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

For two right modules is an essential submodule of (that is ) iff for any submodule Equivalently, it's an essential submodule iff for any there is such that

The singular submodule of an module is the set of all such that A module is singular iff

I have to prove that if an module is singular, then there are two modules such that

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 ( ) such that Let and let be the equivalence class of Then there is such that That means that which in turn means that But this proves nothing, because the important part is that I do know that but this is not enough.

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 not I'll edit this right away.

PS: I should have said that is the __right__ annihilator of

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

Quote:

Originally Posted by

**ymar** For two right

modules

is an essential submodule of

(that is

) iff for any submodule

Equivalently, it's an essential submodule iff for any

there is

such that

The singular submodule

of an

module

is the set of all

such that

A module

is singular iff

I have to prove that if an

module

is singular, then there are two

modules

such that

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

(

) such that

Let

and let

be the equivalence class of

Then there is

such that

That means that

which in turn means that

But this proves nothing, because the important part is that

I do know that

but this is not enough.

here is a hint: every -module is the homomorphic image of a free -module. so you only need to prove this:

if is a free -module and is a submodule of such that is singular, then

by the way, the converse of your problem is also true, i.e. if with then is singular.

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 doesn't have any linearly independent subset when viewed as a submodule of the (free) module 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 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 ? I don't even see this.

OK, so I know that

What I want to know is that

You told me to assume that has a basis. To me it just means, "Write in the basis." OK. Let be a basis of Let

where

If I knew that can "push" the summands into would I know that I can "push" the sum too? If I had such that

for

would I be able to find such that

If the ring were an integral domain, I could use the product but it's not...

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

Quote:

Originally Posted by

**ymar** 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

doesn't have any linearly independent subset when viewed as a submodule of the (free) module

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

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

? I don't even see this.

OK, so I know that

What I want to know is that

You told me to assume that

has a basis. To me it just means, "Write

in the basis." OK. Let

be a basis of

Let

where

If I knew that can "push" the summands into

would I know that I can "push" the sum too? If I had

such that

for

would I be able to find

such that

If the ring were an integral domain, I could use the product

but it's not...

almost! so, as i said, there is an onto homomorphism from some free module to and so if is the kernel, then so is singular and we only need to prove that is an essential extension of first note that because otherwise would be a free singular module and that's not possible (why?). now let be an -basis for and let we need to show that there exists such that we can write, after renaming the indices if necessarily,

where for any let now, since is singular, and so so there exists such that and hence gives us

note that because now, if for all then and we are done. otherwise, after renaming the indices in the sum on the right hand side of if necessary, we may assume that now repeat the above process to get such that and then will give us

the first two terms on the right hand side of are now in and because if we continue this process, we will eventually have a positive integer and such that as desired. note that i wrote instead of because some terms might vanish in the process.

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

to

and so if

is the kernel, then

so

is singular and we only need to prove that

is an essential extension of

first note that

because otherwise

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 cannot be essential in the ring.

Quote:

now let

be an

-basis for

and let

we need to show that there exists

such that

we can write, after renaming the indices if necessarily,

where

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

Thank you very much. I see it now.

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.