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

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

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

    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 $\displaystyle A\subseteq B$ ($\displaystyle A\neq\{0\}$) such that $\displaystyle B/A=M.$ Let $\displaystyle m'\in B\setminus\{0\}$ and let $\displaystyle m\in M$ be the equivalence class of $\displaystyle m'.$ Then there is $\displaystyle r\in R$ such that $\displaystyle r=1\cdot r\in\mathrm{ann}(m)\setminus\{0\}.$ That means that $\displaystyle mr=0,$ which in turn means that $\displaystyle m'r\in A.$ But this proves nothing, because the important part is that $\displaystyle m'r\neq 0.$ I do know that $\displaystyle r\neq 0$ but this is not enough.
    Last edited by ymar; Nov 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 $\displaystyle B/A,$ not $\displaystyle A/B.$ I'll edit this right away.

    PS: I should have said that $\displaystyle \mathrm{ann}(m)$ is the right annihilator of $\displaystyle m.$
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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 $\displaystyle R-$modules $\displaystyle A_R\subseteq B_R,$ $\displaystyle A$ is an essential submodule of $\displaystyle B$ (that is $\displaystyle A\subseteq_e B$) iff for any submodule $\displaystyle X\subseteq B,$ $\displaystyle A\cap X\neq \{0\}.$ Equivalently, it's an essential submodule iff for any $\displaystyle b\in B\setminus\{0\}$ there is $\displaystyle r\in R$ such that $\displaystyle br\in A\setminus\{0\}.$

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

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

    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 $\displaystyle A\subseteq B$ ($\displaystyle A\neq\{0\}$) such that $\displaystyle B/A=M.$ Let $\displaystyle m'\in B\setminus\{0\}$ and let $\displaystyle m\in M$ be the equivalence class of $\displaystyle m'.$ Then there is $\displaystyle r\in R$ such that $\displaystyle r=1\cdot r\in\mathrm{ann}(m)\setminus\{0\}.$ That means that $\displaystyle mr=0,$ which in turn means that $\displaystyle m'r\in A.$ But this proves nothing, because the important part is that $\displaystyle m'r\neq 0.$ I do know that $\displaystyle r\neq 0$ but this is not enough.
    here is a hint: every $\displaystyle R$-module is the homomorphic image of a free $\displaystyle R$-module. so you only need to prove this:
    if $\displaystyle F$ is a free $\displaystyle R$-module and $\displaystyle K$ is a submodule of $\displaystyle F$ such that $\displaystyle F/K$ is singular, then $\displaystyle K \subseteq_e F.$
    by the way, the converse of your problem is also true, i.e. if $\displaystyle M = A/B$ with $\displaystyle B \subseteq_e A,$ then $\displaystyle M$ is singular.
    Last edited by NonCommAlg; Nov 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 $\displaystyle \mathbb{Z}$ doesn't have any linearly independent subset when viewed as a submodule of the (free) module $\displaystyle \left(\mathbb{Z}^2\right)_{\mathbb{Z}^2}.$ 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 $\displaystyle B$ 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 $\displaystyle B$? I don't even see this.

    OK, so I know that

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

    What I want to know is that

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

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

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

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

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

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

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

    If the ring were an integral domain, I could use the product $\displaystyle \prod_{k=1}^{t}s_k,$ 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 $\displaystyle \mathbb{Z}$ doesn't have any linearly independent subset when viewed as a submodule of the (free) module $\displaystyle \left(\mathbb{Z}^2\right)_{\mathbb{Z}^2}.$ 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 $\displaystyle B$ 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 $\displaystyle B$? I don't even see this.

    OK, so I know that

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

    What I want to know is that

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

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

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

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

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

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

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

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

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

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

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

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

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

    the first two terms on the right hand side of $\displaystyle (3)$ are now in $\displaystyle K$ and $\displaystyle xs_1s_2 \neq0$ because $\displaystyle r_2s_1s_2 \neq 0.$ if we continue this process, we will eventually have a positive integer $\displaystyle 1 \leq m \leq n$ and $\displaystyle s = s_1s_2 \cdots s_m \in R$ such that $\displaystyle 0 \neq xs \in K,$ as desired. note that i wrote $\displaystyle m$ instead of $\displaystyle n$ 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 $\displaystyle F$ to $\displaystyle M$ and so if $\displaystyle K$ is the kernel, then $\displaystyle F/K \cong M.$ so $\displaystyle F/K$ is singular and we only need to prove that $\displaystyle F$ is an essential extension of $\displaystyle K.$ first note that $\displaystyle K \neq \{0\}$ because otherwise $\displaystyle M$ 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 $\displaystyle \mathrm{ann}(1)=\{0\}$ cannot be essential in the ring.
    now let $\displaystyle \{x_i: \ i \in I \}$ be an $\displaystyle R$-basis for $\displaystyle F$ and let $\displaystyle 0 \neq x \in F.$ we need to show that there exists $\displaystyle s \in R$ such that $\displaystyle 0 \neq xs \in K.$ we can write, after renaming the indices if necessarily,

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

    where $\displaystyle r_1 \neq 0.$
    Right. This is where I went wrong. I forgot to use the fact that $\displaystyle x\neq 0.$ 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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