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MIT PRIMES/Art of Problem Solving

CROWDMATH 2022: Factorizations in Additive Structures

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julmath   1
N Jan 2, 2023 by felixgotti
Hello! Here I share an example where I am able to find the catenary degree of $\mathbb{N}_0[\alpha]$.
Suppose that the minimal pair of $\alpha>1$ is $(p(x),n)$, where $p(1)<n$ and $p$ is not monic. It is not hard to see that $\mathbb{N}0[\alpha]$ is atomic and $\mathcal{A}(\mathbb{N}0[\alpha])= \{\alpha^n: n \in \mathbb{N}0\}$. Let $x \in \mathbb{N}0[\alpha]$, and $z \in \mathsf{Z}(x)$, $z=\sum_{i=0}^l a_i\alpha^i$. If $a_j\geq n$ for some $j \in [ 0,l] $ then there exists $z_1 \in \mathsf{Z}(x)$ given by
$$z_1=\sum_{i=0}^{j-1}a_i\alpha^i+(a_j-n)\alpha^j+p(\alpha)\alpha^j+\sum_{i=j+1}^la_i\alpha^i $$and $\vert z_1\vert < \vert z \vert$. Also $d(z,z_1)=n$. Following this procedure we can create a finite sequence $z=z_0, z_1,z_2,...,z_k$ such that $d(z_i,z_{i+1})=n$ and $z_k=\sum_{i=0}^mb_i\alpha^i$ satisfies that $b_i<n$ for all $i \in [ 0,m ]$.
Now, assume that $z', z'' \in \mathsf{Z}(x)$, $z'\neq z''$ such that $z'=\sum_{i=0}^sc_i\alpha^i$, $z''=\sum_{i=0}^{s}d_i\alpha^i$ and $c_i,d_i\in [ 0,n-1 ] $ for all $i \in [b 0,s]$. Then
$$\sum_{i=0}^s(c_i-d_i)x^i=[p(x)-n]r(x)$$
where $r\in \mathbb{Z}[x]$. It is clear that $\vert c_i-d_i \vert <n $ for all $i \in [ 0,s]$, but the coefficient of the monomial of least degree in the right side of (1) is a multiple of $n$. This is a contradiction. Then there exists a unique factorization $z^0 \in \mathsf{Z}(x)$ such that every atom appears in $z^0$ less than $n$ times. Also note that $z^0$ is the factorization of minimal length of $x$. We have proved that for every $x \in \mathbb{N}0[\alpha]$ and $z\in \mathsf{Z}(x)$ there exists a n-chain of factorizations that connects $z$ and the unique factorization of minimal length of $x$. It follows that $c(\mathbb{N}0[\alpha])\leq n$. Following the same reasoning as before, we can prove that there is no $z\in \mathsf{Z}(p(\alpha))$ such that $d(z,n\cdot 1)<n$, so $c(p(\alpha))=n$. Then $c(\mathbb{N}_0[\alpha])=n$.
1 reply
julmath
Dec 2, 2022
felixgotti
Jan 2, 2023
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