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
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.
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.
Thanks!
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.
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.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
Thank you very much. I see it now.
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.