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$\newcommand\mee{\mathbin{\text{::}}}\newcommand\moo{\mathbin{\text{#}}}$ 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 $n\notin A\moo B$.

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).


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 $\moo$.

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Can you explain what $A\# B$ means in more detail? As I understand it, the number of pairs $(a,b),(a',b')\in A\times B$ with $n=a+b=a'+b'$ is just $k^2-k$ where $k$ is the number of pairs $(a,b)\in A\times B$ with $n=a+b$, and $k^2-k$ is always even, so $A\# B$ is empty! Or do you mean undordered pairs of tuples $(a,b),(a',b')$? Then it'd be $\frac{k^2-k}{2}$, which you might as well just directly impose as a condition on $k$'s residue mod $4$ instead of speak about unordered pairs of tuples. –  anon Oct 31 '13 at 6:09
    
Secondly, you say outright that $n\in A\# B$ if $n=a+b$ for some $(a,b)\in A\# B$. But if this is so, then how can you further stipulate conditions for $n\not\in A\#B$ if $n\in A\# B$ already! –  anon Oct 31 '13 at 6:13
    
@anon: $\{1,5,6,8\}\moo\{2,3,4,5\}=\{3,4,5,6,7,10,12,13\}$ as $3=1+2$, $4=1+3$, $5=1+4$ $6=1+5$, $7=5+2$, $8=5+3=6+2$ (two pairs: canceled), $9=5+4=6+3$, $10=5+5=6+4=8+2$ (three pairs: not-canceled), $11=6+5=8+3$, $12=8+4$, $13=8+5$. If there is one pair $a\in A,b\in B$ such as $a+b=n$ then $n\in A\moo B$. If there are two: $n\notin A\moo B$, if there are tree in, four out, five in, etc. I'm not counting pairs of pairs $k^2-k$ nor $\frac{k^2-k}2$ but just pairs: $k$. –  Carlos Eugenio Thompson Pinzón Oct 31 '13 at 17:25
    
Clearly you did not understand my comments. –  anon Oct 31 '13 at 21:07
4  
It seems to me that $A\moo B$ is simply the set of $n$ such that $n=a+b$ for an odd number of $(a, b)$ in $A\times B$. –  Jack M Nov 1 '13 at 0:42
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up vote 2 down vote accepted
+50

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 $(\Z/2\Z)(X)$.

share|improve this answer
    
I'm a little unfamiliar with the notation $(\Z/2\Z)[X]$, although if I understand the $[X]$ bit right I think you're correct. –  Carlos Eugenio Thompson Pinzón Nov 5 '13 at 22:36
    
That is the ring of polynomials over the field with two elements. –  Marc van Leeuwen Nov 6 '13 at 6:14
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