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Finite Boolean Ring

Posted: Sat Jun 25, 2016 6:59 am
by Tsakanikas Nickos
Here is a really difficult exercise!

Let \( \displaystyle R \) be a finite boolean ring. Show that there exists a set \( \displaystyle X \) and a ring isomorphism \( \displaystyle R \cong \mathcal{P}(X) \).


HINT
As \( \displaystyle X \) consider the set \( \displaystyle Hom\left( R, \mathbb{Z}_{2} \right) \) of ring homomorphisms \( \displaystyle R \rightarrow \mathbb{Z}_{2}\).

Re: Finite Boolean Ring

Posted: Tue Aug 02, 2016 3:48 pm
by Papapetros Vaggelis
Hi Nickos.

Firstly, if \(\displaystyle{X}\) is a non-empty set, then, the power set \(\displaystyle{\mathbb{P}(X)}\)

is a \(\displaystyle{\rm{Boolean}}\) ring with addition

\(\displaystyle{A+B:=(A-B)\cup(B-A)\,,\forall\,A\,,B\in\mathbb{P}(X)}\)

and multiplication

\(\displaystyle{A\cdot B:=A\cap B\,,\forall\,A\,,B\in\mathbb{P}(X)}\)

and then \(\displaystyle{0_{\mathbb{P}(X)}=\varnothing\,\,\,,1_{\mathbb{P}(X)}=X}\).

Nickos, my idea is to define the map

\(\displaystyle{\Phi:R\to \mathbb{P}(\rm{Hom}(R,\mathbb{Z}_{2}))\,,r\mapsto \Phi(r)=\left\{f\in\rm{Hom}(R,\mathbb{Z}_{2}), f(r)=1\right\}}\)

which is a ring homomorphsim.

I am not sure if \(\displaystyle{\Phi}\) is one to one and onto.

I like very much this question and i gave an answer in order to make some progress.