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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.

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    $\begingroup$ 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 $\endgroup$
    – rschwieb
    Sep 10, 2014 at 13:16

1 Answer 1

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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.

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  • $\begingroup$ Thank you user174627. Really nice proof.Lot easier than my attempt.Thanx again $\endgroup$
    – usermath
    Sep 9, 2014 at 19:18
  • $\begingroup$ You're welcome! $\endgroup$
    – user174627
    Sep 9, 2014 at 19:21

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