I was trying to solve the following problem from Introduction to Commutative Algebra by Atiyah and Macdonald. (It is Problem 6 of Chapter 1.)
While trying to solve the problem, I am facing trouble in the final stage of my attempt which is mentioned after the problem. Please help me to solve it.
A ring $A$ is such that every ideal not contained in the nilradical contains a nonzero idempotent (that is, an element $e$ such that $e^2 = e \ne 0$). Prove that the nilradical and Jacobson radical of $A$ are equal.
I have tried and what I have done so far is the following:
Since one side inclusion ie Nilradical = $N(A)\subseteq J(A)$ = Jacobson radical is obvious so I think it is sufficient to prove that every prime ideal in $A$ is maximal.
To prove so let $P$ be a prime ideal and say $x\in A-P$ (ie $x+P\ne 0+P$) and thus by applying given condition on $<x>$ we have $\exists a\in A $ such that $a\ne0,ax=a^2x^2$ So considering $A/P$(which is integral domain as $P$ is prime ideal) we see $(ax+P)((ax+P)-(1+P))=0+P$ and hence $ax\in P$ or $ax-1\in P$ If the later is true then $(a+P)(x+P)=1+P \Rightarrow A/P$ is field $\Rightarrow P$ maximal.
But I failed to exclude the first case ie I can't prove $ax\notin P$.
Maybe I am missing something very easy or there may be an easier way to solve the problem. Please help me. Thnx in advance.
The nilradical is always in the Jacobson radical.
Now, fact: the Jacobson radical never has non trivial idempotents. Indeed $y$ is in the Jacobson iff $1+xy$ is invertible for all $x$ (indeed if $y$ is in every maximal ideal thus $1+xy$ is 1 mod every maximal ideal so cannot be in any of them that is equivalent to be invertible, conversely if $x$ is not in some maximal ideal m than mod m is invertible, write down the relation xy-1 in m, and you're done). Then if $e$ is an idempotent in the Jacobson radical $1-e$ is invertible. But $(1-e)e=0$ thus $e=0$.
So since by hypothesis every ideal not in the nilradical has non trivial idempotents, and since the Jacobson radical cannot have, the Jacobson radical is in the nilradical finishing the proof.