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# Science, Math, & Technology

STEM_fields++;

## commutative algebra problem of the week finitely generated module and exact sequence

Posted Over 8 Years ago by Yeano

Let A be a ring and M be an abelian group. M is an A-module if there is a scalar multiplication function

μ : A x M → M (denoting μ(a,x) = ax) such that the following properties hold:
(i) (a + b)x = ax + bx for all a, b in A and for all x in M
(ii) a(x + y) = ax + ay for all a in A and for all x, y in M
(iii) a(bx) = (ab)x for all a, b in A and for all x in M
(iv) 1x = x for all x in M

(If you are not comfortable with modules, you can use vector spaces instead of modules for this problem.)

If M and N are A-modules, a function f : M → N is an A-linear map (or an A-module homomorphism) if the following hold:
(i) f(x+y) = f(x) + f(y) for all x, y in M
(ii) f(ax) = af(x) for all a, b in A and for all x in M.

Now, we define an exact sequence. Note that because of notation issues on gtx0, I will use →f to denote the function f between two modules.

A sequence of modules

... →fi-1 Mi-1fi Mifi+1 Mi+1fi+2 ...

is said to be exact at Mi if Im fi = ker fi+1.

A sequence is said to be exact if it is exact at all modules it contains.

In particular, we get what is called a short exact sequence. This is a sequence of the form:

0 → M' →f M →g M'' → 0

Because of how an exact sequence is defined and because the zero module is placed at the beginning and the end, the short exact sequence above forces f to be injective and g to be surjective.

An A-module M is said to be finitely generated if there exist finitely many elements x1, ..., xn so that every element of x of M can be written as a1x1 + ... + anxn for some elements a1, ..., an of A (not necessarily in a unique way!).

So here's the question:

Suppose M' and M'' are finitely generated A-modules, and the following is a short exact sequence of A-modules:

0 → M' →f M →g M'' → 0

Can you show that M is finitely generated as an A-module?

## There are 8 Replies

It seems that M could be generated by the set consisting of the image of the generators of M' and the preimage of the generators of M''.

I don't know if anyone besides you, The Fly, and I are willing and able to try these sorts of problems, but I'll wait a bit to actually attempt a proof just in case.

Over 8 Years ago
EN For example, let Z be the integers and consider the sequence of Z-modules

0 --> Z -->f Z x Z -->g Z --> 0

where f(n) = (n,0) and g(a,b) = b. It's easy to check that this is short exact.

The generating set for Z is {1}. What does f take 1 to? (1,0). What is something that g takes to 1? Well, infinitely many things, but let's choose (0,1) for simplicity. Thus {(1,0), (0,1)} should be a generating set for Z x Z, which we can see is correct.

Over 8 Years ago
EN Indeed, that is a valid method! And you pointed out in your second reply that we need to choose an element in the preimage for each generator of M'' since we are not guaranteed that the entire preimage is finite!

So yes, that's a great way to do it and a great way to avoid a potential trap with the preimage of the generators.

Over 8 Years ago
Yeano

Are you satisfied, Yeano? I think EN has answered it. So we take preimages in M (under g) of a finite generating set in M'', and the image in M (under f) of a finite generating set for M'. The union of these two finite subsets of M would constitute a finite generating set for M. The short exactness, the fact that the kernel of g is equal to image of f, is crucial for getting this to work. I hope that by now you already have the details nailed down! {;)}

PS--the ring doesn't have to be commutative. The result remains true even for modules over noncommutative rings.

Over 8 Years ago
The Fly Oh yes, what EN posted is fine. Both of you can probably figure out the details if you haven't already.

Also, I usually post these after I've completed the respective problem. Usually, I've posted problems that I have considered fun and should be accessible to The Fly and EN, at the very least.

I don't know if I'll be able to post any more of them, though. We've gotten into localization now, and it seems like there may not be any more problems that are at an appropriate level to post here.

Over 8 Years ago
Yeano

I haven't ever done anything with localization, although I think it's covered in one of the later chapters in my algebra book.

I made a startling (to me) algebra discovery yesterday. When I get some free time I'll try to make a post about it.

Over 8 Years ago
EN I just looked up my two algebra reference books -- Hungerford and Cohn -- and I do now recall going thru that construction of getting a local ring out of some non-units of an algebra. But that was long ago and I didn't go into much detail about the study of localization. But now I vaguely remember. I assume you are looking at the commutative rings first, esp in relation to polynomial rings.

Over 8 Years ago
The Fly I am interested in noncommutative localization. 