# Boolean Ring and prime ideals

Referring to the question: Finitely generated ideals in a Boolean ring are principal, why?

How to prove: In every Boolean Ring Does there exist any prime ideal in a Boolean Ring. Only Boolean ring I know is power set any set with symmetric difference and intersection. But there is no prime ideal as much as I can figure out. Only case is $\mathbb Z_{2}$ also there are no prime ideals except $\{0\}$

• So are you studying commutative algebra $?$ See the link . Might help . math.stackexchange.com/questions/877892/… Sep 22, 2015 at 16:44
• I am planning to. But how this links helps. I didn't get any idea about prime ideals in Boolean Ring only fact these are maximal. But are there any prime ideal in a boolean ring. @user118494 Sep 22, 2015 at 16:49
• How did you conclude that the power set ring has no prime ideals? Sep 22, 2015 at 16:53
• If P prime ideal in power set ring of X. Let P contains non empty set. Then let A is in P. Now A.X is in P. Hence X is in P. Hence every subset of X is in P. Then P={X} contradiction. Case where P only contains empty set also gives contradiction you can figure that out. I think. @user26857 Sep 22, 2015 at 16:57
• @user26857 Oh no how totally spoiled things. But what are prime ideals in any power set ring. Can you please help Sep 22, 2015 at 17:02

For concrete examples, just take $\prod \Bbb Z_2$, any number of copies of the field of two elements. You can produce a maximal ideal (many, actually) by picking a particular position and looking at the set of elements which are zero on that position.
Let $$R$$ be boolean, ie $$\forall x \in R, \, x^2 = x$$. Then any quotient of $$R$$ is also boolean. Let $$I$$ be a prime ideal of $$R$$, then $$R / I$$ is an integral domain. We show that $$R/I$$ is a field, whence $$I$$ is maximal.
In fact, every boolean ring $$R$$ that is an integral domain is isomorphic to the field $$\mathbb{F}_2$$ :
let $$x \in R$$. Then $$x (1 - x) = x - x^2 = x - x = 0$$. Since $$R$$ is an integral domain, that means that either $$x=0$$ or $$1 - x=0$$. Hence, $$R = \{0;1\}$$ as a set. The function $$f : R \mapsto \mathbb{F}_2$$ defined by $$f(0) = 0$$ and $$f(1)=1$$ is obviously a ring isomorphism.