Let be a commutative ring. Let and denote the nilradical and the singular ideal respectively. I can prove that In Lam's second book there is an exercise in which the reader is told to show that and to give a counterexample for the equality.
I can't do either. I'm quite sure the first one must be easy, but not for me, as usual. I'll try to think more about it and maybe I'll get it finally. So if it's very, very easy, please don't tell me how to do it. But if you think some hint would be helpful, then please post it.
For the second, I think I know too little to find the example. I know that the example can't be a Noetherian ring. Actually, it can't have the ACC property on annihilators of elements. All non-Noetherian commutative rings I know are polynomial rings in infinitely many variables and power series in infinitely many variables. But I know nothing about them really. In the polynomial ring in infinitely many variables, there is the ideal chain
which doesn't terminate. But these ideals aren't annihilators of elements for any ring, because doesn't annihilate anything non-zero. So this is not what I want. Polynomial rings would be nice though because I know what nilpotents look like there...
Thank you.
I think it's the greatest example I have ever seen. So to the point.for the counterexample, let and put now choose and show that is not nilpotent but
is not nilpotent, because raising to any finite power can annihilate only a finite number of coordinates. is singular because and it is essential in because
OK, I think I understand. But why can't we just take
Let Then so Therefore we have some such that
and
But then so is a non-zero nilpotent as we wanted, right?
oh, ok. well, if that's what the problem wants you to prove, then you're right although the problem is not interesting anymore!
i thought we're looking at as a ring (without identity probably). then, obviously, the nilradical of would be an ideal of my solution proves that the nilradical of is an essential ideal of
by the way, we should assume that is not nonsingular, a condition that should have been mentioned in the problem.
Ahh! Well, Lam's rings all have identity.
OK, I will try to think about it but I can't today because I'm preparing a presentation about all of this for tomorrow's seminar and I'm badly short of time. I don't think I'm going to sleep tonight.i thought we're looking at as a ring (without identity probably). then, obviously, the nilradical of would be an ideal of my solution proves that the nilradical of is an essential ideal of
For which problem do we need nonsingularity? I'm not using nonsingularity in my post 3. Is the proof incorrect?by the way, for all that we need R to be nonsingular, a condition that should have been mentioned in the problem.