4

$\newcommand\mee{\mathbin{\text{::}}}\newcommand\moo{\mathbin{\text{#}}}$ 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 $\moo$.

Carlos Eugenio Thompson Pinzón
  • 2,780
  • 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 06:09
  • 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 06:13
  • @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
  • @anon, from your second comment, $n=a+b$ is necesary but not sufficient. From your first comment, I have expained that I don't mean pairs of pairs $k^2-k$ but just number of pairs $k$. Was it anything else I didn't understood? – Carlos Eugenio Thompson Pinzón Oct 31 '13 at 21:54
  • You said "Where $n\in A#B$ if $n=a+b$ for some $a\in A$ and some $b\in B$" in your question. I am pointing out that this statement contradicts what you actually want. It is important that you do not contradict yourself when defining something. Second, instead of saying "$n=a_i+b_i=a_j+b_j$ for an even number of pairs $(a_i,b_i),(a_j,b_j)$," what you should say to avoid misinterpretation is simply "$n=a+b$ for an even number of pairs $(a,b)\in A\times B$." It is important that when defining something you do not have ambiguity (i.e. did you mean tuples, pairs, pairs of tuples, etc.). – anon Oct 31 '13 at 23:26
  • Second point granted. On the first point there is a comma and a second and, which should mean that conditions are not finnished. I'll fix the pair parts. – Carlos Eugenio Thompson Pinzón Oct 31 '13 at 23:32
  • A comma cannot excuse a contradiction. You wrote a contradiction. I can understand having difficulties with writing in a foreign language though. – anon Oct 31 '13 at 23:33
  • @anon, I hope it is much clear now, however feel free to eidt and correct my language. It should not have been a contradiction but rather a further condition. – Carlos Eugenio Thompson Pinzón Oct 31 '13 at 23:39
  • 5
    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 01 '13 at 00:42
  • A minor point $A # B = A # C \Rightarrow B = C$ only if $A \neq \emptyset$. In general, what you have is an integral domain (a commutative ring without zero divisors). This embeds naturally in its field of fractions, which is what you're trying to construct in your second attempt. – Magdiragdag Nov 12 '13 at 21:13

1 Answers1

4

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

Marc van Leeuwen
  • 115,048