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

• Nov 28th 2011, 12:48 PM
ymar
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.
• Nov 28th 2011, 04:12 PM
ymar
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 $m.$
• Nov 29th 2011, 12:29 AM
NonCommAlg
Re: a singular module is the quotient of a module and its essential submodule
Quote:

Originally Posted by ymar
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.
• Nov 29th 2011, 01:38 PM
ymar
Re: a singular module is the quotient of a module and its essential submodule
Hi, NonCommAlg.

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...
• Nov 29th 2011, 02:54 PM
NonCommAlg
Re: a singular module is the quotient of a module and its essential submodule
Quote:

Originally Posted by ymar
Hi, NonCommAlg.

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.
• Nov 30th 2011, 03:06 AM
ymar
Re: a singular module is the quotient of a module and its essential submodule
Thanks!

Quote:

Originally Posted by NonCommAlg
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.
Quote:

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.
• Nov 30th 2011, 03:50 AM
NonCommAlg
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.