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Math Help - nilradical and singular ideal

  1. #1
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    nilradical and singular ideal

    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...
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    Re: nilradical and singular ideal

    Quote Originally Posted by ymar View Post
    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).
    Last edited by NonCommAlg; December 6th 2011 at 05:04 PM.
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    Re: nilradical and singular ideal

    Thank you.
    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 View Post
    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?
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    Re: nilradical and singular ideal

    Quote Originally Posted by ymar View Post
    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).
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    Re: nilradical and singular ideal

    Quote Originally Posted by NonCommAlg View Post
    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?
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    Re: nilradical and singular ideal

    Quote Originally Posted by ymar View Post
    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.
    Last edited by NonCommAlg; December 7th 2011 at 10:48 AM.
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    Re: nilradical and singular ideal

    Quote Originally Posted by NonCommAlg View Post
    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.

    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.
    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?
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    Re: nilradical and singular ideal

    well, if R is nonsingular, then Z(R) is zero and we can't really talk about "essential" submodules (or ideals) anymore.
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    Re: nilradical and singular ideal

    Quote Originally Posted by NonCommAlg View Post
    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.
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    Re: nilradical and singular ideal

    Quote Originally Posted by ymar View Post
    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.
    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.
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