0
$\begingroup$

I need help to understand where my argument is wrong, because I proved that every boolean ring has only $2$ elements which is wrong.

My proof goes like:

Let $R$ be a boolean ring. Then by definition $x^2=x$ which implies $x^n=x$ for all $n\ge1$. Therefore the only nilpotent element is $0$ hence the nilradical is just $0$. But the nilradical is a prime ideal therefore $R/0=R$ is an integral domain. Now consider an arbitrary element $x\in R$. We have $x^2-x=0$. So $x(x-1)=0$ therefore since $R$ is an integral domain $x$ must either be zero or $1$.

Thanks in advance

$\endgroup$
4
$\begingroup$

Who told you that the nilradical is prime? This is false in general.

In fact the nilradical is the intersection of all minimal primes, i.e. it is prime if and only if there is precisely one minimal prime. Geometrically, this means that $Spec(R)$ is irreducible.

$\endgroup$
  • $\begingroup$ I see I thought the intersection of prime ideals is prime $\endgroup$ – TheGeometer Feb 23 '16 at 11:11
  • $\begingroup$ How did you come up with this thought? The opposite is true: The intersection of two prime ideals is prime if and only if one is contained in the other. $\endgroup$ – MooS Feb 23 '16 at 11:14
  • 1
    $\begingroup$ (I know MooS knows this already but comment follows anyway) For some version of "the opposite" :) The intersection of infinitely many prime ideals can be prime without any of the prime ideals containing each other, but generalizing MooS's example, if an intersection of finitely many primes is prime, then one of the primes is contained in all the others. $\endgroup$ – rschwieb Feb 23 '16 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.