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CROWDMATH 2024: Where Does the Goldbach Conjecture Hold?

Where Does the Goldbach Conjecture Hold? Polymath project forum
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Where Does the Goldbach Conjecture Hold? Polymath project forum
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Problem 2
aeemc2   2
N Nov 9, 2024 by aeemc2
This is my solution of Problem 2, part (a). Please let me know if there are any mistakes.

It was proven in Theorem 3.6 of the paper "A Golbach conjecture for Laurent Series Semidomain" that if $S$ is an additively reduced and additively atomic semidomain, the following conditions are equivalent

(1) $\mathcal{A}(S)=S^{\times}.$

(2) every $f \in S[x^{\pm 1}]$ with |supp$(f)|>1$ can be expressed as the sum of at most 2 irreducibles.

Consider the additive monoid $M = \bigg\langle \bigg(\frac{3}{4}\bigg)^n \, | \, n \in \mathbb{N}_0 \bigg\rangle$. It is well know that $M$ is atomic and $\mathcal{A}(M)= \bigg\{\bigg(\frac{3}{4}\bigg)^n \, | \, n \in \mathbb{N}_0\bigg\} $. Since $M$ is also closed with respect to the usual multiplication of rational numbers, we know that $M$ is a semidomain. Now consider the multiplicative monoid $N= \bigg\langle \frac{3}{4} \bigg\rangle$. It is clear that $N$ is a multiplicative subset of $M$, and then we can consider the localization $N^{-1}M[x]$.

Claim 1: We see that $N^{-1}M= \bigg\langle \bigg(\frac{3}{4}\bigg)^n \, | \, n \in \mathbb{Z} \bigg\rangle$.

Proof: Let us consider an element $r\in N^{-1}M$. This means that there are $m \in M$ and $n \in N$ such that $r=\frac{m}{n}$. Take \[m=\sum_{i=1}^{k}a_i\bigg(\frac{3}{4}\bigg)^{c_i},\]for some $a_1,...,a_k, c_1,..., c_k \in \mathbb{N}_0$, and we can suppose that $n = \bigg(\frac{3}{4} \bigg)^t$ for some $t \in \mathbb{N}_0$. Note that then \[r=\sum_{i=1}^{k}a_i\bigg(\frac{3}{4}\bigg)^{c_i-t}.\]
Since $c_i-t \in \mathbb{Z}$ for every $i \in [[1,k]]$, we can conclude that $r \in \bigg\langle \bigg(\frac{3}{4}\bigg)^n \, | \, n \in \mathbb{Z} \bigg\rangle$.

Suppose now that $r\in \bigg\langle \bigg(\frac{3}{4}\bigg)^n \, | \, n \in \mathbb{Z} \bigg\rangle$. We can write then $r=\sum_{i=1}^{k}a_i\bigg(\frac{3}{4}\bigg)^{z_i}$ for some $a_1,...,a_k \in \mathbb{N}_0$, and some $z_1,...,z_k \in \mathbb{Z}$. For each $z_i$ we can find some natural numbers $z'_i$ and $z''_i$ such that $z_i=z'_i-z''_i$. Then we can write \[r=\sum_{i=1}^{k}a_i \frac{\bigg(\frac{3}{4}\bigg)^{z'_i}}{\bigg(\frac{3}{4}\bigg)^{z''_i}},\]which is clearly an element of $N^{-1}M$.
Let us denote $S=N^{-1}M$. In the paper "Puiseux monoids and transfer isomorphism", Dr. Felix Gotti proved in Proposition 3.5 that if $r\in \mathbb{Q}_{>0}$, then the additive monoid $M=\langle r^n  |  n \in \mathbb{Z} \rangle$ is atomic with $\mathcal{A}(M)=\{r^n | n \in \mathbb{Z}\} $. Then, by the theorem mentioned above, we know that the semidomain $S[x]$ satisfies Goldbach conjecture. However $M[x]$ clearly does not satisfy it. Indeed, for every $n$ the polynomial $f=\sum_{i=0}^{n}\frac{3}{4}x^{k_i}$ cannot be expressed as the sum of $k$ or fewer irreducibles.
2 replies
aeemc2
Nov 3, 2024
aeemc2
Nov 9, 2024
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