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MIT PRIMES/Art of Problem Solving

CROWDMATH 2022: Factorizations in Additive Structures

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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 $M = \mathbb{N}_0[q]$, for rational $q = a/b$:
Lemma 1.
- $M$ is atomic iff $q$ is not the reciprocal of any integer
- If $M$ is atomic, the Betti elements of $M$ are $$\left\{ \frac{\mathsf{n}(q)^{m+1}}{\mathsf{d}(q)^m}\ \Big|\ m \in \mathbb{N}_0\right\}$$- If $M$ is atomic, the set of lengths of any element is an arithmetic progression with difference $a - b$


Lemma 2. Let $q$ be a positive rational, and let $n \geq 2$ such that $\sqrt[n]{q}$ is an irreducible n-th root of $q$. Then the polynomial $x^n - q$ is irreducible in $\mathbb{Q}$.
Proof: This is seen in the textbook Algebra by S. Lang, in page 297.

Corollary. Let $q$ be a positive rational, and let $\alpha \coloneqq \sqrt[n]{q}$ be a positive irreducible n-th root of $q$. Then $1, \alpha, \alpha^2, \dots, \alpha ^{n-1}$ are linearly independent in $\mathbb{Q}$.
Proof: The proof follows from Lemma 2; If these powers of $\alpha$ were not linearly independent, the polynomial $x^n - q$ (whose unique positive real root is $\alpha$) would be reducible in $\mathbb{Q}$.

Claim. Let $q$ be a positive rational that is not the reciprocal of any integer, and let $\alpha \coloneqq \sqrt[n]{q}$ be a positive irreducible n-th root of $q$. Let $f: \mathsf{Z}(\mathbb{N}_0[q])^n \rightarrow \mathsf{Z}(\mathbb{N}_0[\alpha])$ be the map defined by $f(z_1, z_2, \dots, z_n) = z_1 + \alpha z_2 + \alpha^2z_3 + \dots + \alpha^{n-1}z_n$. Then the following statements hold:
$(i)$ $f$ is an isomorphism.
$(ii)$ If $\pi(f(z_1, z_2, \dots, z_n)) = \pi(f(z_1', z_2', \dots, z_n'))$, then $\pi(z_i) = \pi(z_i')$ for all $i \in [1, n ]$. ($\pi$ is the isomorphism mapping a collection of atoms to the element they form.)
Proof. $(i)$ It's easy to check that $f$ 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 $\mathsf{Z}(\mathbb{N}_0[\alpha])$ corresponds to a unique $n$-tuple of factorizations in $\mathsf{Z}(\mathbb{N}_0[q])^n$, showing both surjectivity and injectivity. Since $\mathbb{N}_0[\alpha]$ is generated by $\{\alpha^k \mid k \in \mathbb{N}_0\}$, its set of atoms must be a subset of this generating set. Moreover, as $\alpha^n =  q$, any atom $\alpha^{k'}$ can be written in the form $\alpha^rq^k$, where $k' = kn + r$, $k \geq 0$, and $r \in [0, n-1]$. Thus, any factorization in $z \in \mathsf{Z}(\mathbb{N}_0[\alpha])$ can be written as $$z = \sum_{r = 0}^{n-1}\sum_{k=0}^{\infty}c_{r,k}\alpha^rq^k$$For some coefficients $c_{r,k}\in\mathbb{N}_0$. This can be rewritten further as
$$z=\sum_{r = 0}^{n-1}\alpha^r(\sum_{k=0}^{\infty}c_{r,k}q^k)$$Lemma 1 shows that the set of atoms of $\mathbb{N}_0[q]$ is its generating set. Combining this with the previous equation, it's clear to see that the unique $n$-tuple of factorizations in $\mathsf{Z}(\mathbb{N}_0[q])^n$ that maps to $z$ is $(\sum_{k=0}^{\infty}c_{0,k}q^k, \sum_{k=0}^{\infty}c_{1,k}q^k, \dots, \sum_{k=0}^{\infty}c_{n-1,k}q^k)$. Thus, $f$ is an isomorphism.
$(ii)$ $\pi(f(z_1, z_2, \dots, z_n)) = \pi(f(z_1', z_2', \dots, z_n'))$ implies that $\pi(z_1) + \alpha \pi(z_2) + \dots + \alpha^{n-1}\pi(z_n) = \pi(z_1') + \alpha \pi(z_2') + \dots + \alpha^{n-1}\pi(z_n')$, and the linear independence shown in Corollary 1 shows that $\pi(z_i) = \pi(z_i')$ for all $i \in [1, n]$.

Corollary 1. Let $q$ be a positive rational that is not the reciprocal of any integer, and let $\alpha \coloneqq \sqrt[n]{q}$ be a positive irreducible n-th root of $q$. The set of atoms of $\mathbb{N}_0[\alpha]$ is its generating set, $\{\alpha^k \mid k \in \mathbb{N}_0\}$.
Proof: Any element that is generated by at least two generating elements can not be an atom, so $\mathcal{A}(N_0[\alpha]) \subseteq \{\alpha^k \mid k \in \mathbb{N}_0\}$. On the other hand, by the isomorphism $f$, the factorizations of $\alpha^{k'} = \alpha^rq^k$ (where $k' = kn+r$) are isomorphic to the factorizations of the element $(0, 0, \dots, 0, q^k, 0, \dots, 0)$ in $\mathbb{N}_0[q]^n$, where $q^k$ is at the $r$-th index. As shown in Lemma 1, $q^k$ is an atom of $N_0[q]$, thus $\alpha^{k'}$ must be an atom, hence $\mathcal{A}(N_0[\alpha]) = \{\alpha^k \mid k \in \mathbb{N}_0\}$.

Corollary 2.Let $q$ be a positive rational that is not the reciprocal of any integer, and let $\alpha \coloneqq \sqrt[n]{q}$ be a positive irreducible n-th root of $q$. The set of Betti elements of $\mathbb{N}_0[\alpha]$ is $$\left\{ \frac{\mathsf{n}(q)^{m+1}}{\mathsf{d}(q)^m}\alpha^r\ \Big|\ m \in \mathbb{N}_0\ \text{and}\ r \in [ 0, n-1 ]\right\}.$$Proof: For a fixed $r \in [0, n-1]$, we identify all Betti elements of the form $t\alpha^r$, where $t \in \mathbb{Q}_{>0}$. By the mapping $f$, the factorizations of $t\alpha^r$ are isomorphic to the factorizations of the element $(0, 0, \dots, 0, t, 0, \dots, 0)$ in $\mathbb{N}_0[q]^n$, where $t$ is at the $r$-th index. [Previous result] shows that the only Betti elements of $\mathbb{N}_0[q]$ are $\left\{ \frac{\mathsf{n}(q)^{m+1}}{\mathsf{d}(q)^m}\ \Big|\ m \in \mathbb{N}_0 \right\}$. Thus, by isomorphism, the only Betti elements of this form are $$\left\{ \frac{\mathsf{n}(q)^{m+1}}{\mathsf{d}(q)^m}\alpha^r\ \Big|\ m \in \mathbb{N}_0\right\}$$
Lastly, we prove that all elements of $\mathbb{N}_0[\alpha]$ that involve at least two different (linearly independent) powers of $\alpha$ can not be Betti elements. Let $a = t_0 + t_1\alpha + t_2\alpha^2 + \dots + t_{n-1}\alpha^{n-1}$ be such an element, where $t_i \in \mathbb{Q}_{\geq0}$, and at least two of the $t_i$'s are nonzero. By the mapping $f$, the factorizations of $a$ are isomorphic to the factorizations of $(t_0, t_1, \dots, t_{n-1})$ in $\mathbb{N}_0[q]^n$. Let $z$ and $z'$ be two arbitrary factorizations of $a$, and let $(z_0, z_1, \dots, z_{n-1})$ and $(z_0', z_1', \dots, z_{n-1}')$ be the $n$-tuples of factorizations corresponding to them by $f$. $\pi(z) = \pi(z') = a$, and so by property $(ii)$ in Claim 1, $\pi(z_i) = \pi(z_i')$ for all $i$. As a result, $(z_0, z_1, z_2 \dots, z_{n-1}) \rightarrow (z'_0, z_1, z_2 \dots, z_{n-1}) \rightarrow (z'_0, z'_1, z_2 \dots, z_{n-1}) \rightarrow \dots \rightarrow (z'_0, z'_1, z'_2, \dots, z'_{n_1})$ is a sequence consisting of factorizations of $(t_0, t_1, \dots, t_{n-1})$. 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 $\mathcal{G}((t_0, t_1, \dots, t_{n-1}))$. Therefore, $(t_0, t_1, \dots, t_{n-1})$ is not a Betti element, and by isomorphism, $a$ is not a Betti element.

Corollary 3. Let $q$ be a positive rational that is not the reciprocal of any integer, and let $\alpha \coloneqq \sqrt[n]{q}$ be a positive irreducible n-th root of $q$. For any $a \in \mathbb{N}_0[\alpha]$, $\mathsf{L}(a)$ is an arithmetic progression with difference $|\mathsf{n}(q) - \mathsf{d}(q)|$.
Proof: Let $a = t_0 + t_1\alpha + t_2\alpha^2 + \dots + t_{n-1}\alpha^{n-1}$, where $t_i \in \mathbb{Q}_{\geq0}$ for all $i$. By the mapping $f$, the factorizations of $a$ are isomorphic to the factorizations of $T = (t_0, t_1, \dots, t_{n-1})$ in $\mathbb{N}_0[q]^n$. The length of a certain factorization $(z_0, z_1, \dots, z_{n-1})$ of $T$ is equal to $\sum|z_i|$. Let $d = |\mathsf{n}(q) - \mathsf{d}(q)|$. From Lemma 1, the length of the factorization $z_i$ of $t_i$ can take any value in $\{\min\mathsf{L}(t_i) + dk | k \in [0, \frac{\max\mathsf{L}(t_i) - \min\mathsf{L}(t_i)}{d}]\}$. Moreover, since the terms of the $n$-tuple are independent, $\min\mathsf{L}(T) = \sum\min\mathsf{L}(t_i)$. Thus, $|(z_0, z_1, \dots, z_{n-1})|$ can take any value in

$$\{\min\mathsf{L}(T) + dk | k \in [0, \frac{\max\mathsf{L}(T) - \min\mathsf{L}(T)}{d}]\}$$.

Hence, $\mathsf{L}(T)$ is an arithmetic progression with difference $d$, and by isomorphism, the same applies to $\mathsf{L}(a)$.

Please let me know if anything is unclear!
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gourmet_salad
Jan 2, 2023
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The last problem of CrowdMath 2022 has been posted!
felixgotti   0
Jan 1, 2023
Hi everyone! Check out the tab of problems. The last problem of CrowdMath 2022 (Problems 7), which has several parts, has just been posted. The problem is aimed to better understand the sets of lengths, catenary degrees, and the sets of Betti elements of the monoids we are studying in our research project. There is no resource corresponding to this last problem, but I have stated the new and needed definitions as part of the problem statement. Feel free of course to post any question you may have. Enjoy this last problem, and Happy New Year!!!
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felixgotti
Jan 1, 2023
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