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.


  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?

1 Answer 1


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.


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