Boolean commutative ring

• Apr 7th 2011, 01:26 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 $R$ 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?
• Apr 7th 2011, 01:47 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 $R$ 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?
• Apr 7th 2011, 01:59 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, 02:08 PM
DanielThrice
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
• Apr 7th 2011, 02:09 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 8th 2011, 12:01 AM
Deveno
F2 is the finite field of 2 elements. Z2 is another name for it, yes.