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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
exercise 1.3
aeemc2   2
N Nov 3, 2024 by aeemc2
One of the irreducible criteria is well-known for polynomials with integers coefficients and also holds naturally for polynomials of $\mathbb{N}_0[x]$ is Eisenstein's irreducible criterion that states
Criterion 1 (Eisenstein's irreducible criterion).
Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 \in \mathbb{N}_0[x] $. If there exists a prime number $p$ satisfying the following three conditions:

- $p$ divides each $a_i$ for $0 \leq i < n$.
-$p$ does not divide $a_n$, and
-$p^2$ does not divide $a_0$.

Then the polynomial $p(x)$ is irreducible.

We have another criterion, which is even easier to check.

Criterion 2: Let $p(x) \in  \mathbb{N}_0[x]$. Suppose that $p(0) \neq 0$ and $p(1)$ is a prime number, then $p(x)$ is irreducible.

Proof
Indeed, suppose that $p(x)=q(x)r(x)$ for some $q(x), r(x) \in    \mathbb{N}_0[x]$. Note that $p(1)=q(1)r(1)$, and since $p(1)$ is prime, we know that either $q(1)=1$ or $r(1)=1$. WLOG, assume that $r(1)=1$, then $r(x)=x^n$ for some $n \in \mathbb{N}_0$. Note that if $n > 0$, then $r(0)=0$, and then $p(0)=0$, which contradicts the hypothesis. Therefore $r(x)=1$, and this implies that $p(x)$ is irreducible.

2 replies
aeemc2
Nov 1, 2024
aeemc2
Nov 3, 2024
Exercise 0.2
aeemc2   4
N Oct 1, 2024 by felixgotti
First, suppose that $n=1$. Note that we can find a polynomial of $\mathbb{N}_0[x]$ that is not irreducible. We can consider, for example, $f(x)=x^3+5x=x(x^2+5)$. Fix $n \in \mathbb{N}_{>1}$ and consider the polynomial $p(x)=x^{n}+1$. Suppose towards a contradiction that $p(x)=\sum_{i=1}^n a_i(x)$ with $a_i(x) \in \mathcal{A}(\mathbb{N}_0[x])$. Note that we have at least one $a_i(x)$ such that $ord(a_i(x)) \neq 0$. Indeed, if $ord(a_i(x))=0$ for every $i \in [[1,n]]$, then the constant coefficient of $p(x)$ is at least $n$. However if $ord(a_i(x) \neq 0$, then we have that $x \mid_{\mathbb{N}_0[x]} a_i(x)$, which is a contradiction with the irreducibility of $a_i(x)$. This proves the result.
4 replies
aeemc2
Sep 7, 2024
felixgotti
Oct 1, 2024
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