It will be useful to have other ways of showing that a given function 
between boolean rings is a morphism. The definition, of course, is that 
and 
, since 
and 
here denote our addition and multiplication, respectively.
It’s also sufficient to show that 
and 
. Indeed, we can build both and from 
and 
:
So if and are preserved, then so are and .
A little more subtle is the fact that for surjections we can preserve order in both directions. That is, if 
is equivalent to 
for a surjection 
from one underlying set onto the other, then is a homomorphism of Boolean rings. We first show that preserves by using its characterization as a least upper bound: 
, 
, and if 
and 
then 
.
So, since preserves order, we know that 
, and that 
. We can conclude from this that 
, and we are left to short that 
. But f(E)\cup f(F)=f(G)$ for some 
, since is surjective. Since 
, we must have , and similarly . But then , and so 
, as we wanted to show.
We thus know that preserves the operation , and we can similarly show that preserves , using the dual universal property. In order to build and from and , we need complements. But complements satisfy a universal property of their own: 
, and if 
then 
.
First, I want to show that 
by showing it is below all elements of 
. Indeed, by the surjectivity of , every element of is the image of some element 
. Thus we want to show that 
. But this is true because 
for all , and so
Now given a set 
and its complement 
, we know that . Since preserves , we must have 
. Let’s say 
is another set so that 
. By the surjectivity of , we have 
for some 
. Thus 
, and thus . This tells us that , and thus 
. And so 
.
Therefore, since preserves , , and complements, it preserves and , and thus is a morphism of Boolean rings.
August 11, 2010
Posted by John Armstrong |
Algebra, Ring theory |
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The Unapologetic Mathematician 