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 $\{0\}$

  • $\begingroup$ So are you studying commutative algebra $?$ See the link . Might help . math.stackexchange.com/questions/877892/… $\endgroup$
    – user118494
    Commented Sep 22, 2015 at 16:44
  • $\begingroup$ 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 $\endgroup$
    – Sushil
    Commented Sep 22, 2015 at 16:49
  • $\begingroup$ How did you conclude that the power set ring has no prime ideals? $\endgroup$
    – user26857
    Commented Sep 22, 2015 at 16:53
  • $\begingroup$ 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 $\endgroup$
    – Sushil
    Commented Sep 22, 2015 at 16:57
  • $\begingroup$ @user26857 Oh no how totally spoiled things. But what are prime ideals in any power set ring. Can you please help $\endgroup$
    – Sushil
    Commented Sep 22, 2015 at 17:02

2 Answers 2


Every ring with identity has a maximal ideal and every maximal ideal is prime. You can find several more proofs of these statements on the site, as well as online, or in any algebra book.

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.


I suppose your question is "why is any prime ideal in a boolean ring maximal?". Here's a proof :

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.


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