# Atiyah-Macdonald, Problem 6 of Chapter 1

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 $$$$ we have $$\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.

• Hmm, at first I agreed that it was a good idea to pursue "prime implies maximal" but actually I think it will not work. That would be equivalent to $R/nil(R)$ being von Neumann regular, something which is going to be strictly stronger than this condition on idempotents. It was a nice idea, though Sep 10, 2014 at 13:16

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 $e=0$.