Timeline for Is there some example that nicely extends the multiplication of natural numbers?
Current License: CC BY-SA 4.0
10 events
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| Feb 19, 2022 at 11:13 | comment | added | Salvo Tringali | Fair enough. So it's kind of natural, I think, to (re)start with the following subquestion: Let $f: \mathbb N \to S$ be a unit-preserving semiring embedding s.t. $S$ is a commutative semidomain and every non-zero non-unit of $S$ factors into a product of primes (which, by my other post, is equivalent to saying that every non-zero non-unit has an essentially unique factorization into atoms). Is it possible for the set of primes of $S$ to be "additively smoother" than the set of primes of $\mathbb N$? And more importantly, what does it mean to be "additively smoother"? | |
| Feb 19, 2022 at 10:54 | comment | added | Zelox | Yes, it is true. The algebraic system $S$ is still a free ab monoid. But we have additive structure now. So we are trying to find a regularity of atoms using the addition operation, that is something cannot be done in the setting of multiplicative monoid. | |
| Feb 19, 2022 at 10:50 | comment | added | Salvo Tringali | @Zelox We are back to my previous post because some of the conclusions I drew there for a monoid embedding $(\mathbb N^+, \cdot) \to H$ are independent from the fact that $H$ is the multiplicative monoid of the non-zero elements of a (commutative) semidomain. (By the way, $\alpha^2$ will never be an atom in $S_\alpha$ and hence the set of atoms of $S_\alpha$ will never be the closed interval $[\alpha, \alpha^2]$.) | |
| Feb 19, 2022 at 10:44 | comment | added | Zelox | I admit that my usage of "any" is wrong. I should use "almost all": for those number $x = abcd...$ with any two of $a,b,c,d,...$ locates at the middle of an interval of atoms. | |
| Feb 19, 2022 at 10:42 | comment | added | Zelox | I'm kind of confused. Your previous post is talking about monoid, but now we are talking about semiring. So why we are back to the previous point? | |
| Feb 19, 2022 at 10:34 | comment | added | Salvo Tringali | @Zelox If $f \colon \mathbb N \to S$ is a unit-preserving semiring embedding such that $S$ is a commutative semidomain and every non-zero non-unit element of $S$ has an essentially unique atomic factorization, then we are back to a point I've already made in my other post mathoverflow.net/a/415988/16537: $S \setminus \{0_S\}$ is basically a free abelian monoid under multiplication. Also, there is something wrong in your argument (or in the way I'm interpreting your "any"): If $\alpha = \sqrt{2}$, then $2$ has an essentially unique atomic factorization in $S_\alpha$. | |
| Feb 19, 2022 at 10:06 | comment | added | Zelox | Extend to the semiring $\mathbb N \cup \mathbb R_{\geq \alpha}$ is certainly one impressive way to extend $\mathbb N^+$. The set of atoms are the interval $[\alpha, \alpha^2]$. However, as long as the set of atoms has an interval as its subset, the factorization of any element cannot be unique: let $abcd$ be a factorization of some element, then $((1+x)\cdot a)(\frac{1}{1+x} \cdot b)cd$ is also a factorization for a sufficient small $x>0$. The simplicity of atoms and the uniqueness of factorization cannot be satisfied simultaneously, so to speak. | |
| Feb 17, 2022 at 13:10 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
shown that S_\alpha is not, in general, FF
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| Feb 17, 2022 at 10:42 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
added 9 characters in body
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| Feb 17, 2022 at 10:34 | history | answered | Salvo Tringali | CC BY-SA 4.0 |