# Boolean commutative ring

• Apr 7th 2011, 12:26 PM
DanielThrice
Boolean commutative ring
I'm just starting work on Boolean rings. I understand so far that if we have some commutative Boolean ring $\displaystyle$ R$$with unity, that for every \displaystyle \alpha\in R we have \displaystyle \alpha^2 = \alpha. I would think then that some ideal \displaystyle I in \displaystyle R would make the factor ring \displaystyle R/I also boolean. How can I prove that this factor ring \displaystyle R/I is also boolean? As for maximals, how is it we would prove that every proper nontrivial prime ideal \displaystyle I<R is maximal? • Apr 7th 2011, 12:47 PM Drexel28 Quote: Originally Posted by DanielThrice I'm just starting work on Boolean rings. I understand so far that if we have some commutative Boolean ring \displaystyle  R$$ with unity, that for every $\displaystyle \alpha\in R$ we have $\displaystyle \alpha^2 = \alpha$.

I would think then that some ideal $\displaystyle I$ in $\displaystyle R$ would make the factor ring $\displaystyle R/I$ also boolean. How can I prove that this factor ring $\displaystyle R/I$ is also boolean?

As for maximals, how is it we would prove that every proper nontrivial prime ideal $\displaystyle I<R$ is maximal?

These questions seem to be coming from overarching question, did you forget to include it?
• Apr 7th 2011, 12:59 PM
Deveno
if R is boolean, then for any quotient R/I, we have: (α + I)(α + I) = α^2 + I = α + I, so R/I is boolean.

suppose I is a proper non-trivial prime ideal. then ab in I implies either a in I, or b in I, so R/I is an integral domain. but R/I is also boolean, so if a ≠ 0 in R/I,

then a = a^2 = a^3. thus a(1) = a(a^2), so a^2 = 1. hence a = 1 or -1. but -1 = (-1^2) = 1, so R/I = {0,1}.

from distributivity we have (1 + 1)(1 + 1) = 1 + 1 + 1 + 1, from the fact that R/I is boolean, we have (1 + 1)(1 + 1) = 1 + 1.

thus 1 + 1 = 0, so R/I is isomorphic to F2. since F2 is a field, I must be maximal.
• Apr 7th 2011, 01:08 PM
DanielThrice
By F2 do you mean $\displaystyle \mathbb Z_2$ or just some generic field?

Thank you, that's a pretty clear way of looking at it
• Apr 7th 2011, 01:09 PM
DanielThrice
Quote:

Originally Posted by Drexel28
These questions seem to be coming from overarching question, did you forget to include it?

And no this is a basic theory kinda problem, no specifics
• Apr 7th 2011, 11:01 PM
Deveno
F2 is the finite field of 2 elements. Z2 is another name for it, yes.