$\mathbb Q/\mathbb Z$ and $\mathbb R/\mathbb Z$ are injective cogenerators for $\mathbb{Z}\textbf{-Mod}$

Question

I wish to show that

The abelian groups $\mathbb Q/\mathbb Z$ and $\mathbb R/\mathbb Z$ are injective cogenerators for the category $\mathbb{Z}\textbf{-Mod}$.

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Recall that

An object $E$ in a category $\mathscr C$ is said to be a cogenerator of $\mathscr C$ if for any $f,g: X\to Y$, if $f\ne g$, then there exists $h: Y\to E$ such that $hf\ne hg$.

and

An object ${\displaystyle Q}$ in a category ${\displaystyle \mathscr {C}}$ is said to be injective if for every monomorphism ${\displaystyle f:X\to Y}$ and every morphism ${\displaystyle g:X\to Q}$ there exists a morphism ${\displaystyle h:Y\to Q}$ extending ${\displaystyle g}$ to ${\displaystyle Y}$, i.e. such that ${\displaystyle h\circ f=g}$.

First, let us work with $\mathbb Q/\mathbb Z$, and hopefully, the proof for $\mathbb R/\mathbb Z$ will be similar. Note that injective objects in $\mathbb{Z}\textbf{-Mod}$ are just the injective modules. Are $\mathbb Q/\mathbb Z$ and $\mathbb R/\mathbb Z$ injective $\Bbb Z$-modules?

Further, to show that $\mathbb Q/\mathbb Z$ is a cogenerator in $\mathbb{Z}\textbf{-Mod}$, suppose $f,g: X\to Y$ where $f\ne g$. How should I construct an $h: Y\to \Bbb Q/\Bbb Z$ such that $hf\ne hg$?

At this point, I'm very new to these definitions, and I'm unable to use them well. I'd appreciate any help and clarifications! I've also edited my question to focus it to $\mathbb{Z}\textbf{-Mod}$ instead of $\textbf{Ab}$, which is an isomorphic category (see @Mindlack's comments).

Answer 1

Let me try to give you a detailed proof. I will assume the following well-known facts about $\mathbb Z$:

1. $\mathbb Z$ is a PID, that is, $\mathbb Z$ is an integral domain and the ideals of $\mathbb Z$ are all principal (in fact, they are all of the form $n\mathbb Z$, for some $n\in \mathbb Z$).
2. The prime ideals of $\mathbb Z$ are the following: the trivial ideal $0=\{0\}$ and those of the form $p\mathbb Z$ for $p$ a prime. Furthermore, the prime ideals of the form $p\mathbb Z$ (for $p\neq 0$ a prime) are exactly the maximal ideals in $\mathbb Z$.
3. As consequence of part 2, the simple objects in $\mathrm{Mod}(\mathbb Z)$ are of the form $\mathbb Z(p):=\mathbb Z/p\mathbb Z$.

I will also need the following well-known fact (of which I'm including an almost complete proof):

BAER'S CRITERION (for PIDs): Let $R$ be a PID. An $R$-module $E$ is injective if and only if it is divisible, that is, if and only if, for all $0\neq r\in R$ and $e\in E$, there is $e'\in E$ such that $re'=e$.

Proof (sketch). If $E$ is injective, let $e\in E$ and $0\neq r\in R$. Then, there is a homomorphism $\phi\colon rR\to E$ mapping $rs\mapsto se$, for all $s\in R$. Consider now the obvious inclusion $rR\hookrightarrow R$ and, using injectivity, extend $\phi$ to a morphism $\phi'\colon R\to E$ such that $\phi'_{\restriction rR}=\phi$. Then, letting $e':=\phi'(1_R)$, we have that $$re'=r\phi'(1_R)=\phi'(r\cdot 1_R)=\phi(r\cdot 1_R)=1_R\cdot e=e,$$ showing that $E$ is divisible. For the converse, let $Y$ be an $R$-module and $X\leq Y$ a submodule. By transfinite induction one can construct a sequence $(X_\alpha)_{\alpha<\sigma}$ for some ordinal $\sigma$, with the following properties:

(P1) $X_0=X$, $X_\alpha<X_\beta\leq Y$ for all $\alpha<\beta<\sigma$, and $\bigcup_{\alpha<\sigma}X_\alpha=Y$;

(P2) if $\alpha<\alpha+1<\sigma$, then $X_{\alpha+1}=X_\alpha+Rx_\alpha$ for some $x_\alpha\in X_{\alpha+1}$;

(P3) if $\lambda<\sigma$ is a limit ordinal, then $X_\lambda=\bigcup_{\alpha<\lambda}X_\alpha$.

Suppose now that $E$ is divisible and take a morphism $\phi\colon X\to E$. To extend $\phi$ to a morphism $Y\to E$ one can proceed inductively: if $\alpha=0$, then $X_0=X$ and $\phi_0:=\phi$. If $\alpha<\alpha+1<\sigma$ and we already have $\phi_\alpha\colon X_\alpha\to E$, choose $x_\alpha\in X_{\alpha+1}$ such that $X_{\alpha+1}=X_\alpha+Rx_\alpha$ and verify that $I:=\{r\in R:rx_\alpha\in X_\alpha\}\leq R$ is an ideal, so that $I=s_\alpha R$ for some $s_\alpha\in R$ as $R$ is a PID. As $E$ is divisible, there is $e_\alpha\in E$ such that $s_\alpha e_\alpha=\phi_\alpha(s_\alpha x_\alpha)$ (in case $s_\alpha=0$, just take $e_\alpha=0$). One checks that $\phi_{\alpha+1}\colon X_{\alpha+1}\to E$ such that $\phi_{\alpha+1}(x+rx_\alpha):=\phi_\alpha(x)+re_\alpha$ is a well-defined homomorphism such that $(\phi_{\alpha+1})_{\restriction X_\alpha}=\phi_\alpha$. If $\lambda<\sigma$ is a limit, define $\phi_\lambda\colon X_\lambda\to E$ as $\phi_\lambda(x):=\phi_\alpha(x)$, where $\alpha<\lambda$ is the smallest ordinal for which $x\in X_\alpha\leq X_\lambda$; one checks that $\phi_\lambda$ is a well-defined homomorphism such that $(\phi_\lambda)_{\restriction X_\alpha}=\phi_\alpha$ for all $\alpha<\lambda$.

Finally, define a function $\psi\colon Y\to E$ as $\psi(y):=\phi_\alpha(y)$, where $\alpha<\sigma$ is the smallest ordinal for which $y\in X_\alpha$. To conclude one just needs to check that $\psi$ is a well-defined homomorphism: it is then clear by construction that $\psi_{\restriction X}=\phi$, as desired.\\\

As a consequence of the above result, one gets:

COROLLARY: Let $R$ be a PID. Then the following statements hold true:

(1) If $E$ is an injective $R$-module and $K\leq E$, the $E/K$ is injective (that is, quotients of injectives are injective);

(2) If $\{E_i:i\in I\}$ is a (possibly infinite) family of injectives, then $\bigoplus_IE_i$ is injective (direct sum of injectives is injectives).

Proof. (1): We have to show that $E/K$ is divisible. Indeed, given $e\in E$ and $0\neq r\in R$, use the divisibility of $E$ to get $e'\in E$ such that $re'=e$. This shows that, in $E/K$, $r\cdot (e'+K)=e+K$.

(2): We have to show that $\bigoplus_IE_i$ is divisible. Indeed, given $(e_i)_I\in \bigoplus_IE_i$ and $0\neq r\in R$, use, for each $i\in I$, the divisibility of $E_i$ to get $e_i'\in E_i$ such that $re_i'=e_i$. This shows that, in $\bigoplus_IE_i$, $r\cdot (e'_i)_I=(re_i')_I=(e_i)_I$.\\\

Specializing the above result to $R=\mathbb Z$, we obtain the following:

COROLLARY: An Abelian group $E$ (that is, a $\mathbb Z$-module), is injective if and only if it is divisible. In particular, $E$ is injective if and only if, for all $0\neq n\in \mathbb Z$ and $e\in E$, there exists $e'\in E$ such that $ne'=e$. Furthermore, the class of injective Abelian groups is closed under arbitrary quotients and direct sums.

Using the above description of the injective Abelian groups, we can easily deduce that both $\mathbb Q/\mathbb Z$ and $\mathbb R/\mathbb Z$ are injective:

EXAMPLE: The following are examples of injective Abelian groups:

• first of all, $\mathbb Q$ is clearly divisible and so, by the above corollary, it is also injective;
• by the above corollary, quotients of injective Abelian groups are injective and, therefore, also $\mathbb Q/\mathbb Z$ is injective;
• now consider $\mathbb R$. This is clearly divisible, so again it is injective and its quotient $\mathbb R/\mathbb Z$ is injective as well;
• we can also proceed slightly differently to show the injectivity of $\mathbb R$ and $\mathbb R/\mathbb Z$: one can see that $\mathbb R\cong \mathbb Q^{(|\mathbb R|)}$, that is, $\mathbb R$ is a $\mathbb Q$-vector space of dimension $|\mathbb R|$ (the dimension is forced by cardinality reasons). In particular, $\mathbb R$ is an infinite direct sum of copies of the injective Abelian group $\mathbb Q$ and it is therefore injective itself. As a consequence of this description, one can see that $$ \mathbb R/\mathbb Z\cong \mathbb Q/\mathbb Z\oplus \mathbb Q^{(|\mathbb R|)}\cong \mathbb Q/\mathbb Z\oplus\mathbb R $$ so $\mathbb R/\mathbb Z$ is a direct sum of two injective Abelian groups and, therefore, it is injective.

We now know all we need to known about injective Abelian groups and we can turn our attention to those injectives that are also cogenerators:

PROPOSITION: The following are equivalent for an injective Abelian group $E$:

(1) $E$ is a cogenerator;

(2) $\mathrm{Hom}(X,E)\neq 0$, whenever $X$ is a non-trivial Abelian group;

(3) $\mathrm{Hom}(\mathbb Z/n\mathbb Z,E)\neq 0$, for all $0\neq n\in \mathbb Z$, and $\mathrm{Hom}(\mathbb Z,E)\neq 0$;

(4) $\mathrm{Hom}(\mathbb Z(p),E)\neq 0$, for each prime $p\in \mathbb Z$;

(5) $\mathbb Z(p)\leq E$ for all $p\in \mathbb Z$ prime.

Proof. The implications ''(2)$\Rightarrow$(3)$\Rightarrow$(4)'' and ''(5)$\Rightarrow$(4)'' are trivial.

(1)$\Rightarrow$(2): If $X\neq 0$, then $\mathrm{id}_X\neq 0\colon X\rightrightarrows X$. Therefore, by definition of cogenerator, there exists $h\in \mathrm{Hom}(X,E)$ such that $h=h\circ \mathrm{id}_X\neq h\circ 0=0$. Hence, $\mathrm{Hom}(X,E)\neq 0$.

(4)$\Rightarrow$(5): note that any non-trivial homomorphism $f\colon\mathbb Z(p)\to E$ is injective. In fact, as $\mathbb Z(p)$ is simple, either $\mathrm{Ker}(f)=0$ or $\mathrm{Ker}(f)=\mathbb Z(p)$. As the latter option is excluded (because we have assumed that $f$ is non-trivial), we deduce that $\mathrm{Ker}(f)=0$, that is, $f$ is injective. But then, $\mathrm{Im}(f)\cong \mathbb Z(p)$ is a subgroup of $E$, as desired.

(5)$\Rightarrow$(3): given a positive integer $n$, write $n=p_1^{a_1}\cdot \ldots\cdot p_n^{a_n}$, where $p_1,\dots,p_n$ are distinct primes and $a_1,\dots, a_n$ are non-zero positive integers. Then, $n\mathbb Z\subseteq p_1\mathbb Z$ and, therefore, there is a non-trivial morphism $f\colon \mathbb Z/n\mathbb Z\to \mathbb Z(p_1)$ that sends $x+n\mathbb Z\mapsto x+p_1\mathbb Z$ (e.g., use the Third Isomorphism Theorem to deduce that $\mathbb Z(p_1)\cong(\mathbb Z/n\mathbb Z)/(p_1\mathbb Z/n\mathbb Z)$ is a quotient of $\mathbb Z/n\mathbb Z$, where $f$ is the canonical projection onto the quotient). Composing $f$ with the inclusion $\iota_{p_1}\colon\mathbb Z(p_1)\to E$, we deduce that $\mathrm{Hom}(\mathbb Z(n),E)\neq 0$ since it contains $\iota_{p_1}\circ f\neq 0$. Finally, to show that $\mathrm{Hom}(\mathbb Z,E)\neq 0$ just take a prime $p$, then the obvious projection $\pi_p\colon\mathbb Z\to \mathbb Z(p)$ is non-trivial, so $\iota_p\circ\pi_p\colon \mathbb Z\to E$ is non-trivial.

(3)$\Rightarrow$(2): Let $X$ be a non-trivial Abelian group, that is, there is $x\in X$ such that $x\neq 0$. Consider the subgroup $\mathbb Zx:=\{nx:n\in \mathbb Z\}\leq X$ generated by $x$ in $X$. The canonical group homomorphism $\pi_x\colon \mathbb Z\to \mathbb Zx$, sending $n\mapsto nx$ is clearly surjective and so $\mathbb Zx\cong \mathbb Z/\mathrm{Ker}(\pi_x)$ (by the First Isomorphism Theorem). In particular, as $\mathrm{Ker}(\pi_x)$ is a subgroup of $\mathbb Z$, we have that either $\mathrm{Ker}(\pi_x)=0$ or $\mathrm{Ker}(\pi_x)=n\mathbb Z$ for some $0\neq n\in \mathbb Z$. In the former case $\mathbb Zx\cong \mathbb Z$, while in the latter $\mathbb Zx\cong \mathbb Z/n\mathbb Z$. In either case, our hypothesis tells us that there is a non-trivial morphism $f\colon \mathbb Zx\to E$. Finally, since $E$ is injective, we can extend $f$ through the inclusion $\varepsilon_x\colon \mathbb Zx\to X$ to get a non-trivial morphism $f'\colon X\to E$ such that $f'\circ \varepsilon_x=f$. In particular, $\mathrm{Hom}(X,E)\neq 0$, as desired.

(2)$\Rightarrow$(1): Let $f\neq g\colon X\rightrightarrows Y$ be two different homomorphisms of Abelian groups and let us find some $h\colon Y\to E$ such that $h\circ f\neq h\circ g$. Indeed, consider the morphism $f-g\colon X\to Y$, sending $x\mapsto f(x)-g(x)$ and identify $\bar X:=\mathrm{Im}(f-g)\cong X/\mathrm{Ker}(f-g)$ with a subgroup of $Y$. Note that, since $f\neq g$, we have that $\mathrm{Ker}(f-g)\neq X$ and, therefore, $0\neq \bar X\leq Y$. By (2), there is a non-trivial morphism $h'\colon \bar X\to E$ and, using the injectivity of $E$, we can extend $h'$ to a morphism $h\colon Y\to E$ such that $h\restriction_{\bar X}\ =h'$. Let us verify that $h\circ f\neq h\circ g$, indeed, choose $\bar x\in \bar X$ such that $h'(\bar x)\neq 0$ (it exists as $h'\neq 0$ by construction) and note that $\bar x= (f-g)(x)$ for some $x\in X$. Thus, $$ h(f(x))-h(g(x))=h((f-g)(x))=h(\bar x)=h'(\bar x)\neq 0 $$ so, in particular, $h(f(x))\neq h(g(x))$ and $h\circ f\neq h\circ g$.\\\

Note that we have the following easy corollary of the above characterization:

COROLLARY: Let $E$ be an injective cogenerator in $\mathrm{Mod}(\mathbb Z)$. If $E'$ is an injective Abelian group that contains $E$ as a subgroup, then also $E'$ is an injective cogenerator.

Proof. One can use the equivalent characterizations given in the above proposition, e.g., note that each simple $\mathbb Z(p)\leq E\leq E'$ is a subgroup of $E'$.\\\

In particular, as $\mathbb Q/\mathbb Z\leq \mathbb R/\mathbb Z$, if we can prove that $\mathbb Q/\mathbb Z$ is an injective cogenerator, then the above corollary tells us that also $\mathbb R/\mathbb Z$ is an injective cogenerator. Finally, to see that $\mathbb Q/\mathbb Z$ is an injective cogenerator, it is enough to find a copy of each simple $\mathbb Z(p)$ embedded in $\mathbb Q/\mathbb Z$. For this, just note that: $$ \mathbb Z(p)\cong\{n/p:n\in \mathbb Z\}/\mathbb Z\leq \mathbb Q/\mathbb Z $$ To show the above isomorphism, consider the following homomorphism of Abelian groups: $$ \pi_p:\mathbb Z\to \mathbb Q/\mathbb Z\quad\text{such that}\quad \pi_p(n):=n/p+\mathbb Z. $$ Clearly, the image of $\pi_p$ is a subgroup $\mathrm{Im}(\pi_p)\leq \mathbb Q/\mathbb Z$ and, by the First Isomorphism Theorem, this subgroup is isomorphic to $\mathbb Z/\mathrm{Ker}(\pi_p)$. To conclude, just note that $$ \mathrm{Ker}(\pi_p)=\{n\in \mathbb Z:\pi_p(n)=0\}=\{n\in \mathbb Z:n/p\in \mathbb Z\}=\{n\in \mathbb Z:p|n\}=p\mathbb Z. $$

This concludes the proof that $\mathbb Q/\mathbb Z$ and $\mathbb R/\mathbb Z$ are both injective cogenerators in the category of Abelian groups, so answering your question. Let me just add a couple of final observations:

(Obs.1) It is true for a general ring $R$ that a right $R$-module $E$ is an injective cogenerator if and only if $E$ is injective and it contains a copy of all the simple right $R$-modules;

(Obs.2) For a general ring $R$, and for a right $R$-module $M$, let $E(M)$ be the injective envelope of $M$ (this is an injective module that contains $M$ as a submodule and which is "minimal", in a sense that can be made precise, with respect to this property). Of course, if $E$ is an injective module and $M\leq E$, then $E(M)$ is a summand of $E$. Therefore, one can reformulate (Obs.1) by saying that an injective right $R$-module $E$ is a cogenerator if, and only if, $E(S)$ is a summand of $E$, for every simple right $R$-module $S$;

(Obs.3) One can improve even further the above observations as follows: let $R$ be a ring and let $\{S_i:i\in I\}$ be a family of representatives of the isomorphism classes of simple right $R$-modules (e.g., one can take $\{R/\mathbb m:\mathbb m\text{ maximal right ideal of $R$}\}$). Then, an injective right $R$-module $E$ is a cogenerator if and only if $E(\bigoplus_IS_i)$ is a summand of $E$. For this reason, the right $R$-modules isomorphic to $E(\bigoplus_IS_i)$ are said to be minimal injective cogenerators. Note in particular that any other injective cogenerator $E$ is of the form $E\cong E_{min}\oplus E'$, where $E_{min}$ is a minimal injective cogenerator and $E'$ is any injective right $R$-module;

(Obs.4) If $R$ is a right Noetherian ring, then infinite direct sums of injectives are injective and, in the notation of (Obs.3), it is not difficult to se that $E(\bigoplus_IS_i)\cong \bigoplus_IE(S_i)$, that is, for a right Noetherian ring the minimal injective cogenerators are just those modules which are isomorphic to the direct sum of the injective envelopes of the simples;

(Obs.5) If $R$ is a commutative Noetherian ring, let $\mathrm{Spec}(R)$ and $\mathrm{MSpec}(R)$ be the sets of prime and of maximal ideals of $R$, respectively. It is then known that an injective $R$-module $E$ can be written uniquely, up to isomorphism, as $E\cong \bigoplus_{\mathbb p\in \mathrm{Spec}(R)}E(R/\mathbb p)^{(I_{\mathbb p}(E))}$. In particular, such an $E$ is an injective cogenerator if and only if $I_{\mathbb p}(E)\neq \emptyset$ for all $\mathbb p\in \mathrm{MSpec}(R)$. Furthermore, $E$ is a minimal injective cogenerator if, and only if, $E\cong \bigoplus_{\mathbb m\in \mathrm{MSpec}(R)}E(R/\mathbb m)$.

Finally, specializing the above observations to $R=\mathbb Z$, and noting that $\mathbb Z$ is commutative Noetherian, one can use the above observations to obtain all the injective cogenerators (up to isomorphism):

COROLLARY: For each prime $p$, consider the following subgroup of $\mathbb Q/\mathbb Z$: $$ \mathbb Z(p^\infty):=\{m/p^n:m\in \mathbb Z,n\in \mathbb N\}/\mathbb Z\leq \mathbb Q/\mathbb Z. $$ Then, $\mathbb Z(p^\infty)$ is the injective envelope of the simple $\mathbb Z(p)$. Furthermore, $\mathbb Q/\mathbb Z\cong \bigoplus_{p}\mathbb Z(p^{\infty})$ is a minimal injective cogenerator in $\mathrm{Mod}(\mathbb Z)$.

In general, any injective Abelian group $E$ can be written uniquely as $$ E\cong \mathbb Q^{(I_{0}(E))}\oplus \bigoplus \mathbb Z(p^{\infty})^{(I_p(E))} $$ and $E$ is an injective cogenerator if and only if $I_p(E)\neq \emptyset$ for all $p$ prime.

Let me conclude with the following two examples:

• $\mathbb Q$ and $\mathbb R\cong \mathbb Q^{(\mathbb R)}$ are both injective Abelian groups but they are not cogenerators (as $I_p(\mathbb Q)=\emptyset=I_p(\mathbb R)$ for all $p\neq 0$);
• both $\mathbb Q/\mathbb Z$ and $\mathbb R/\mathbb Z\cong \mathbb Q^{(\mathbb R)}\oplus \mathbb Q/\mathbb Z$ arę injective cogenerators, as they contain the minimal injective cogenerator $\mathbb Q /\mathbb Z$ as a summand. In fact, $|I_p(\mathbb Q/\mathbb Z)|=1=|I_p(\mathbb R/\mathbb Z)|$ for all $p$ prime (while $|I_0(\mathbb Q/\mathbb Z)|=0$ and $|I_0(\mathbb R/\mathbb Z)|=|\mathbb R|$).

To convince yourself that $\mathbb R$ (or $\mathbb Q$) is not a cogenerator in the category of Abelian groups (as claimed above), just take the two morphisms $0\colon \mathbb Z(p)\to \mathbb Z(p)$ and $\mathrm{id}_{\mathbb Z(p)}\colon \mathbb Z(p)\to \mathbb Z(p)$ for some prime $p$. Then, clearly $0\neq \mathrm{id}_{\mathbb Z(p)}$. On the other hand, $\mathrm{Hom}(\mathbb Z(p),\mathbb R)=0$ (and, for similar reasons, $\mathrm{Hom}(\mathbb Z(p),\mathbb Q)=0$), in fact, given a morphism $f\colon \mathbb Z(p)\to \mathbb R$, we have that $p x=0$ for all $x\in \mathrm{Im}(f)$ and, inside $\mathbb R$, $p x=0$ if and only if $x=0$, that is, $\mathrm{Im}(f)=0$, proving that the unique morphism $h\colon \mathbb Z(p)\to \mathbb R$ is the $0$-morphism $h=0$. But then,
$$ h\circ \mathrm{id}_{\mathbb Z(p)}=0=h\circ 0\qquad\text{ for all $h\in \mathrm{Hom}(\mathbb Z(p),\mathbb R)$}, $$ so that $\mathbb R$ (and, similarly, $\mathbb Q$) cannot be a cogenerator.