There is no natural polynomial map (other than $1$) that can eventually leave the $k$-semiprimes behind.

Question

Let $F(S)$ be the free commutative monoid on countably many symbols $S$. Then it's obvious that $F(S) = \{1\} \uplus S \uplus S^2 \uplus \dots$

One can take $S = $ the prime numbers in $\Bbb{N}$ in which case $F(S) = \Bbb{N}$ itself.

It's also obvious that $S^i S^j$ done elementwise is equal to $S^{i +j}$ for all $i, j \in \Bbb{N} \cup \{0\}$ if we let $S^0 = \{1\}$.

Suppose that $R = F(S) \cup \{0\}$ has the additional structure of a semiring in which $+$ is compatible with the free monoid $\cdot$ in a distributive way as in $\Bbb{N} \cup \{0\}$.

Let $f : R \to R$ be any semiring polynomial such as $f(n) = n + 1$.

Then consider the set $X_i$ of any such $f$ such that $f$ is eventually free of smaller sets $S^k, k=1..i$. That is to say that $f \in X_i \iff $ there exists $n \in \Bbb{N}$ such that $m \geq n \implies f(m) \notin S^k, k=1..i$.

We do not know if $f(n) = n(n+2)$ is in the set $X_2$. If it is in $X_2$, then clearly twin primes is false since that would mean that $n(n+2) \notin S^2$.

We do know, however, that $1(n) = 1$ is in each $X_i$ since $1 \notin $ any $S^i$ automatically. Similarly we know that if $f \in X_i$ and $i \geq 2$ that $f \in X_{i-1}$ as anything eventually outside of a larger set is eventually outside any smaller set. Thus we have:

$$ X_1 \supset X_2 \supset \dots $$

What must be at the end of this chain? It must be the trivial monoid $X_{\infty} = \{1(n)\}$. This is because there are no numbers outside of all numbers that are not $1$.

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Each $X_i$ forms a monoid. Let $f(n) \in X_i, g(n) \in X_j$. Then there exists $n \in \Bbb{N}$ such that $m \geq n \implies f(m) \in \{1\} \cup S^{i+1} \cup S^{i+2} \cup \dots$ and similarly for $n'$ and $g$. Then by nature of polynomial maps, $f(n) \neq 1$ eventually. That means that $f(n) \in S^{i+1} \cup S^{i+2} \cup \dots$ eventually and similarly goes for $g$. Together this means that $f(n) g(n) \in S^{i + j + 2} \cup S^{i + j + 3} \cup S^{i + j + 4 } \dots \subset \cup S^{i + j + 1} \cup S^{i + j + 2} \cup \dots$

Clearly this means that $fg \in X_{i+j} \subset X_i, X_j$. Thus, all $X_i$ are monoids.

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But since $X_i \ni 1$ we have that $X_i = X_i X_i \subset X_{2i}$. $X_{1} \subset X_2$ when we know $X_2 \subset X_1$ already, thus $X_1 = X_2$. Similarly $X_2 \subset X_2 X_1 \subset X_3$ and so on... Therefore all these monoids equal $\{1(n)\}$ the trivial monoid!

Therefore, the twin prime conjecture is true by construction.

$f(n) = n(n+2)$ cannot eventually leave behind $S^2 = $ the products of two primes, because the only polynomial natural map that does that is $1(n)$!

Answer 1

The problem is here:

Then by nature of polynomial maps, $f(n) \neq 1$ eventually.

This isn't true for $1(n)=1$.

Answer 2

The flaw is. Let $k \in \Bbb{N}$ be greater than $1$.

$f(n) = n^k$ eventually leaves all $S^1 \cup \dots \cup S^{k-1}$. Thus we must admit powers of $n$ into $X_i$. Surely, this invalidates the proof for now.