2
$\begingroup$

Consider two exact sequences $0\rightarrow N\rightarrow P\rightarrow A\rightarrow 0$ and $0\rightarrow M\rightarrow Q\rightarrow A\rightarrow 0$, where $P,Q$ are projective modules. The exercise(pg 26, 4.5) asked me to prove that $P\oplus M\cong Q\oplus N$. I do not know how to proceed.

Certainly the map $P\rightarrow A$ is surjective, and since we have the trivial map $A\rightarrow A$identify $A$ with itself, any map $Q\rightarrow A$ can be extended to the composition $Q\rightarrow P\rightarrow A$. Similarly any map $Q\rightarrow A$ can be extended to $P\rightarrow Q\rightarrow A$. I do not know how to proceed further from here; certainly the fact that any map $P\rightarrow A$ can be factored through $Q$ as $P\rightarrow Q\rightarrow P\rightarrow A$ is interesting, but this does not help me to prove the statement.

The author give the hint that I should look over Exercise 3.4(page 22), but I found it to be quite irrelevant as 3.4 was concerning a commutative diagram of two exact sequences, while this problem does not admit any arrow from $N$ to $M$ as only $P$ and $Q$ are supposed to be projective. So I got stuck. I think I need some help as the question is quite trivial.

Another trivial question is assume $A$ to be a finite abelian group, prove $\operatorname{Hom}_{\mathbb{Z}}(A,\mathbb{Q}/\mathbb{Z})$ is isomorphic to $A$. This would be trivial if I can prove it by $\mathbb{Z}$ and $\mathbb{Z}_{p^{n}}$ respectively. But plainly I do not see how $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z},\mathbb{Q}/\mathbb{Z})$ can be cyclic, since any such $\mathbb{Z}$ homomorphism must map $1$ to some element in $\mathbb{Q}/\mathbb{Z}$, and such choice can be totally arbitrarily among $\mathbb{Q}/\mathbb{Z}$. But we know that $\mathbb{Q}/\mathbb{Z}$ is not finitely generated, and it cannot be cyclic. I do not know where my reasoning got wrong, so I hope someone may help me point out my mistake and give me a simple proof on this.

  • 2
    The hint is not irrelevant at all: look up the proof of [Schanuel's lemma](http://en.wikipedia.org/wiki/Schanuel's_lemma) and find the short exact sequence they're talking about... A LaTeX thing: don't write `Hom$_{\mathbb{Z}}(...)$` rather use `$\operatorname{Hom}_{\mathbb{Z}}$`2011-12-22
  • 2
    The content of the exercise *is* Schanuel's lemma :)2011-12-22
  • 0
    By projectivity of $P$ you have a homomorphism $P\to Q$. Then you can always find a homomorphism $N\to M$ and thus extend the diagram while preserving the commutativity.2011-12-22
  • 0
    @Damian: Well, that's true but it won't help proving Schanuel's lemma.2011-12-22
  • 0
    Okay, this extension homomorphism $\alpha':N\to M$ can be found as follows. Let $x\in N$. Then $\mu(x)\in\text{ker}(\epsilon)$, where $\mu:N\to P,\epsilon:P\to A$. By commutativity of the diagram: $\alpha(\mu(x))\in\text{ker}(\epsilon')$, where $\alpha:P\to Q,\epsilon':Q\to A$. So there is a unique $y\in M: \mu'(y)=\alpha(\mu(x))$, where $\mu':M\to Q$. Now take $\alpha'(x):=y$. You should check that $\alpha'$ is a homomorphism.2011-12-22
  • 0
    @commenter: it will. There is a hint to use another exercise which joins these two exact sequences with another one, from which the Schanuel's lemma follows easily.2011-12-22
  • 0
    @Damian: no, that's not what the hint is about. You should take the pull-back of $P \to A$ and $Q \to A$. Then you get a commutative square with corner $X$. The two maps $X \to P $ and $X \to Q$ are both surjective, hence they split and their kernels are $M$ and $N$, respectively (that's what the exercise says). This then shows that $X$ is isomorphic to $P \oplus M$ and $Q \oplus N$.2011-12-22
  • 0
    @commenter: are you talking about hints and exercises from the book of Hilton & Stammbach? I am just right now looking into the book and the exercise is about obtaining some concrete exact seqence from these two beginning exact sequences (ex. 4.5 with the hint: ex. 3.4 following 3.3). But in general it is similar to your way of reasoning.2011-12-22
  • 0
    @Damian: I am, but I mistook exercise 3.3 for exercise 3.4, that's why I was so insistent. Now I understood how your argument works, nice! I haven't seen that way of putting it, I always did it the other way around. Apologies for the somewhat harshly formulated previous comments...2011-12-22
  • 0
    @commenter: it's ok, they weren't harsh. Greetz!2011-12-22
  • 0
    For the second problem: $\mathbb{Z}$ is not a finite abelian group, so the conclusion doesn't need to hold for it (in fact, $\operatorname{hom}_{\mathbb{Z}}(\mathbb{Z}, \mathbb{Q/Z})$ is isomorphic to $\mathbb{Q/Z}$). As to how to prove the result, note that $\operatorname{hom}$ commutes with finite coproducts (*technically the result is a product, but products/coproducts coincide in the finite case...). Coupling this with the structure theorem for finite abelian groups reduces the problem to considering only cyclic groups (of prime power order).2011-12-22
  • 0
    @Riley E: It is embarassing that I ignored the "finite" condition in the statement..thank you.2011-12-23
  • 0
    @ChangweiZhou: A cool thing to note is that, although the finiteness condition is necessary for the exercise to be true, ignoring it _does_ lead to some interesting things. For example, $\operatorname{hom}_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Q/Z}) \simeq \mathbb{R}$. More generally, $\operatorname{hom}_{\mathbb{Z}}(-,\mathbb{Q/Z})$ tells a lot about $\operatorname{Ext}_{\mathbb{Z}}(-,\mathbb{Z})$ (cf. section III.6)2011-12-23
  • 0
    @Riley E: I can see the cardinality of Hom$_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q}/\mathbb{Z})$ be $c$, but I could not tell why it is isomorphic to $\mathbb{R}$. I have not reached section III yet..:(2011-12-24
  • 0
    @ChangweiZhou: The (unfortunately unsatisfying) reason is that $\operatorname{hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q/Z})$ is a $\mathbb{Q}$-module (it's both divisible and torsionfree) that has cardinality $c$.2011-12-24

2 Answers 2

2

If you want find the detail of proof of Schanuel's lemma, you can see the note:

http://www.math.ku.edu/~dlk/Lecture07.pdf (by Daniel Katz)

Do the dual version of Schanuel's lemma ( $i.e.$ injective version) to make sure you really go through the proof of Schanuel's lemma.

Note that for injective version, you can copy all steps in projective version. The only one step you need to be careful is the construction of module homomorphism ("+" or "-").

1

Using exercise 3.4, we can get $0\to N\to P\oplus M\to Q\to 0$. $Q$ is projective, so we get the exact sequence splits and hence the result.