# Proving the Dual of Schanuel's Lemma

• A
• TMO

#### TMO

Given: the short exact sequences 0 → M → E → K → 0 and 0 → M → E' → K' → 0 where M is a left R-module and E and E' are injective left R-modules. Prove: E K' ≅ E' K.

First, let f be the morphism represented by M → E and g be the morphism represented by M → E'. Therefore we can construct a pushforward which is the object X together with the morphism i1 : E → X and i2 : E' → X

How do I proceed from there? I need to then somehow chain together the morphisms from this pushout to get two short exact sequences, show that they split, and then show that they are equal each to the direct sum.

I understand how he concluded that the morphism h existed (An R-module E is injective iff. for all R-module homomorphisms ϕ : M → N and ψ : M → E where ϕ is injective, there exists an R-linear homomorphism θ : N → E such that θ ◦ ϕ = ψ). But how exactly did he conclude that the morphism k existed when it isn't explicitly stated in our hypotheses that Q, Q' are injective?

what have you tried? ( use the fact that Q is the cokernel of the map i, i.e. "diagram chasing".)

what have you tried? ( use the fact that Q is the cokernel of the map i, i.e. "diagram chasing".)

First of all, I cannot use the fact that Q is the cokernel of the map because I do not know how he is even defining k to begin with. Are you allowed to take two objects in a diagram and say "let <foo> be a morphism between <ma> and <mi>" without giving evidence to suggest such a morphism should exist to begin with?

A cokernel has by definition a universal mapping property also, in this case we can state it more simply using the fact that as the cokernel of M-->E, Q is isomorphic to E/M. Hence to define a map out of Q, all we need is a map out of E that equals zero when restricted to M. Do you see such a map?

Last edited:
@TMO: I am not sure why you are reading Rotman, maybe as an assigned course book, but I want to suggest it may not be a good resource. I looked at it and found it very terse and hard to learn from myself. It seems to assume a lot of familiarity with basic module theory that it does not review thoroughly. I.e. before attempting homological algebra, I would recommend thorough grounding in module theory per se, especially submodules and quotient modules and their mapping properties, direct sums and direct products, hom and tensor products, as well as perhaps exterior products. Categorical fundamentals like Yoneda's lemma are also recommended, as well as practice using modules and module homomorphisms to prove concrete results like canonical forms of matrices over a field using structure of finitely generated modules over a polynomial ring.

My free web notes (for the course 845-1, 845-2, 845-3, with parts 1 and 3 doing the more general stuff), cover this stuff, spending 100-150 pages on the material, without yet beginning projectives and injectives or any more abstract topics. I cannot really recommend my own notes, since I am not objective, but i do offer them free.

http://alpha.math.uga.edu/~roy/

I tried for some time to find another source to recommend but could not find one that treated modules in a way I thought sufficient, but i will try a bit more. I do think there are some good sources out there on the web

...(later)

Well I have not found a free one, but the book Groups, Rings, and Modules, by Auslander and Buchsbaum, does at least treat basic properties of modules and module maps thoroughly. The property we need here is in theorem 4.8 pages 200-201.

Basically the cokernel of a module map f:M-->N, is a map g: N-->Q such that gf = 0, and is universal for that property, i.e. any other map h:N-->P such that hf = 0, factors through a unique map r:Q-->P, such that rg = h.

the point here is that the definition of a cokernel tells us that under certain conditions, there exists a map out of that cokernel. Thus it is custom made to answer a question like where does the map k come from? Thus I recommend geting a thorough grounding in this sort of thing before reading any book on homological algebra. And when the time comes I would choose maybe Northcott, or even Cartan - Eilenberg.

If you read somewhere a statement about the exactness properties of the functor Hom( , M), in particular that it transforms an exact sequence of form A-->B-->C-->0, into an exact sequence of form:

0-->Hom(C,M) -->Hom(B,M)-->Hom(A,M),

perhaps you can translate this into the statement we want, that under the given hypothesis, maps out of C are precisely those maps out of B that become zero as maps out of A. If this is not clear, I offer it as inducement to acquire thorough familiarity with the several "isomorphism theorems" as a solid foundation to study of homological algebra. good luck and godspeed.

Last edited:
well here is a link to the book of Robert Ash that I was looking for,

https://faculty.math.illinois.edu/~r-ash/Algebra.html

but in perusing it I was reminded of something that guided me when I wrote my notes, namely I tried not to say, in proving a theorem on modules e.g., "well this is exactly like the proof i gave for abelian groups so go back and look at that one and use that". I.e. I tried to give the proof over again, since I knew that the student did not really remember it as well as I did and needed it repeated. However, I am afraid that after a while I got tired of doing this myself, and began to think that the reader remembered every argument as well as I did at the time I was writing the book, and so did commit exactly this crime myself. So check it out if you wish. It may be that I only repeated those things that were hard for me, like unique factorization in polynomial rngs, but when it came to module arguments that I understood better I may have skimped. E.g. I wrote out the argument for the polynomials over the integers being a ufd, and then later wrote it out again for polynomials over any ufd. But I am not sure about say diagonalizing a matrix over a euclidean domain after writing it out for the integers. My excuse is that here the argument really is the same, whereas for the ufd case it is not, since there are more units in a general ufd than in the integers, hence the definition of a gcd is more complicated, so the whole argument is more complicated. anyway...