Papers
Topics
Authors
Recent
Assistant
AI Research Assistant
Well-researched responses based on relevant abstracts and paper content.
Custom Instructions Pro
Preferences or requirements that you'd like Emergent Mind to consider when generating responses.
Gemini 2.5 Flash
Gemini 2.5 Flash 70 tok/s
Gemini 2.5 Pro 48 tok/s Pro
GPT-5 Medium 27 tok/s Pro
GPT-5 High 24 tok/s Pro
GPT-4o 75 tok/s Pro
Kimi K2 175 tok/s Pro
GPT OSS 120B 447 tok/s Pro
Claude Sonnet 4.5 Pro
2000 character limit reached

Arithmetic properties encoded in undermonoids (2412.11199v1)

Published 15 Dec 2024 in math.AC

Abstract: Let $M$ be a cancellative and commutative monoid. A submonoid $N$ of $M$ is called an undermonoid if the Grothendieck groups of $M$ and $N$ coincide. For a given property $\mathfrak{p}$, we are interested in providing an answer to the following main question: does it suffice to check that all undermonoids of $M$ satisfy $\mathfrak{p}$ to conclude that all submonoids of $M$ satisfy $\mathfrak{p}$? In this paper, we give a positive answer to this question for the property of being atomic, and then we prove that if $M$ is hereditarily atomic (i.e., every submonoid of $M$ is atomic), then $M$ must satisfy the ACCP, proving a recent conjecture posed by Vulakh and the first author. We also give positive answers to our main question for the following well-studied factorization properties: the bounded factorization property, half-factoriality, and length-factoriality. Finally, we determine all the monoids whose submonoids/undermonoids are half-factorial (or length-factorial).

Summary

We haven't generated a summary for this paper yet.

Lightbulb Streamline Icon: https://streamlinehq.com

Continue Learning

We haven't generated follow-up questions for this paper yet.

List To Do Tasks Checklist Streamline Icon: https://streamlinehq.com

Collections

Sign up for free to add this paper to one or more collections.

X Twitter Logo Streamline Icon: https://streamlinehq.com

Tweets

This paper has been mentioned in 1 post and received 0 likes.