# Math Help - Boolean commutative ring

1. ## Boolean commutative ring

I'm just starting work on Boolean rings. I understand so far that if we have some commutative Boolean ring $R$ with unity, that for every $\alpha\in R$ we have $\alpha^2 = \alpha$.

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

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

2. Originally Posted by DanielThrice
I'm just starting work on Boolean rings. I understand so far that if we have some commutative Boolean ring $R$ with unity, that for every $\alpha\in R$ we have $\alpha^2 = \alpha$.

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

As for maximals, how is it we would prove that every proper nontrivial prime ideal $I is maximal?
These questions seem to be coming from overarching question, did you forget to include it?

3. 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.

4. By F2 do you mean $\mathbb Z_2$ or just some generic field?

Thank you, that's a pretty clear way of looking at it

5. 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

6. F2 is the finite field of 2 elements. Z2 is another name for it, yes.