# What are the prime elements of the ring of finite natural subsets?

Let consider $\def\NN{\mathbb N}\def\PP{\mathbb P}\def\eqset{\mathrel{:=}}\PP$ the set of all finite subsets of the natural numbers (with zero). $\PP=\{A\subset\NN\mathrel{:}\#A\in\NN\}$.

Let's define an “addition” the symmetric difference:$$\begin{array}{r@{\ }c@{\,}c@{\,}l}+:&\PP\times\PP&\to&\PP\\&(A,B)&\mapsto&A+B\\&&&\eqset\big\{n\in\NN\mathrel:(n\in A\mathrel\wedge n\notin B)\mathrel\vee(n\notin A\mathrel\wedge n\in B)\big\}.\end{array}$$ We can show that this addition is internal, associative, commutative, has an identity element: $\{\}$, and is inversive: $A+A=\{\}$.

Let's define a multiplication as: $$A\times B=\begin{cases} \{\},&B=\{\};\\ \big\{n\in\NN\mathrel:n-m\in A\big\},&B=\{m, and \{m,n\}=\};\\ \big(A\times\{m\}\big)+\big(A\times(B\setminus\{m\})\big),&m\in B. \end{cases}$$ We can check that multiplication is well defined, it is internal, associative, commutative, and has an identity element: $\{0\}$. For simplification, we can notate $AB\eqset A\times B$, and this notation has precedence over addition: $A+BC=A+(B\times C)$.

We can call it a multiplication because it distributes addition: $A(B+C)=AB+AC$.

$\PP$ also has an preorder. Let's say that $A>B$ if $\max(A)>\max(B)$. We can show that $AB>A$ for $B\neq\{\}$ and $B\neq\{0\}$.

Let's define power as repeated multiplication:$$\begin{array}{r@{\ }c@{\,}c@{\,}l}\text{pow}:&\NN\times\PP&\to&\PP\\&(n,A)&\mapsto&A^n\eqset\begin{cases}\{0\},&n=0\\A^k\times A,&n=k+1\end{cases}\end{array}$$

We will say that $P\in\PP$ is irreducible if there are no $A,B\in\PP$, such as $AB=P$, except for $P$ and $\{0\}$. We will say that $P$ is prime if $P$ is irreducible and $P\neq\{0\}$.

So $\{1\}$ is prime. $\{2\}$ is not as $\{1\}^2=\{2\}$. And there is no other unitary set in $\PP$ that is prime: $\{n\}=\{1\}^n$.

$\{0,1\}$ is prime. $\{0,2\}$ is not: $\{0,2\}=\{0,1\}^2$. $\{1,2\}$ is neither prime: $\{1,2\}=\{0,1\}\times\{1\}$. Less obvious, there is no other set with two elements that is prime. $\{0,n\}=\{k\mathrel:k<n\}\times\{0,1\}$.

With three elements: $\{0,1,2\}$, $\{0,1,3\}$, $\{0,2,3\}$, $\{0,1,4\}$, $\{0,3,4\}$ are prime. But $\{0,2,4\}=\{0,1,2\}^2$ is not prime.

So.

1. How can we predict which other prime sets are in $\PP$? Are there rules to determinate which sets are likely to be prime or not?
2. How can we prove that prime decomposition is unique?

Your ring is much better known as the polynomial ring $\mathbb{F}_2[x]$. Indeed, given a finite subset $A\subset\mathbb{N}$, you can associate to $A$ the polynomial $\sum_{n\in A}x^n\in\mathbb{F}_2[x]$, and this is an isomorphism from your ring $\mathbb{P}$ to $\mathbb{F}_2[x]$.
So your ring has unique factorization since it is a polynomial ring in one variable over a field (in particular, it is a PID). The "prime" elements are just irreducible polynomials over $\mathbb{F}_2$. As far as I know there is no particularly simple way to determine whether a polynomial over $\mathbb{F}_2$ is irreducible, but algorithms for factorization of polynomials over finite fields have been studied a lot: see https://en.wikipedia.org/wiki/Factorization_of_polynomials_over_finite_fields for a brief overview.