Ring structure on subsets of the natural numbers Let $$\mathcal{N}=\{\{k_1,\ldots,k_s\}:\ s>0,\ \mbox{and the}\ k_i\ \mbox{are non-negative and pairwise different integers}\}\cup\{\emptyset\}.$$ Note that there is a bijection with the naturals,
$$
\begin{array}{rccc}
B:&\mathcal{N}&\longrightarrow &\mathbb{N}\\
  &\emptyset& \longmapsto & 0\\
  &\{k_1,\ldots,k_s\}& \longmapsto & 2^{k_1}+\cdots+2^{k_s}
\end{array}.
$$
For $K,L\in\mathcal{N}$ define $$K\oplus L=(K\cup L)\setminus(K\cap L),$$ $$K\otimes L=\bigoplus_{k\in K,\ l\in L}\{k+l\}.$$
The definition of the product doesn't make sense if either $K$ or $L$ is empty. In that case take the product to be $\emptyset$.
Note that the sum I defined is just the Nim sum (justifying the CGT tag) but the product is not the usual Nim product. In particular I'm using the associativity of $\oplus$ to define $\otimes$.
Here are my questions:
1) If I'm not mistaken $(\mathcal{N},\oplus,\otimes)$ is a commutative ring with unit. Can anyone confirm this? In principle the distributive property is the only tricky one.
2) If we do have a ring indeed, what is it known about it?

Your ring is isomorphic to $\Bbb F_2[X]$.
For $k \ge 0$, define $1_k : \mathcal N \to \Bbb F_2$ by  $1_k(A) = 1$ if $k\in A$, and $0$ otherwise.
Then define $f : \mathcal N \to \Bbb F_2[X]$ by $f(A) = \sum_{k \ge 0} 1_k(A) X^k$.
You can check that $f(A)+f(B) = \sum_{k \ge 0} (1_k(A)+1_k(B)) X^k = \sum_{k \ge 0} l_k(A \oplus B) X^k = f(A \oplus B)$.
And $f(A)f(B) = (\sum_{k \ge 0} 1_k(A)x^k)(\sum_{l \ge 0} 1_l(A)X^l) = \sum_{k,l \ge 0} 1_k(A)1_l(B)X^{k+l} = \sum_{k,l \ge 0} 1_k(A)1_k(B)f(\{k+l\}) = \sum_{k \in A, l \in B} f(\{k+l\}) = f(\bigoplus_{k \in A, l \in B} \{k+l\}) = f(A \otimes B)$
