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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 .
This post has been edited 3 times. Last edited by GPZ, Apr 6, 2023, 4:56 AM