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 Post subject: Finite Boolean Ring Posted: Sat Jun 25, 2016 6:59 am
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 314
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

Top   Post subject: Re: Finite Boolean Ring Posted: Tue Aug 02, 2016 3:48 pm
 Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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.

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