Happy Thanksgiving! Please note that there are no classes November 25th-December 1st.

MIT PRIMES/Art of Problem Solving

CROWDMATH 2023: Arithmetic of Power Monoids

G
Topic
First Poster
Last Poster
A result regarding problem 2
GPZ   7
N Oct 26, 2023 by felixgotti
Let $M$ be an atomic Puiseux monoid. We will split the proof of the problem into the following two cases.

Case 1: $M$ is a MCD monoid. Suppose, by way of contradiction, that $\mathcal{P}_{\text{fin}}(M)$ is not atomic, then there exists some non-atomic element $A$ of $\mathcal{P}_{\text{fin}}(M)$. This implies that for every factorization of $A$, one of the factors must be a non-atomic element. Consider the set $L:=\{|B|, \text{ where } B \text{ is a non-atomic divisor of } A\}$. $L\subset \mathbb{N}$ so it has a minimum, namely $k$. See that $k>1$ because every element of $M$ is an atomic element. Now consider one element $B_k=\{b_1,b_2,\ldots,b_k\}$ of $L$. See that $|B_k|=k$. Now, if we take $d=\text{mcd}(b_1,b_2,\ldots,b_k)$ we have that $b_n-d$ belongs to $M$ for all $n\in[[1,k]]$, so $B_d:=\{b_1-d,b_2-d,\ldots,b_k-d\}$ is an element of $\mathcal{P}_{\text{fin}}(M)$. Now we have that $B_k=B_d+\{d\}$. Clearly $\{d\}$ is an atomic element of $\mathcal{P}_{\text{fin}}(M)$. Let us prove that $B_d$ is also an atomic element of $\mathcal{P}_{\text{fin}}(M)$. Let $C+D$ be an arbitrary factorization of $B_d$. If $|C|<|B_d|$ and $|D|<|B_d|$, then none of them belongs to $L$ and that implies that $B_d$ is an atomic element. So we will assume WLOG that $|C|=|B_d|$, but this implies that $|D|=1$, so $D=\{0\}$, because 0 is the only common divisor of the set $B_d$. So $B_d$ is an atomic element, implying that $B_k$ is an atomic element, which is a contradiction. Notice that if $M$ is a $k$-MCD atomic cancellative monoid, then we have proved that every element $B$ in $\mathcal{P}_{\text{fin}}(M)$ such that $|B|\le k$ is an atomic element of $\mathcal{P}_{\text{fin}}(M)$.

Case 2: $M$ is not a 2-MCD monoid. In this case, we know that there exists two elements of $M$, namely $a$ and $b$, such that $\text{mcd}(a,b)$ does not exist. Let us prove that the element $\{a,b\}$ is not an atomic element of $\mathcal{P}_{\text{fin}}(M)$. Suppose, by way of contradiction, that $\{a,b\}=\sum\limits_{k=1}^{n}A_n$, where $A_n$ is an atom of $\mathcal{P}_{\text{fin}}(M)$ for all $n\in \mathbb{N}$. We can tell WLOG that $|A_1|=2$ and $|A_n|=1$ for all $n\ge2$. But, since $A_1=\{a,b\}-\sum\limits_{k=2}^{n}A_n=\{a,b\}-\{c\}$, and $\{c\}\neq\text{mcd}(a,b)$, it must exists some non-invertible element $C$ of $\mathcal{P}_{\text{fin}}(M)$ such that $C|_{\mathcal{P}_{\text{fin}}(M)}A_1$ 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 $\mathcal{P}_{\text{fin}}(M)$ and we could conclude that in general the property of being atomic does not ascend from $M$ to $\mathcal{P}_{\text{fin}}(M)$.
7 replies
GPZ
Apr 6, 2023
felixgotti
Oct 26, 2023
No more topics!
a