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Math Help - a singular module is the quotient of a module and its essential submodule

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    a singular module is the quotient of a module and its essential submodule

    For two right R-modules A_R\subseteq B_R, A is an essential submodule of B (that is A\subseteq_e B) iff for any submodule X\subseteq B, A\cap X\neq \{0\}. Equivalently, it's an essential submodule iff for any b\in B\setminus\{0\} there is r\in R such that br\in A\setminus\{0\}.

    The singular submodule Z(M) of an R-module M is the set of all m\in M such that \mathrm{ann}(m)\subseteq_e R_R. A module M is singular iff Z(M)=M.

    I have to prove that if an R-module M is singular, then there are two R- modules A\subseteq_e B such that B/A=M.

    I have no idea how to go about it. I'm completely new to module theory and I don't have much intuition regarding it. How does one find two modules whose quotient is given but which have to satisfy a certain condition?

    I was thinking this:

    Let's take any two modules A\subseteq B ( A\neq\{0\}) such that B/A=M. Let m'\in B\setminus\{0\} and let m\in M be the equivalence class of m'. Then there is r\in R such that r=1\cdot r\in\mathrm{ann}(m)\setminus\{0\}. That means that mr=0, which in turn means that m'r\in A. But this proves nothing, because the important part is that m'r\neq 0. I do know that r\neq 0 but this is not enough.
    Last edited by ymar; November 28th 2011 at 03:13 PM.
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    Re: a singular module is the quotient of a module and its essential submodule

    I have just noticed that I wrote all the quotients incorrectly. It should be B/A, not A/B. I'll edit this right away.

    PS: I should have said that \mathrm{ann}(m) is the right annihilator of m.
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    Re: a singular module is the quotient of a module and its essential submodule

    Quote Originally Posted by ymar View Post
    For two right R-modules A_R\subseteq B_R, A is an essential submodule of B (that is A\subseteq_e B) iff for any submodule X\subseteq B, A\cap X\neq \{0\}. Equivalently, it's an essential submodule iff for any b\in B\setminus\{0\} there is r\in R such that br\in A\setminus\{0\}.

    The singular submodule Z(M) of an R-module M is the set of all m\in M such that \mathrm{ann}(m)\subseteq_e R_R. A module M is singular iff Z(M)=M.

    I have to prove that if an R-module M is singular, then there are two R- modules A\subseteq_e B such that B/A=M.

    I have no idea how to go about it. I'm completely new to module theory and I don't have much intuition regarding it. How does one find two modules whose quotient is given but which have to satisfy a certain condition?

    I was thinking this:

    Let's take any two modules A\subseteq B ( A\neq\{0\}) such that B/A=M. Let m'\in B\setminus\{0\} and let m\in M be the equivalence class of m'. Then there is r\in R such that r=1\cdot r\in\mathrm{ann}(m)\setminus\{0\}. That means that mr=0, which in turn means that m'r\in A. But this proves nothing, because the important part is that m'r\neq 0. I do know that r\neq 0 but this is not enough.
    here is a hint: every R-module is the homomorphic image of a free R-module. so you only need to prove this:
    if F is a free R-module and K is a submodule of F such that F/K is singular, then K \subseteq_e F.
    by the way, the converse of your problem is also true, i.e. if M = A/B with B \subseteq_e A, then M is singular.
    Last edited by NonCommAlg; November 30th 2011 at 02:37 AM.
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    Re: a singular module is the quotient of a module and its essential submodule

    Hi, NonCommAlg.

    Thank you for your answer. I know about the converse and it's easy to prove.

    I've been thinking about your hint the whole day today and I'm afraid I'm still very far from seeing it. I noticed instead that free modules are very different from vector spaces. I was shocked to notice that \mathbb{Z} doesn't have any linearly independent subset when viewed as a submodule of the (free) module \left(\mathbb{Z}^2\right)_{\mathbb{Z}^2}. This is awful.

    The first thing I would like to ask is if it's obvious for some reason that taking the free module as B has to work given that we know that the theorem is true. In other words, why should I have thought of the free module when I was trying to prove it?

    The second thing is, is it enough to prove this for the elements of a basis of B? I don't even see this.

    OK, so I know that

    \left(\forall m'\in B\right) \left(\forall r\in R\setminus\{0\}\right) \left(\exists s\in R\right) \; m'rs\in A.

    What I want to know is that

    \left(\forall m'\in B\setminus\{0\}\right) \left(\exists s\in R\right) \; m's\in A \mbox{ and } m's\neq 0.

    You told me to assume that B has a basis. To me it just means, "Write m' in the basis." OK. Let X=\left\{x_i\right\}_{i\in I} be a basis of B. Let

    m'=\sum_{k=1}^t x_{i_k}a_k, where a_k\in R.

    If I knew that can "push" the summands into A\setminus\{0\} would I know that I can "push" the sum too? If I had s_1,...,s_t\in R such that

    x_{i_k}a_k s_k\in A\setminus\{0\} for k=1,..,t,

    would I be able to find S\in R such that

    m's\in A\setminus\{0\}?

    If the ring were an integral domain, I could use the product \prod_{k=1}^{t}s_k, but it's not...
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    Re: a singular module is the quotient of a module and its essential submodule

    Quote Originally Posted by ymar View Post
    Hi, NonCommAlg.

    Thank you for your answer. I know about the converse and it's easy to prove.

    I've been thinking about your hint the whole day today and I'm afraid I'm still very far from seeing it. I noticed instead that free modules are very different from vector spaces. I was shocked to notice that \mathbb{Z} doesn't have any linearly independent subset when viewed as a submodule of the (free) module \left(\mathbb{Z}^2\right)_{\mathbb{Z}^2}. This is awful.

    The first thing I would like to ask is if it's obvious for some reason that taking the free module as B has to work given that we know that the theorem is true. In other words, why should I have thought of the free module when I was trying to prove it?

    The second thing is, is it enough to prove this for the elements of a basis of B? I don't even see this.

    OK, so I know that

    \left(\forall m'\in B\right) \left(\forall r\in R\setminus\{0\}\right) \left(\exists s\in R\right) \; m'rs\in A.

    What I want to know is that

    \left(\forall m'\in B\setminus\{0\}\right) \left(\exists s\in R\right) \; m's\in A \mbox{ and } m's\neq 0.

    You told me to assume that B has a basis. To me it just means, "Write m' in the basis." OK. Let X=\left\{x_i\right\}_{i\in I} be a basis of B. Let

    m'=\sum_{k=1}^t x_{i_k}a_k, where a_k\in R.

    If I knew that can "push" the summands into A\setminus\{0\} would I know that I can "push" the sum too? If I had s_1,...,s_t\in R such that

    x_{i_k}a_k s_k\in A\setminus\{0\} for k=1,..,t,

    would I be able to find S\in R such that

    m's\in A\setminus\{0\}?

    If the ring were an integral domain, I could use the product \prod_{k=1}^{t}s_k, but it's not...
    almost! so, as i said, there is an onto homomorphism from some free module F to M and so if K is the kernel, then F/K \cong M. so F/K is singular and we only need to prove that F is an essential extension of K. first note that K \neq \{0\} because otherwise M would be a free singular module and that's not possible (why?). now let \{x_i: \ i \in I \} be an R-basis for F and let 0 \neq x \in F. we need to show that there exists s \in R such that 0 \neq xs \in K. we can write, after renaming the indices if necessarily,

    x = \sum_{i=1}^n x_ir_i, \ \ \ \ \ \ \ \ \ \ (1)

    where r_1 \neq 0. for any y \in F, let \overline{y}=y+K \in F/K. now, since F/K is singular, \text{ann}(\overline{x_1}) \subseteq_e R and so \text{ann}(\overline{x_1}) \cap r_1R \neq \{0\}. so there exists s_1 \in R such that r_1s_1 \neq 0 and x_1r_1s_1 \in K. hence (1) gives us

    xs_1= x_1r_1s_1 +\sum_{i=2}^n x_ir_is_1. \ \ \ \ \ \ \ \ \ (2)

    note that xs_1 \neq 0 because r_1s_1 \neq 0. now, if r_is_1 = 0 for all 2 \leq i \leq n, then xs_1 =x_1r_1s_1 \in K and we are done. otherwise, after renaming the indices in the sum on the right hand side of (2) if necessary, we may assume that r_2s_1 \neq 0. now repeat the above process to get s_2 \in R such that r_2s_1s_2 \neq 0 and x_2r_2s_1s_2 \in K. then (2) will give us

    xs_1s_2 = x_1r_1s_1s_2 + x_2r_2s_1s_2 + \sum_{i=3}^n x_ir_is_1s_2. \ \ \ \ \ \ \ (3)

    the first two terms on the right hand side of (3) are now in K and xs_1s_2 \neq0 because r_2s_1s_2 \neq 0. if we continue this process, we will eventually have a positive integer 1 \leq m \leq n and s = s_1s_2 \cdots s_m \in R such that 0 \neq xs \in K, as desired. note that i wrote m instead of n because some terms might vanish in the process.
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    Re: a singular module is the quotient of a module and its essential submodule

    Thanks!

    Quote Originally Posted by NonCommAlg View Post
    almost! so, as i said, there is an onto homomorphism from some free module F to M and so if K is the kernel, then F/K \cong M. so F/K is singular and we only need to prove that F is an essential extension of K. first note that K \neq \{0\} because otherwise M would be a free singular module and that's not possible (why?).
    Because a submodule a of a singular module must be singular and a free module contains the ring as its submodule. A ring cannot be singular because \mathrm{ann}(1)=\{0\} cannot be essential in the ring.
    now let \{x_i: \ i \in I \} be an R-basis for F and let 0 \neq x \in F. we need to show that there exists s \in R such that 0 \neq xs \in K. we can write, after renaming the indices if necessarily,

    x = \sum_{i=1}^n x_ir_i, \ \ \ \ \ \ \ \ \ \ (1)

    where r_1 \neq 0.
    Right. This is where I went wrong. I forgot to use the fact that x\neq 0. Eh. Just another clue that I have chosen the wrong studies.

    Thank you very much. I see it now.
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    Re: a singular module is the quotient of a module and its essential submodule

    you asked why i thought of free modules. we have two dual concepts: quotient and submodule. every module is a quotient of some free module and every module is a submodule of some injective module. so, when dealing with modules in general, that's probably the most natural way to try.
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