG let's set Now we have Since , we clearly aren't done yet, so let's add some other to any of the sets, WLOG Hence so so inspection reveals that must be in order for to be In that case our singular set form is However, this is just a unique case of the one from Case 4a.
To conclude, if , we have significant singular form:
Finally, we tackle the case where which implies that (This is by far the most complicated case.) Once again, use our - construction is and we also have two big cases.
Case 5a
If , let WLOG. Then, at most four elements of are known, so we must add some element to one of the . WLOG, add to , so that Then Then exactly one of and is in ; otherwise, would be Inspection reveals that the only possibilities for are , , and , so the possible constructions of are and However, if we let and , note that so the latter two cases above are the same. Hence we have two structures here for : and
Case 5b
If , things get a whole lot more complicated. Specifically, we have so must include which WLOG we let be equal to Then we use the same strategy: add some to some , WLOG Then there are two more cases based on whether .
Case 5bi
If , then so Since , so WLOG let If so, so it follows that and But this is exactly what we found in 5a, so this isn't anything new.
Case 5bii
Here Thus so so WLOG let If so, we must add a new element to either set. If we add it to , then so , and it is easy to see by inspection that , so that Similarly, if we add to , then giving Again, inspection gives , so Hence in both cases, which again is just a special case of 5a. Hence, again we have nothing new.
Thus, the only two distinct general forms for are and
I feel like the process for findings these forms are recursive in some way. Maybe the number of unique forms as increases will follow a linear recursions, such as that of the Fibonacci sequence. But I don't know yet, since I'm still thinking about the which is even more complicated than before.
Part III, computations
Here I was able to use the results from Part II to bash out the first values of by hand, for positive integers The method I used was basically just noting that and , and brute-forcing everything else when The results I got were
which seems to agree with the conjecture of unimodality. If I get to I might be able to get a gauge on the bound for Problem 1.1 found by , as
Part IV, finite differences???
This is one final whimsical thought that I had. The idea is that if the are unimodal, then interpreting this as a function of for a fixed will give a concave-down graph. If so, then the second difference of the must be nonpositive, much like for concave-down functions. If so, the second finite difference is I'm not sure how to continue with this though. I'll keep trying this method, though I'm doubtful of how useful this strategy is.
--WJH
Part I, notation
For no particular reason, let us call some singular if it is noninvertible and not an atom. Because is just , it follows that is singular iff we can write it as where the both have at least a nonzero element.
Because , we must have Hence, if , then implying that Similarly It follows that
Part II, classification of singular sets (small cases)
What I did next was to find possible "structures" of singular for cases where was small.
So if , then because and thus contains . But is invertible, so is neither an atom nor is singular.
If , then However, if such that is a nonzero element of , then so this implies that as so that But this is impossible, so can't be singular if
Now consider , so Then we can write where the represents some unknown element. Then all the elements must also be in , as established previously. Hence WLOG we let be in the first component set : Then, let : Then, so we must have in order to satisfy Thus one singular form would be
Next, consider , in which We use the same - setup as before: Here we have two cases.
Case 4a
If , then WLOG let as Then and clearly , , , and must all be distinct. Thus and since Thus another singular form here is
Case 4b
If , then is in both , so clearly Thus, WLOG let's set Now we have Since , we clearly aren't done yet, so let's add some other to any of the sets, WLOG Hence so so inspection reveals that must be in order for to be In that case our singular set form is However, this is just a unique case of the one from Case 4a.
To conclude, if , we have significant singular form:
Finally, we tackle the case where which implies that (This is by far the most complicated case.) Once again, use our - construction is and we also have two big cases.
Case 5a
If , let WLOG. Then, at most four elements of are known, so we must add some element to one of the . WLOG, add to , so that Then Then exactly one of and is in ; otherwise, would be Inspection reveals that the only possibilities for are , , and , so the possible constructions of are and However, if we let and , note that so the latter two cases above are the same. Hence we have two structures here for : and
Case 5b
If , things get a whole lot more complicated. Specifically, we have so must include which WLOG we let be equal to Then we use the same strategy: add some to some , WLOG Then there are two more cases based on whether .
Case 5bi
If , then so Since , so WLOG let If so, so it follows that and But this is exactly what we found in 5a, so this isn't anything new.
Case 5bii
Here Thus so so WLOG let If so, we must add a new element to either set. If we add it to , then so , and it is easy to see by inspection that , so that Similarly, if we add to , then giving Again, inspection gives , so Hence in both cases, which again is just a special case of 5a. Hence, again we have nothing new.
Thus, the only two distinct general forms for are and
I feel like the process for findings these forms are recursive in some way. Maybe the number of unique forms as increases will follow a linear recursions, such as that of the Fibonacci sequence. But I don't know yet, since I'm still thinking about the which is even more complicated than before.
Part III, computations
Here I was able to use the results from Part II to bash out the first values of by hand, for positive integers The method I used was basically just noting that and , and brute-forcing everything else when The results I got were
which seems to agree with the conjecture of unimodality. If I get to I might be able to get a gauge on the bound for Problem 1.1 found by , as
Part IV, finite differences???
This is one final whimsical thought that I had. The idea is that if the are unimodal, then interpreting this as a function of for a fixed will give a concave-down graph. If so, then the second difference of the must be nonpositive, much like for concave-down functions. If so, the second finite difference is I'm not sure how to continue with this though. I'll keep trying this method, though I'm doubtful of how useful this strategy is.
--WJH
21 replies
A result regarding problem 2
GPZ 7
N
Oct 26, 2023
by felixgotti
Let be an atomic Puiseux monoid. We will split the proof of the problem into the following two cases.
Case 1: is a MCD monoid. Suppose, by way of contradiction, that is not atomic, then there exists some non-atomic element of . This implies that for every factorization of , one of the factors must be a non-atomic element. Consider the set . so it has a minimum, namely . See that because every element of is an atomic element. Now consider one element of . See that . Now, if we take we have that belongs to for all , so is an element of . Now we have that . Clearly is an atomic element of . Let us prove that is also an atomic element of . Let be an arbitrary factorization of . If and , then none of them belongs to and that implies that is an atomic element. So we will assume WLOG that , but this implies that , so , because 0 is the only common divisor of the set . So is an atomic element, implying that is an atomic element, which is a contradiction. Notice that if is a -MCD atomic cancellative monoid, then we have proved that every element in such that is an atomic element of .
Case 2: is not a 2-MCD monoid. In this case, we know that there exists two elements of , namely and , such that does not exist. Let us prove that the element is not an atomic element of . Suppose, by way of contradiction, that , where is an atom of for all . We can tell WLOG that and for all . But, since , and , it must exists some non-invertible element of such that which is a contradiction.
I have not find an atomic Puiseux monoid non 2-MCD, but if this monoid exists, we will be able to find a non-atomic element in and we could conclude that in general the property of being atomic does not ascend from to .
Case 1: is a MCD monoid. Suppose, by way of contradiction, that is not atomic, then there exists some non-atomic element of . This implies that for every factorization of , one of the factors must be a non-atomic element. Consider the set . so it has a minimum, namely . See that because every element of is an atomic element. Now consider one element of . See that . Now, if we take we have that belongs to for all , so is an element of . Now we have that . Clearly is an atomic element of . Let us prove that is also an atomic element of . Let be an arbitrary factorization of . If and , then none of them belongs to and that implies that is an atomic element. So we will assume WLOG that , but this implies that , so , because 0 is the only common divisor of the set . So is an atomic element, implying that is an atomic element, which is a contradiction. Notice that if is a -MCD atomic cancellative monoid, then we have proved that every element in such that is an atomic element of .
Case 2: is not a 2-MCD monoid. In this case, we know that there exists two elements of , namely and , such that does not exist. Let us prove that the element is not an atomic element of . Suppose, by way of contradiction, that , where is an atom of for all . We can tell WLOG that and for all . But, since , and , it must exists some non-invertible element of such that which is a contradiction.
I have not find an atomic Puiseux monoid non 2-MCD, but if this monoid exists, we will be able to find a non-atomic element in and we could conclude that in general the property of being atomic does not ascend from to .
7 replies
Resource 0 Questions
DouDragon 2
N
Oct 11, 2023
by PiGuy3141592
Hi, I'm just reading over the resources. I don't quite understand what a unit of is. Can someone explain?
2 replies
Casework
CharmaineMa07292010 1
N
Sep 13, 2023
by WilsonLiuwz
What I've tried so far:
I've tried to consider a few cases in the set N ( positive integers) and showed that it was true. I then moved on to show that this is true for k integers in each set.
<Describe what you have tried so far here. That way, we can do a better job helping you!>
Where I'm stuck:
I don't know how I can prove that the statement is true for any set S.
<Describe what's confusing you, or what your question is here!>
I've tried to consider a few cases in the set N ( positive integers) and showed that it was true. I then moved on to show that this is true for k integers in each set.
<Describe what you have tried so far here. That way, we can do a better job helping you!>
Where I'm stuck:
I don't know how I can prove that the statement is true for any set S.
<Describe what's confusing you, or what your question is here!>
1 reply
Welcome to CrowdMath 2023!
felixgotti 7
N
Aug 13, 2023
by felixgotti
Hi everyone!
Welcome to CrowdMath 2023!
CrowdMath is a collaborative and inclusive research program welcoming participants from all cultural background and region in the planet. With this in mind, we expect all participants to be respectful with their peers and mentors.
Research Project
Our research project this year will be focused on additive combinatorics and power monoids. We will initially consider certain open questions related the set of all finite subsets consisting of nonnegative integers under the Minkowski addition (see https://en.wikipedia.org/wiki/Minkowski_addition). Following our motivating paper (see https://arxiv.org/abs/1701.09152), we call this the power monoid of . Then keeping Minkowski addition in mind, we will study generalized versions of the power monoid of . Although the algebraic (ideal theoretical) and arithemetic properties of power monoids have been systematically investigated only recently, there has been a significant activity in this research area during the last five years. This last statement will be ratified by the CrowdMath resources we will be posting throughout the next few months.
Problems
The section Problems contains the main open problems pertaining the CrowdMath 2023 project. The problems will become systematically available during the coming months as we progress and learn the necessary background and tools needed to understand and work on them. All of the proposed open problems will be thematically linked to the 2023 CrowdMath research project. Most of the problems in the section Problems will have an associated resource, which can be found in the section Resource.
Resources
The content of each resource is directly connected and relevant to the CrowdMath 2023 project. Indeed, each resource contains the notions, definitions, and tools the participants will be learning as the primary sources to understand and make progress on the coming list of open problems. Resources also comes with practice exercises for participants to train. We highly encourage CrowdMath participants to try to solve all the practice exercises, as this will help to deeper understand the new concepts and to get a better insight of the overall project. Participants can discuss the exercises by clicking either the `View Discussions' or `Start New Topic' button.
Intellectual Property
As CrowdMath is a massive collaborative project, sometimes it is not possible to determine the lines between one participant's work and the rest of the group. In such cases, the results created must be attributed to all CrowdMath contributors.
Welcome to CrowdMath 2023!
CrowdMath is a collaborative and inclusive research program welcoming participants from all cultural background and region in the planet. With this in mind, we expect all participants to be respectful with their peers and mentors.
Research Project
Our research project this year will be focused on additive combinatorics and power monoids. We will initially consider certain open questions related the set of all finite subsets consisting of nonnegative integers under the Minkowski addition (see https://en.wikipedia.org/wiki/Minkowski_addition). Following our motivating paper (see https://arxiv.org/abs/1701.09152), we call this the power monoid of . Then keeping Minkowski addition in mind, we will study generalized versions of the power monoid of . Although the algebraic (ideal theoretical) and arithemetic properties of power monoids have been systematically investigated only recently, there has been a significant activity in this research area during the last five years. This last statement will be ratified by the CrowdMath resources we will be posting throughout the next few months.
Problems
The section Problems contains the main open problems pertaining the CrowdMath 2023 project. The problems will become systematically available during the coming months as we progress and learn the necessary background and tools needed to understand and work on them. All of the proposed open problems will be thematically linked to the 2023 CrowdMath research project. Most of the problems in the section Problems will have an associated resource, which can be found in the section Resource.
Resources
The content of each resource is directly connected and relevant to the CrowdMath 2023 project. Indeed, each resource contains the notions, definitions, and tools the participants will be learning as the primary sources to understand and make progress on the coming list of open problems. Resources also comes with practice exercises for participants to train. We highly encourage CrowdMath participants to try to solve all the practice exercises, as this will help to deeper understand the new concepts and to get a better insight of the overall project. Participants can discuss the exercises by clicking either the `View Discussions' or `Start New Topic' button.
Intellectual Property
As CrowdMath is a massive collaborative project, sometimes it is not possible to determine the lines between one participant's work and the rest of the group. In such cases, the results created must be attributed to all CrowdMath contributors.
7 replies
Exercise 3.2
smileapple 6
N
Jun 15, 2023
by felixgotti
& here's a solution to Exercise 3.2. as usual, pls let me know if there are any errors in my reasoning
Problem: Construct a monoid that is almost atomic but not nearly atomic.
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Solution: Let contain the reciprocals of the powers of ; that is Now let In other words, if , we can write -----
Now we determine Clearly, if it must be in the generating set of ; that is, either or However, the former case is invalid, as It is easy to see that the latter case works. Thus
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Note that the some of the on the right hand side must terminate at some point; for instance, we cannot say that is in ; we can only conclude that there exist an infinitude of elements in that approach Under this logic, it follows that there must exist some such that for all We can thus write for some Thus, if , then we can write It is easy to see that all elements of this form must work as well.
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We claim that is almost atomic. Specifically, for let us define so that which is truly atomic, reusing the same notation from 3.1.
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However, we will also prove that is not nearly atomic. Suppose by contradiction that there exists some such that is truly atomic for all However, setting we have Since is odd and is even, it is clear that cannot be written in the form for some and thus is not truly atomic, a contradiction. Thus our works by construction.
Problem: Construct a monoid that is almost atomic but not nearly atomic.
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Solution: Let contain the reciprocals of the powers of ; that is Now let In other words, if , we can write -----
Now we determine Clearly, if it must be in the generating set of ; that is, either or However, the former case is invalid, as It is easy to see that the latter case works. Thus
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Note that the some of the on the right hand side must terminate at some point; for instance, we cannot say that is in ; we can only conclude that there exist an infinitude of elements in that approach Under this logic, it follows that there must exist some such that for all We can thus write for some Thus, if , then we can write It is easy to see that all elements of this form must work as well.
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We claim that is almost atomic. Specifically, for let us define so that which is truly atomic, reusing the same notation from 3.1.
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However, we will also prove that is not nearly atomic. Suppose by contradiction that there exists some such that is truly atomic for all However, setting we have Since is odd and is even, it is clear that cannot be written in the form for some and thus is not truly atomic, a contradiction. Thus our works by construction.
6 replies
Exercise 3.3
aeemc2 11
N
Jun 12, 2023
by smileapple
Here`s a solution to exercise 3.2. Please let me know if there are any errors.
Define . Now let The set K is an infinite subset of . Then it is countable, and we can construct a bijection f between and where is the set of prime numbers greater than 3. Define the monoid where . Firstly let's prove that . We can write Then any number k such that is not an atom. We can prove that the elements are atoms. Suppose that where a and b are elements of M. Then we can write these elements as sum of generators. Let and with all and different than k. Then we have but in this equality of the left side is -1 and for the right side 0. This contradiction proves the result. M is not atomic. Indeed 1/2 cannon be written as a sum of atoms. Suppose that . If does not divide then of the right side is -1 and for the left side 0, contradiction. Then for all i=1,2,...,n. Then we have that where . If 3 does not divide then of the right side is -1 and for the left side 0. Then but this implies that , a contradiction. Let`s prove that M is nearly atomic. Firstly we have that . We can write an arbitrary element of M as where and are atoms of M. Then we have that . Then M is nearly atomic.
Define . Now let The set K is an infinite subset of . Then it is countable, and we can construct a bijection f between and where is the set of prime numbers greater than 3. Define the monoid where . Firstly let's prove that . We can write Then any number k such that is not an atom. We can prove that the elements are atoms. Suppose that where a and b are elements of M. Then we can write these elements as sum of generators. Let and with all and different than k. Then we have but in this equality of the left side is -1 and for the right side 0. This contradiction proves the result. M is not atomic. Indeed 1/2 cannon be written as a sum of atoms. Suppose that . If does not divide then of the right side is -1 and for the left side 0, contradiction. Then for all i=1,2,...,n. Then we have that where . If 3 does not divide then of the right side is -1 and for the left side 0. Then but this implies that , a contradiction. Let`s prove that M is nearly atomic. Firstly we have that . We can write an arbitrary element of M as where and are atoms of M. Then we have that . Then M is nearly atomic.
11 replies
A result regarding problem 3
aeemc2 0
May 18, 2023
Hello everyone. Here I present some ideas related to Problem 3 that could be helpful. Please let me know if there are any mistakes.
Let M be an atomic (and therefore nearly atomic) MCD monoid. Then as was shown in the Problem 2, is atomic and consequently nearly atomic. Then we have to analyze two cases: M is not an MCD monoid or M is not atomic. I`m working in the first case.
Let M be an atomic monoid that is not a 2-MCD monoid as in exercise 2. Clearly, M is nearly atomic. I`m trying to find a monoid that satisfies these properties and the following: For every set such that exists, there also exists and, consequently they are equal. If such a monoid exists, then mcd is an associative operation, and the following reasoning is valid.
Lemma 1: If , then
Proof: Clearly, if exists, it must be at least y because for all i. Since , then the set does not have a non-invertible common divisor and for all Clearly, that implies that divides all the elements for all i, and does not have a non-invertible common divisor. That proves that
Lemma 2:
Let and . , and . If and Then if ,
Proof: In this proof, we use the fact that gcd (and consequently mcd in this case) is an associative operation.By induction on that is the cardinality of the set . For the result is elemental and it can be deduced from lemma 1. For , suppose that and . Now by lemma 1, it`s clear that . Similarly . Now we know that
Suppose the proposition holds for and consider such that . Then we have . We also have that . See that is the sum of the sets and by induction hypothesis where . Finally applying the same procedure, we get that . Since is an arbitrary number, the lemma holds for the sum of two sets of any cardinality.
Lemma 3: If and with and . If , then
Proof: The proof is very easy by induction on n.
This lemma proves that if an element B of has a decomposition of the form and the mcd of the elements of B does not exist, then for one of the sets does not exist the mcd of its elements. Indeed if the mcd of the elements of each set exists, then by lemma 3 there exists the mcd of the elements of B.
Suppose that we have a monoid M that is not a 2-MCD monoid. Then does not exist for some . Then there is not exists . Then we can construct an element of any cardinality such that there`s not exists the mcd of its elements. Let B be such an element. By lemma 3, if we write B as , then it does not exist the mcd for the elements of one of the set , then this element is not an atom. Then it is impossible to write B as a sum of atoms in .
Let C be an arbitrary element of , , where . Firstly, suppose that does not exist. Let an arbitrary non-invertible element of with cardinality equals 1. And consider Suppose that exist, then it is equal to by the definition of our monoid. Since , let for some , since for all i, we have that for all i. Then e is a common divisor of the elements of C and, by definition does not have a non-invertible common divisor, then a contradiction. Then does not exist. Now suppose exists and let such that does not exist . Then is an element of such that mcd of its elements does not exist. Indeed by lemma one and the associative property if which not exists as we proved above. If it`s enough to take and then is again an element of such that mcd of its elements does not exist. We have proved that for any element of we can find another non-invertible element B such that B+C is not atomic. That proves that is not nearly atomic.
Therefore, nearly atomicity does not ascend from M to
Part b)
Let M be a Puiseux monoid that is not a -MCD monoid and satisfies the above-mentioned property. Then there exists such that does not exist. Consider the non-invertible element and let be an atomic element of , then exists and we have that for all C, and does not exist. Then by the previous part of the exercise is not atomic for all atomic elements C . Then is not almost atomic, and this concludes the proof.
I`m investigating if there exists a monoid that satisfies the properties mentioned, but if such a monoid exists then we can conclude that nearly and almost atomicity do not ascend from M to
Let M be an atomic (and therefore nearly atomic) MCD monoid. Then as was shown in the Problem 2, is atomic and consequently nearly atomic. Then we have to analyze two cases: M is not an MCD monoid or M is not atomic. I`m working in the first case.
Let M be an atomic monoid that is not a 2-MCD monoid as in exercise 2. Clearly, M is nearly atomic. I`m trying to find a monoid that satisfies these properties and the following: For every set such that exists, there also exists and, consequently they are equal. If such a monoid exists, then mcd is an associative operation, and the following reasoning is valid.
Lemma 1: If , then
Proof: Clearly, if exists, it must be at least y because for all i. Since , then the set does not have a non-invertible common divisor and for all Clearly, that implies that divides all the elements for all i, and does not have a non-invertible common divisor. That proves that
Lemma 2:
Let and . , and . If and Then if ,
Proof: In this proof, we use the fact that gcd (and consequently mcd in this case) is an associative operation.By induction on that is the cardinality of the set . For the result is elemental and it can be deduced from lemma 1. For , suppose that and . Now by lemma 1, it`s clear that . Similarly . Now we know that
Suppose the proposition holds for and consider such that . Then we have . We also have that . See that is the sum of the sets and by induction hypothesis where . Finally applying the same procedure, we get that . Since is an arbitrary number, the lemma holds for the sum of two sets of any cardinality.
Lemma 3: If and with and . If , then
Proof: The proof is very easy by induction on n.
This lemma proves that if an element B of has a decomposition of the form and the mcd of the elements of B does not exist, then for one of the sets does not exist the mcd of its elements. Indeed if the mcd of the elements of each set exists, then by lemma 3 there exists the mcd of the elements of B.
Suppose that we have a monoid M that is not a 2-MCD monoid. Then does not exist for some . Then there is not exists . Then we can construct an element of any cardinality such that there`s not exists the mcd of its elements. Let B be such an element. By lemma 3, if we write B as , then it does not exist the mcd for the elements of one of the set , then this element is not an atom. Then it is impossible to write B as a sum of atoms in .
Let C be an arbitrary element of , , where . Firstly, suppose that does not exist. Let an arbitrary non-invertible element of with cardinality equals 1. And consider Suppose that exist, then it is equal to by the definition of our monoid. Since , let for some , since for all i, we have that for all i. Then e is a common divisor of the elements of C and, by definition does not have a non-invertible common divisor, then a contradiction. Then does not exist. Now suppose exists and let such that does not exist . Then is an element of such that mcd of its elements does not exist. Indeed by lemma one and the associative property if which not exists as we proved above. If it`s enough to take and then is again an element of such that mcd of its elements does not exist. We have proved that for any element of we can find another non-invertible element B such that B+C is not atomic. That proves that is not nearly atomic.
Therefore, nearly atomicity does not ascend from M to
Part b)
Let M be a Puiseux monoid that is not a -MCD monoid and satisfies the above-mentioned property. Then there exists such that does not exist. Consider the non-invertible element and let be an atomic element of , then exists and we have that for all C, and does not exist. Then by the previous part of the exercise is not atomic for all atomic elements C . Then is not almost atomic, and this concludes the proof.
I`m investigating if there exists a monoid that satisfies the properties mentioned, but if such a monoid exists then we can conclude that nearly and almost atomicity do not ascend from M to
0 replies
Exercise 3.3 [but its wrong]
smileapple 3
N
May 14, 2023
by smileapple
Hi! here's my thoughts/tentative solution on Exercise 3.3. it feels a little sus somehow, mainly because using the function sounds like cheating, so please let me know if there are any logical errors. thanks!
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Problem: Construct a monoid that is nearly atomic but not atomic.
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Solution: Consider the anomalous addition function defined by if and otherwise. Also, define the parity addition function such that we have and Let be the operation Then, we let , where -----
Using an argument similar to the one in 3.2, we find the atoms of Clearly, they must be in but since it follows that It thus easily follows that
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The above implies that is not atomic: specifically, elements of the form with are not truly atomic, again using the same terms from 3.1.
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However, is nearly atomic: letting and , we have that but implies that the above sum is , which is truly atomic as Otherwise, if , then is or ; since , we must have that is invertible. Thus is nearly atomic but not atomic, so we are done.
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Problem: Construct a monoid that is nearly atomic but not atomic.
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Solution: Consider the anomalous addition function defined by if and otherwise. Also, define the parity addition function such that we have and Let be the operation Then, we let , where -----
Using an argument similar to the one in 3.2, we find the atoms of Clearly, they must be in but since it follows that It thus easily follows that
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The above implies that is not atomic: specifically, elements of the form with are not truly atomic, again using the same terms from 3.1.
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However, is nearly atomic: letting and , we have that but implies that the above sum is , which is truly atomic as Otherwise, if , then is or ; since , we must have that is invertible. Thus is nearly atomic but not atomic, so we are done.
3 replies
A result regarding problem 2
GPZ 7
N
Oct 26, 2023
by felixgotti
Let be an atomic Puiseux monoid. We will split the proof of the problem into the following two cases.
Case 1: is a MCD monoid. Suppose, by way of contradiction, that is not atomic, then there exists some non-atomic element of . This implies that for every factorization of , one of the factors must be a non-atomic element. Consider the set . so it has a minimum, namely . See that because every element of is an atomic element. Now consider one element of . See that . Now, if we take we have that belongs to for all , so is an element of . Now we have that . Clearly is an atomic element of . Let us prove that is also an atomic element of . Let be an arbitrary factorization of . If and , then none of them belongs to and that implies that is an atomic element. So we will assume WLOG that , but this implies that , so , because 0 is the only common divisor of the set . So is an atomic element, implying that is an atomic element, which is a contradiction. Notice that if is a -MCD atomic cancellative monoid, then we have proved that every element in such that is an atomic element of .
Case 2: is not a 2-MCD monoid. In this case, we know that there exists two elements of , namely and , such that does not exist. Let us prove that the element is not an atomic element of . Suppose, by way of contradiction, that , where is an atom of for all . We can tell WLOG that and for all . But, since , and , it must exists some non-invertible element of such that which is a contradiction.
I have not find an atomic Puiseux monoid non 2-MCD, but if this monoid exists, we will be able to find a non-atomic element in and we could conclude that in general the property of being atomic does not ascend from to .
Case 1: is a MCD monoid. Suppose, by way of contradiction, that is not atomic, then there exists some non-atomic element of . This implies that for every factorization of , one of the factors must be a non-atomic element. Consider the set . so it has a minimum, namely . See that because every element of is an atomic element. Now consider one element of . See that . Now, if we take we have that belongs to for all , so is an element of . Now we have that . Clearly is an atomic element of . Let us prove that is also an atomic element of . Let be an arbitrary factorization of . If and , then none of them belongs to and that implies that is an atomic element. So we will assume WLOG that , but this implies that , so , because 0 is the only common divisor of the set . So is an atomic element, implying that is an atomic element, which is a contradiction. Notice that if is a -MCD atomic cancellative monoid, then we have proved that every element in such that is an atomic element of .
Case 2: is not a 2-MCD monoid. In this case, we know that there exists two elements of , namely and , such that does not exist. Let us prove that the element is not an atomic element of . Suppose, by way of contradiction, that , where is an atom of for all . We can tell WLOG that and for all . But, since , and , it must exists some non-invertible element of such that which is a contradiction.
I have not find an atomic Puiseux monoid non 2-MCD, but if this monoid exists, we will be able to find a non-atomic element in and we could conclude that in general the property of being atomic does not ascend from to .
7 replies