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About the ring: $R$ is a commutative boolean ring with $1$ in which for each $x \in R$, $x + x = 0$. There is a partial order on elements of $R$ written $x \leqslant y \iff xy = y$. The ordering respects multiplication and addition, i.e. $\forall x,y,z \in R$, $x \leq y \implies z x \leq z y$ and $z + x \leq z + y$.

I am modelling my ring after Symmetric Difference Ring on the Power Set.

Knowing that, clearly I want to end up with $x + y = \overline{x}y + x \overline{y}$. Where $(\overline{x}y)(x\overline{y}) = 0$.

There is notion of simple element in the ring. These turn out to be integer intervals $(i,j)$ in my setting where $0 \leqslant i \lt j \leqslant N$ for some fixed $N \in \Bbb{N}$. Any result that leads to $j \leqslant i$ is identified with $0 \in R$. The other elements can be represented as sets $x \equiv \{x^t, x_1, \dots, x_n\}$ where $x_i \leqslant x^t$ for each $i=1,\ldots,n$, and $x^t$ If you have an element in which the $x_i$ are not all contained in some larger element $x^t$ then we say $x^t = 0$. We always take $x^t$ to be a simple element. Thus we include $0$ in the simple elements, along with basic intervals. The $t$ in $x^t$ stands for "top-level". Since $x^2 = x$ for all $x \in R$, we don't mind occupying the superscript position.

The representation described above has additional constraints, which make it into a unique representation of any element. In our computations we ensure that $x_i x_j = 0$ for all $i \neq j; i,j = 1,\ldots,n$. Also, $n$ is the largest such $n$. That makes the representation unique. Essentially the $x_i^t$ are all of the largest simple elements contained in $x^t$. Any time $0 \in x$ we can remove it and the two versions are identified. Thus $0$ itself is represented as $\{\}$.

You may be confused now, but just note that the representation forms a tree (it's recursive).

A simple element $x$ is largest in the the simple element $y$ when $x \leqslant y$ and whenever $x \leqslant z \leqslant y$ either $z = x$ or $z = y$.

I have defined $xy := \{x^t y^t, x_iy_j : i=1,\ldots,n, j=1,\ldots,m\}$. If you instead take the elementwise product of two representations, although you might be able to sort it out, it becomes very messy. So I've essentially taken that elementwise product and removed some of the mess-making elements. This is okay, since all we're trying to do is achieve the definition of a ring. As long as every object we're interested is an element in the ring, we don't really care what $\cdot$ is precisely. This definition of $\cdot$ applies recursively until you reach two simple elements $x= (i,j), y = (a,b)$ in which case $xy := (\max(i,a), \min(j,b))$. Note, that our definition respects the unique representation. I.e. check that $(x_i y_j)(x_a y_b) = 0$ for unequal elements and that all of the $x_iy_j$ are contained in $x^ty^t$.


Question

I'm wondering, is there an elegant discription of complement such as (something like) "The complement of $x \in R$ is the unique unary operator that sends largest elements in $x$ to smallest elements and vise versa". Also, how would we write it down formally w.r.t. our recursive representation of an element?

1 Answers1

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I don't think I understand the construction of your ring, but it is well known that every boolean ring can be turned into a boolean algebra.

To (hopefully) answer your question, we can set $$\overline{x} = x+1$$

Then $$\overline{x}y + x\overline{y} = (x+1)y + x(y+1) = xy + y + xy + x = x+y$$ (since $xy + xy = 0$ and $y+x = x+y$)


I hope this helps ^_^

HallaSurvivor
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  • I've actually tried that before, and so maybe I'll go back to it. Problem is, how would you compute $x + 1$ if you don't yet have a definition for $x + y$ in general? – Daniel Donnelly Oct 18 '19 at 16:56
  • Maybe $1 + (i,j) := {(0,i), (j, N)}$ but then we may need a "glue union" for more complicated elements. – Daniel Donnelly Oct 18 '19 at 16:58
  • Can you tell me what parts you're not understanding? That will greatly guide me in writing a paper. – Daniel Donnelly Oct 18 '19 at 17:00
  • I guess what I need to know are what additional properties "adding $1$" or $x+1$ must have. For example it has to be true that $x + 1 + 1 = x$. If I can do that, then I can go ahead and define a recursive function, which may not really have an elegant formula on paper. – Daniel Donnelly Oct 18 '19 at 17:27
  • I don't understand any of the construction, unfortunately... Perhaps if you have an example? Also, if $x+y$ is undefined, how do you know that $R$ is a commutative boolean ring? – HallaSurvivor Oct 18 '19 at 19:07