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I have not learned anything about 'injective objects', though I got familiar with divisible groups while I was trying to understading Let $G$ be any abelian group and $a\in{G}$. Show there exists a homomorphism $f:G\rightarrow{\mathbb{Q}/\mathbb{Z}}$ such that $f(a)\neq{0}$.. Can someone help me prove this statement just using 'basic' module theorems? Or maybe by explicitly defining a homomorphism from a $\mathbb{Z}$-module M to $\mathbb{Q}/\mathbb{Z}$? From the link, I had a hard time understanding that we can 'define' an arbitrary, nontrivial homomorphism from $<a>$ to $\mathbb{Q}/\mathbb{Z}$, and that we can extend to a subset $U$ containing $<a>$. Could someone add some details here? Thanks.

Edit: I have something further in mind; Is it true that $Hom(M,\mathbb{Q}/\mathbb{Z})\cong M$? If so why? Can we use this fact to prove the above? If it is true, I guess so, since we have assumed that M is not a zero set. I think that maybe by proving this for a cyclic subgroup $<a>$ for an arbitrary $a \in M$, then using something like the Fundamental Theorem for Abelian Groups. But my guess is that it is not true for 'any' abelian group, since the FTAG applies only for finite abelian groups. Any details would be appreciated!

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    $\begingroup$ Abelian groups and $\Bbb Z$-modules are the same thing. $\endgroup$ – ÍgjøgnumMeg Oct 11 '18 at 17:39
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    $\begingroup$ The proof requires the axiom of choice, so it's not generally possible to exhibit an explicit nonzero homomorphism $M\to\mathbb{Q}/\mathbb{Z}$. $\endgroup$ – egreg Oct 11 '18 at 17:53
  • $\begingroup$ @ÍgjøgnumMeg Yup, I'm aware of that. But I'm making no progress! $\endgroup$ – 김현빈 Oct 11 '18 at 18:00
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A basic module theorem is Baer’s lemma.

Baer’s lemma. Let $R$ be a ring and $E$ a left $R$-module. Then $K$ is injective if and only if, for every left ideal $I$ of $R$ and every homomorphism $f\colon I\to E$, there exist $g\colon R\to E$ such that $g|_I=f$.

Corollary. Every divisible $\mathbb{Z}$-module is injective.

In particular $\mathbb{Q}/\mathbb{Z}$ is injective; if $M$ is an abelian group (or $\mathbb{Z}$-module, it's the same thing) and $x\in M$, $x\ne0$, we can show that there exists $g\colon M\to\mathbb{Q}/\mathbb{Z}$ such that $g(x)\ne0$.

Indeed, $\mathbb{Z}x\ne0$, so it has a maximal submodule $L$ and the quotient $\mathbb{Z}x/L$ embeds in $\mathbb{Q}/\mathbb{Z}$. Call $f$ the composition of this embedding with the projection $\mathbb{Z}x\to\mathbb{Z}x/L$. Then $f(x)\ne0$. Since $\mathbb{Q}/\mathbb{Z}$ is injective, there exists $g\colon M\to\mathbb{Q}/\mathbb{Z}$ such that $g|_{\mathbb{Z}x}=f$.

You find the proof of Baer’s lemma in any book on module theory.

A sketch is as follows. One direction is obvious. For the other, suppose $L$ is a submodule of $M$ and take a homomorphism $f\colon L\to E$.

Consider the set $\mathscr{F}$ of all pairs $(K,k)$, where $K$ is a submodule of $M$ with $L\subseteq K$ and $k\colon K\to E$ satisfies $K|_L=f$. Order $\mathscr{F}$ by decreeing that $(K,k)\le(K',k')$ if and only if $K\subseteq K'$ and $k'|_{K}=k$.

Prove that $\mathscr{F}$ satisfies the hypotheses of Zorn's lemma. Take $(H,h)$ maximal in $\mathscr{F}$ and prove that $H=M$. Indeed if $H\ne M$, there is $y\in M$, $y\notin H$. Let $I=\{r\in R:ry\in H\}$. Then $I$ is a left ideal and $Iy$ is a submodule of $H$. This allows to build a homomorphism $I\to E$ and its extension to $R$ provides the contradiction.


Is it true that $\operatorname{Hom}(M,\mathbb{Q}/\mathbb{Z})\cong M$? Of course not, the case of a free module with a countable basis should be easy to see.

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