8
$\begingroup$

1) Prove that $\operatorname{Hom}_{\Bbb{Z}}(\Bbb{Q},\Bbb{Z}) = 0$.

2) Show that $\Bbb{Q}$ is not a projective $\Bbb{Z}$-module.

1) We know that $\Bbb{Q}$ is an injective $\Bbb{Z}$-module.

This implies that every short exact sequence $$0 \rightarrow \Bbb{Q} \xrightarrow{f} A \xrightarrow{g} B \rightarrow 0$$ is split exact.

In particular,

$$0 \rightarrow \Bbb{Q} \xrightarrow{f} \Bbb{Z} \xrightarrow{g} A \rightarrow 0$$

is split exact.

But this implies that we have a homomorphism $k: \Bbb{Z} \rightarrow \Bbb{Q}$ such that $kf = 1_{\Bbb{Q}}$.

So we need to look at what possible homomorphisms we could have from $\Bbb{Z}$ to $\Bbb{Q}$. Since $\Bbb{Z}$ is cyclic, the homomorphism $k$ is determined by where it sends $1$. Suppose that we send $1$ to $q \in \Bbb{Q}$ such that $q \not= 1$ and $q \not= 0$. Let $m \in \Bbb{Z}$. Then

$$k(m) = mk(1) = mq$$

But then

$$qmq = k(1)k(m) \not= k(m) = mq$$

So either $1 \in \Bbb{Z}$ must be sent to $1 \in \Bbb{Q}$ or $k$ can be the trivial homomorphism.

But if $1_{\Bbb{Z}}$ is sent to $1_{\Bbb{Q}}$, then $k$ must be the inclusion map. However in order to for $kf = 1_{\Bbb{Q}}$ to hold, $f$ must also sent $z$ to $z$ for all $z \in \Bbb{Q}$. But if that's the case, then for any $a/b \in \Bbb{Q}$ where $b \not= 0, 1$, we have (for $n \in \Bbb{Z}$) $$f(a/b) = n = f(n)$$

and this contradicts the fact that $f$ must be injective (since the sequence is exact).

So the only possible $\Bbb{Z}$-module map from $\Bbb{Z}$ to $\Bbb{Q}$ is $0$.

Do you think my answer is correct?

2) I was wondering if anybody could give a hint on this one, because I couldn't really get started.

Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ Hint: Take a [projective basis][1]. See t.b's answer for more detail. [1]: math.stackexchange.com/questions/22395/… $\endgroup$ – Bombyx mori Sep 26 '13 at 21:40
  • 4
    $\begingroup$ For the first let $f$ be a homomorphism from $\mathbb{Q}$ to $\mathbb{Z}$. Let $q\in \mathbb{Q}$, and let $f(q)=x$. Let $n$ be a positive integer big enough so that $n\not\mid x$. Then, $x=f(q)=f(q/n+...+q/n)=nf(q/n)$, but since $n\not\mid x$, we have that $f(q/n)=0$ and $x=0$. Hence, $f$ is the $0$ map. $\endgroup$ – Daniel Montealegre Sep 26 '13 at 21:48
11
$\begingroup$

1) I think you made this much harder than it has to be. $\mathbb{Q}$ is a divisible $\mathbb{Z}$-module. This implies that if $f:\mathbb{Q}\to\mathbb{Z}$, then $f(\mathbb{Q})$ is a divisible $\mathbb{Z}$-submodule of $\mathbb{Z}$. How many of those do you know?

2) Hint: PIDs have the nice property that a submodule of a free module is free. With this, what can you say about projective $\mathbb{Z}$-modules? Namely note that they are all (BLANK). After you've done this, explain why $\mathbb{Q}$-can't be (BLANK) by 1) and thus can't be projective.

$\endgroup$
  • 2
    $\begingroup$ Thanks a lot. There is a theorem in Hungerford that says that P is projective if and only if "there is a free module $F$ and an $R$-module $K$ such that $F \cong K \oplus P$. So if M is a projective $\Bbb{Z}$-module, then it needs to be a submodule of a free $\Bbb{Z}$-module. Since $\Bbb{Z}$ is a PID, it follows that $M$ must be free. So all projective $\Bbb{Z}$-modules are free. I'll continue below. $\endgroup$ – user58289 Sep 26 '13 at 22:54
  • 1
    $\begingroup$ Now assume, for contradiction, that $\Bbb{Q}$ is a projective $\Bbb{Z}$-module. Then it must be a submodule of a free $\Bbb{Z}$-module by the theorem from Hungerford, which means that $\Bbb{Q}$ must be free as well. But it isn't, and so we have a contradiction. Do you think my answer is correct? $\endgroup$ – user58289 Sep 26 '13 at 22:57
  • 2
    $\begingroup$ @Artus It's not free precisely because if $A$ is free and non-zero, and $B$ non-zero then $\text{Hom}(A,B)$ is non-zero! $\endgroup$ – Alex Youcis Sep 26 '13 at 22:58
  • 2
    $\begingroup$ @Artus You didn't say why it wasn't free, you just said it isn't. There are, in fact, many ways to show that $\mathbb{Q}$ is not free, one of them is part a). $\endgroup$ – Alex Youcis Sep 26 '13 at 23:03
  • 1
    $\begingroup$ @Artus Yes, but you already have! You just need to remark why 1) shows this! $\endgroup$ – Alex Youcis Sep 26 '13 at 23:05
4
$\begingroup$

2) If $\mathbb{Q}$ is a projective $\mathbb{Z}$-module, then it is direct summand of a free module, which is isomorphic to $\mathbb{Z}^{(I)}:=\oplus_{i\in I}\mathbb{Z}$; for some set $I$.

In this case there is a injective homomorphism $\iota:\mathbb{Q}\hookrightarrow \mathbb{Z}^{(I)}$. But note that, for each $i\in I$, if $\pi_i:\mathbb{Z}^{(I)}\to\mathbb{Z}$ is the $i$-ith projection, $\pi_i\iota:\mathbb{Q}\to\mathbb{Z}$ is a homomorphism and, therefore it is the null homomorphism.

Thus, for any $x\in \mathbb{Q}$, $\pi_{i}\iota(x)=0$ for every $i\in I$. Then $\iota(x)=0$. Therefore $\iota$ is also the null map - which contradicts the hypothesis that it is a injective homomorphism.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy