• December 6th 2011, 12:04 PM
ymar
Let $R$ be a commutative ring. Let $Nil(R)$ and $Z(R)$ denote the nilradical and the singular ideal respectively. I can prove that $Nil(R)\subseteq Z(R).$ In Lam's second book there is an exercise in which the reader is told to show that $Nil(R)\subseteq_e Z(R)$ 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

$(X_1)\subseteq (X_1,X_2)\subseteq ...$

which doesn't terminate. But these ideals aren't annihilators of elements for any ring, because $X_i$ 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...
• December 6th 2011, 03:47 PM
NonCommAlg
Quote:

Originally Posted by ymar
Let $R$ be a commutative ring. Let $Nil(R)$ and $Z(R)$ denote the nilradical and the singular ideal respectively. I can prove that $Nil(R)\subseteq Z(R).$ In Lam's second book there is an exercise in which the reader is told to show that $Nil(R)\subseteq_e Z(R)$ 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

$(X_1)\subseteq (X_1,X_2)\subseteq ...$

which doesn't terminate. But these ideals aren't annihilators of elements for any ring, because $X_i$ 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 $a \in Z(R)$. if $a$ is nilpotent, we're done. otherwise, we have $\text{ann}(a) \cap Ra^2 \neq (0).$ thus ...

for the counterexample, let $R_i = \mathbb{Z}/2^i \mathbb{Z}, \ i \geq 1,$ and put $R = \prod_{i=1}^{\infty} R_i.$ now choose $a_i = 2 + 2^i \mathbb{Z} \in R_i, \ i \geq 1,$ and $a = (a_i) \in R.$ show that $a$ is not nilpotent but $a \in Z(R).$
• December 7th 2011, 05:10 AM
ymar
Thank you.
Quote:

for the counterexample, let $R_i = \mathbb{Z}/2^i \mathbb{Z}, \ i \geq 1,$ and put $R = \prod_{i=1}^{\infty} R_i.$ now choose $a_i = 2 + 2^i \mathbb{Z} \in R_i, \ i \geq 1,$ and $a = (a_i) \in R.$ show that $a$ is not nilpotent but $a \in Z(R).$
I think it's the greatest example I have ever seen. So to the point.

$a$ is not nilpotent, because raising $a$ to any finite power can annihilate only a finite number of coordinates. $a$ is singular because $\mathrm{ann}(a) = \{0, (2^0,2^1,2^2,...)\}$ and it is essential in $R$ because $\{0,2^{i-1}\}\subseteq_e R_i.$

Quote:

Originally Posted by NonCommAlg
let $a \in Z(R)$. if $a$ is nilpotent, we're done. otherwise, we have $\text{ann}(a) \cap Ra^2 \neq (0).$ thus ...

OK, I think I understand. But why can't we just take $Ra?$

Let $0\neq a\in Z(R)$ Then $Ra\neq\{0\},$ so $\mathrm{ann}(a)\cap Ra\neq \{0\}.$ Therefore we have some $r\in R$ such that

$ra\neq 0$ and $ra^2=0.$

But then $(ra)^2=r(ra^2)=0,$ so $ra$ is a non-zero nilpotent as we wanted, right?
• December 7th 2011, 08:03 AM
NonCommAlg
Quote:

Originally Posted by ymar
OK, I think I understand. But why can't we just take $Ra?$

Let $0\neq a\in Z(R)$ Then $Ra\neq\{0\},$ so $\mathrm{ann}(a)\cap Ra\neq \{0\}.$ Therefore we have some $r\in R$ such that

$ra\neq 0$ and $ra^2=0.$

But then $(ra)^2=r(ra^2)=0,$ so $ra$ is a non-zero nilpotent as we wanted, right?

no, we want to prove that the nilradical is essential in $Z(R)$ and your $r$ is not necessarily in $Z(R).$
• December 7th 2011, 09:02 AM
ymar
Quote:

Originally Posted by NonCommAlg
no, we want to prove that the nilradical is essential in $Z(R)$ and your $r$ is not necessarily in $Z(R).$

I don't understand. :( I think we want to prove that $Nil(R)\subseteq_e Z(R)$ as $R-$modules. So the scalars are from $R,$ no?

By definition, if $M$ and $N$ are $R-$modules, then $M\subseteq_e N$ iff for any $0\neq n\in N$ there is $r\in R$ such that $0\neq nr\in M.$ So $r$ is from $R.$ What am I missing?
• December 7th 2011, 09:15 AM
NonCommAlg
Quote:

Originally Posted by ymar
I don't understand. :( I think we want to prove that $Nil(R)\subseteq_e Z(R)$ as $R-$modules. So the scalars are from $R,$ no?

By definition, if $M$ and $N$ are $R-$modules, then $M\subseteq_e N$ iff for any $0\neq n\in N$ there is $r\in R$ such that $0\neq nr\in M.$ So $r$ is from $R.$ 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 $Z(R)$ as a ring (without identity probably). then, obviously, the nilradical of $R$ would be an ideal of $Z(R).$ my solution proves that the nilradical of $R$ is an essential ideal of $Z(R).$

by the way, we should assume that $R$ is not nonsingular, a condition that should have been mentioned in the problem.
• December 7th 2011, 09:34 AM
ymar
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 $Z(R)$ as a ring (without identity probably). then, obviously, the nilradical of $R$ would be an ideal of $Z(R).$ my solution proves that the nilradical of $R$ is an essential ideal of $Z(R).$
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?
• December 7th 2011, 09:38 AM
NonCommAlg
well, if $R$ is nonsingular, then $Z(R)$ is zero and we can't really talk about "essential" submodules (or ideals) anymore.
• December 7th 2011, 09:41 AM
ymar
Quote:

Originally Posted by NonCommAlg
well, if $R$ is nonsingular, then $Z(R)$ 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 $R$ is not nonsigular, right? But I think $\{0\}\subseteq_e\{0\}$ trivially because the condition is vacuously satisfied, so it's all fine.
• December 7th 2011, 09:46 AM
NonCommAlg
Oh, so you were saying that we need to assume that $R$ is not nonsigular, right? But I think $\{0\}\subseteq_e\{0\}$ trivially because the condition is vacuously satisfied, so it's all fine.