nilradical and singular ideal

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...

Re: nilradical and singular ideal

Quote:

Originally Posted by

**ymar** 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...

let . if is nilpotent, we're done. otherwise, we have thus ...

for the counterexample, let and put now choose and show that is not nilpotent but

Re: nilradical and singular ideal

Re: nilradical and singular ideal

Quote:

Originally Posted by

**ymar** 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?

no, we want to prove that the nilradical is essential in and your is not necessarily in

Re: nilradical and singular ideal

Re: nilradical and singular ideal

Quote:

Originally Posted by

**ymar** I don't understand. :( I think we want to prove that

as

modules. So the scalars are from

no?

By definition, if

and

are

modules, then

iff for any

there is

such that

So

is from

What am I missing?

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.

Re: nilradical and singular ideal

Quote:

Originally Posted by

**NonCommAlg** oh, ok. well, if that's what the problem wants you to prove, then you're right although the problem is not interesting anymore! :)

Ahh! Well, Lam's rings all have identity.

Quote:

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

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.

Quote:

by the way, for all that we need R to be nonsingular, a condition that should have been mentioned in the problem.

For which problem do we need nonsingularity? I'm not using nonsingularity in my post 3. Is the proof incorrect?

Re: nilradical and singular ideal

well, if is nonsingular, then is zero and we can't really talk about "essential" submodules (or ideals) anymore.

Re: nilradical and singular ideal

Quote:

Originally Posted by

**NonCommAlg** well, if

is nonsingular, then

is zero and we can't really talk about "essential" submodules (or ideals) anymore.

Oh, so you were saying that we need to assume that is **not** nonsigular, right? But I think trivially because the condition is vacuously satisfied, so it's all fine.

Re: nilradical and singular ideal

Quote:

Originally Posted by

**ymar** Oh, so you were saying that we need to assume that

is

**not** nonsigular, right? But I think

trivially because the condition is vacuously satisfied, so it's all fine.

haha, that's, of course, what i meant. well, usually essential submodules are assumed to be nonzero. maybe Lam doesn't put this condition, i don't know.