The Unapologetic Mathematician

Mathematics for the interested outsider

Specifying Morphisms Between Boolean Rings

It will be useful to have other ways of showing that a given function f:\mathcal{S}\to\mathcal{T} between boolean rings is a morphism. The definition, of course, is that f(E\Delta F)=f(E)\Delta f(F) and f(E\cap F)=f(E)\cap f(F), since \Delta and \cap here denote our addition and multiplication, respectively.

It’s also sufficient to show that f(E\setminus F)=f(E)\setminus f(F) and f(E\cup F)=f(E)\cup f(F). Indeed, we can build both \Delta and \cap from \setminus and \cup:

\displaystyle\begin{aligned}E\Delta F&=(E\setminus F)\cup(F\setminus E)\\E\cap F&=E\setminus(E\setminus F)=F\setminus(E\setminus F)\end{aligned}

So if \setminus and \cup are preserved, then so are \Delta and \cap.

A little more subtle is the fact that for surjections we can preserve order in both directions. That is, if E\subseteq F is equivalent to f(E)\subseteq f(F) for a surjection f from one underlying set onto the other, then f is a homomorphism of Boolean rings. We first show that f preserves \cup by using its characterization as a least upper bound: E\subseteq E\cup F, F\subseteq E\cup F, and if E\subseteq G and F\subseteq G then E\cup F\subseteq G.

So, since f preserves order, we know that f(E)\subseteq f(E\cup F), and that f(F)\subseteq f(E\cup F). We can conclude from this that f(E)\cup f(F)\subseteq f(E\cup F), and we are left to short that f(E\cup F)\subseteq f(E)\cup f(F). But f(E)\cup f(F)=f(G)$ for some G\in\mathcal{S}, since f is surjective. Since f(E)\subseteq f(G), we must have E\subseteq G, and similarly F\subseteq G. But then E\cup F\subseteq G, and so f(E\cup F)\subseteq f(G)=f(E)\cup f(F), as we wanted to show.

We thus know that f preserves the operation \cup, and we can similarly show that f preserves \cap, using the dual universal property. In order to build \setminus and \Delta from \cup and \cap, we need complements. But complements satisfy a universal property of their own: E\cap E^c=\emptyset, and if E\cap F=\emptyset then F\subseteq E^c.

First, I want to show that f(\emptyset)=\emptyset by showing it is below all elements of \mathcal{T}. Indeed, by the surjectivity of f, every element of \mathcal{T} is the image of some element E\in\mathcal{S}. Thus we want to show that f(\emptyset)\subseteq f(E). But this is true because \emptyset\subseteq E for all E\in\mathcal{S}, and so f(\emptyset)=\emptyset

Now given a set E and its complement E^c=X\setminus E, we know that E\cap E^c=\emptyset. Since f preserves \cap, we must have f(E)\cap f(E^c)=f(\emptyset)=\emptyset. Let’s say G\in\mathcal{T} is another set so that f(E)\cap G=\emptyset. By the surjectivity of f, we have G=f(F) for some F\in\mathcal{S}. Thus f(E\cap F)=f(E)\cap f(F)=\emptyset=f(\emptyset), and thus E\cap F=\emptyset. This tells us that F\subseteq E^c, and thus G=f(F)\subseteq f(E^c). And so f(E^c)=f(E)^c.

Therefore, since f preserves \cup, \cap, and complements, it preserves \cup and \setminus, and thus is a morphism of Boolean rings.

August 11, 2010 Posted by | Algebra, Ring theory | Leave a comment