Extending algebra into field

Question

$\newcommand\mee{\mathbin{\text{::}}}\newcommand\moo{\mathbin{\text{#}}}$ Let $\mathcal U$ be the collection of all finite subsets of $\mathbb N$.

Let $\mee$ be a binary operation defined as: $$\begin{matrix} \mee:&\mathcal U\times\mathcal U&\to&\mathcal U \\ &(A,B)&\mapsto&A\mee B&\mathrel{:=}\{n\in\mathbb N: n\in A\cup B,\wedge,n\notin A\cap B\} \end{matrix}$$ (basically $\mee$ is the symetric difference and we know it forms a commutative group in $\mathcal U$ with module $\emptyset$)

Let $\moo$ be a binary operation defined as: $$\begin{matrix} \moo:&\mathcal U\times\mathcal U&\to&\mathcal U \\ &(A,B)&\mapsto&A\moo B \end{matrix}$$ Where $n\in A\moo B$ granting two conditions:

1. $n=a+b$ for some $a\in A$ and some $b\in B$ (necesarily but not sufficiently), and
2. if $n=a_i+b_i=a_j+b_j$ for an even number of pairs $(a,b)\in A\times B$ then $n\notin A\moo B$.

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Some trivial properties that should not be proven: ($\forall A,B,C\in\mathcal U$) \begin{align} A\moo B &= B\moo A \\ A\moo\emptyset &= \emptyset \\ A\moo\{0\} &= A \\ \end{align} A little less trivial: \begin{align} (A\moo B)\moo C &= A\moo(B\moo C) \\ A\moo B=A\moo C &\Rightarrow B=C \\ A\moo(B\mee C) &= (A\moo B)\mee(A\moo C) \end{align}

This means that $\moo$ is a commutative product that distributes the addition $\mee\,$. However $\moo$ does not have inverse (except for $\{0\}$ itself).

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The question Is there a natural way to construct a set $\mathcal V$ with structures $\oplus$ and $\otimes$, with a morphism $\langle\mathcal U,\mee,\moo\rangle\to\langle\mathcal V,\oplus,\otimes\rangle$ (at the point that we can identify $\mee\equiv\oplus,\moo\equiv\otimes,\mathcal U\subset\mathcal V$), but where $\otimes$ has inverse in $\mathcal V$?

Attempts

1. Defining $\mathcal V$ as a collection of subset of $\mathbb Z$, defining $\oplus$ and $\otimes$ similarly as $\mee$ and $\moo$, and having $A\in\mathcal U\mapsto B\in\mathcal V\iff A=B$ (identifying $\mathbb N$ as a subset of $\mathbb Z$).

   The main problem is that I seem to need sets with infinite negatives to define some inverses, breaking the symmetry of finite positives of $\mathcal U$.

2. Defining $\mathcal V$ as equivalent classes from $\mathcal U\times(\mathcal U\setminus\{\emptyset\})$, where $\langle(A,B)\rangle=\langle(C,D)\rangle\iff A\moo D=B\moo C$.

   We define $A\in\mathcal U\mapsto\langle(A,\{0\})\rangle$, and we define $\langle(A,B)\rangle\otimes\langle(C,D)\rangle=\langle(A\moo C,B\moo D)\rangle$ and $\langle(A,B)\rangle\oplus\langle(C,D)\rangle=\langle(A\moo D\mee B\moo C,B\moo D)\rangle$.

   We should be able to show that these operations are well defined and that are equivalent to $\mee$ and $\moo$.

Answer 1

As far as I can tell, the map $\def\U{\mathcal U}\def\Z{\Bbb Z}\U\to(\Z/2\Z)[X]$ sending $A\mapsto\sum_{a\in A}X^a$ is an isomorphism of rings. The smallest field into which $(\Z/2\Z)[X]$ embeds it its field of fractions $(\Z/2\Z)(X)$.