# Two trivial questions on projective/injective modules from Hilton-Stammbach

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.

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The hint is not irrelevant at all: look up the proof of Schanuel's lemma and find the short exact sequence they're talking about... A LaTeX thing: don't write Hom$_{\mathbb{Z}}(...)$ rather use $\operatorname{Hom}_{\mathbb{Z}}$ –  t.b. Dec 22 '11 at 15:12
The content of the exercise is Schanuel's lemma :) –  Mariano Suárez-Alvarez Dec 22 '11 at 15:18
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. –  Damian Sobota Dec 22 '11 at 15:31
@Damian: Well, that's true but it won't help proving Schanuel's lemma. –  commenter Dec 22 '11 at 15:43
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. –  Damian Sobota Dec 22 '11 at 15:50

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 "-").

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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.

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