nLab
injective module

Definition

For $R$ a ring, let $R$Mod be the category of $R$-modules.

An injective module over $R$ is an injective object in $R\mathrm{Mod}$.

This is the dual notion of a projective module.

Properties

Equivalent characterizations

Let $R$ be a commutative ring and $C=R\mathrm{Mod}$ the category of $R$-modules. We discuss injective modules over $R$ (see there for more).

Proposition

(Baer's criterion)

If the axiom of choice holds, then a module $Q\in R\mathrm{Mod}$ is an injective module precisely if for $I$ any left $R$-ideal regarded as an $R$-module, any homomorphism $g:I\to Q$ in $C$ can be extended to all of $R$ along the inclusion $I\hookrightarrow R$.

Sketch of proof

Let $i:M\hookrightarrow N$ be a monomorphism in $R\mathrm{Mod}$, and let $f:M\to Q$ be a map. We must extend $f$ to a map $h:N\to Q$. Consider the poset whose elements are pairs $(M\prime ,f\prime )$ where $M\prime$ is an intermediate submodule between $M$ and $N$ and $f\prime :M\prime \to Q$ is an extension of $f$, ordered by $(M\prime ,f\prime )\le (M″,f″)$ if $M″$ contains $M\prime$ and $f″$ extends $f\prime$. By an application of Zorn's lemma, this poset has a maximal element, say $(M\prime ,f\prime )$. Suppose $M\prime$ is not all of $N$, and let $x\in N$ be an element not in $M\prime$; we show that $f\prime$ extends to a map $M″=⟨x⟩+M\prime \to Q$, a contradiction.

The set $\{r\in R:rx\in M\prime \}$ is an ideal $I$ of $R$, and we have a module homomorphism $g:I\to Q$ defined by $g(r)=f\prime (rx)$. By hypothesis, we may extend $g$ to a module map $k:R\to Q$. Writing a general element of $M″$ as $rx+y$ where $y\in M\prime$, it may be shown that

$$f″(rx+y)=k(r)+g(y)$$f''(r x + y) = k(r) + g(y)

is well-defined and extends $f\prime$, as desired.

Corollary

Assume that the axiom of choice holds.

Let $R$ be a Noetherian ring, and let $\{Q_j{\}}_{j\in J}$ be a collection of injective modules over $R$. Then the direct sum $Q={⨁}_{j\in J}{Q}_j$ is also injective.

Proof

By Baer’s criterion, it suffices to show that for any ideal $I$ of $R$, a module homomorphism $f:I\to Q$ extends to a map $R\to Q$. Since $R$ is Noetherian, $I$ is finitely generated as an $R$-module, say by elements $x_1,\dots ,x_n$. Let $p_j:Q\to Q_j$ be the projection, and put $f_j=p_j\circ f$. Then for each $x_i$, $f_j(x_i)$ is nonzero for only finitely many summands. Taking all of these summands together over all $i$, we see that $f$ factors through

$$\prod _{j\in J\prime }{Q}_j=\underset{j\in J\prime }{⨁}{Q}_j\hookrightarrow Q$$\prod_{j \in J'} Q_j = \bigoplus_{j \in J'} Q_j \hookrightarrow Q

for some finite $J\prime \subset J$. But a product of injectives is injective, hence $f$ extends to a map $R\to {\prod }_{j\in J\prime }{Q}_j$, which completes the proof.

Proposition

Conversely, $R$ is a Noetherian ring if direct sums of injective $R$-modules are injective.

This is due to Bass and Papp. See (Lam, Theorem 3.46).

Existence of enough injectives

We discuss that in the presence of the axiom of choice at least, the category $R$Mod has enough injectives in that every module is a submodule of an injective one. We first consider this for $R=\mathbb{Z}$. We do assume prop. 4, which may be proven using Baer's criterion.

Proposition

Assuming the axiom of choice, the category $\mathbb{Z}$Mod $\simeq$ Ab has enough injectives.

Proof

By prop. 4 an abelian group is an injective $\mathbb{Z}$-module precisely if it is a divisible group. So we need to show that every abelian group is a subgroup of a divisible group.

To start with, notice that the group $\mathbb{Q}$ of rational numbers is divisible and hence the canonical embedding $\mathbb{Z}\hookrightarrow \mathbb{Q}$ shows that the additive group of integers embeds into an injective $\mathbb{Z}$-module.

Now by the discussion at projective module every abelian group $A$ receives an epimorphism $({\oplus }_{s\in S}\mathbb{Z})\to A$ from a free abelian group, hence is the quotient group of a direct sum of copies of $\mathbb{Z}$. Accordingly it embeds into a quotient $\tilde{A}$ of a direct sum of copies of $\mathbb{Q}$.

$$\begin{array}{ccc}\ker & \stackrel{=}{\to }& \ker \\ \downarrow & & \downarrow \\ ({\oplus }_{s\in S}\mathbb{Z})& \hookrightarrow & ({\oplus }_{s\in S}\mathbb{Q})\\ \downarrow & & \downarrow \\ A& \hookrightarrow & \tilde{A}\end{array}$$\array{ ker &\stackrel{=}{\to}& ker \\ \downarrow && \downarrow \\ (\oplus_{s \in S} \mathbb{Z}) &\hookrightarrow& (\oplus_{s \in S} \mathbb{Q}) \\ \downarrow && \downarrow \\ A &\hookrightarrow& \tilde A }

Here $\tilde{A}$ is divisible because the direct sum of divisible groups is again divisible, and also the quotient group of a divisible groups is again divisble. So this exhibits an embedding of any $A$ into a divisible abelian group, hence into an injective $\mathbb{Z}$-module.

Proposition

Assuming the axiom of choice, for $R$ a ring, the category $R$Mod has enough injectives.

The proof uses the following lemma.

Write $U:R\mathrm{Mod}\to \mathrm{Ab}$ for the forgetful functor that forgets the $R$-module structure on a module $N$ and just remembers the underlying abelian group $U(N)$.

Lemma

The functor $U:R\mathrm{Mod}\to \mathrm{Ab}$ has a right adjoint

$$R_{*}:\mathrm{Ab}\to R\mathrm{Mod}$$R_* : Ab \to R Mod

given by sending an abelian group $A$ to the abelian group

$$U(R_{*}(A))≔\mathrm{Ab}(U(R),A)$$U(R_*(A)) \coloneqq Ab(U(R),A)

equipped with the $R$-module struture by which for $r\in R$ an element $(U(R)\stackrel{f}{\to }A)\in U(R_{*}(A))$ is sent to the element $rf$ given by

$$rf:r\prime \mapsto f(r\prime \cdot r).$$r f : r' \mapsto f(r' \cdot r) \,.

This is called the coextension of scalars along the ring homomorphism $\mathbb{Z}\to R$.

The unit of the $(Udash{R}_{*})$ adjunction

$${ϵ}_N:N\to R_{*}(U(N))$$\epsilon_N : N \to R_*(U(N))

is the $R$-module homomorphism

$${ϵ}_N:N\to {\mathrm{Hom}}_{\mathrm{Ab}}(U(R),U(N))$$\epsilon_N : N \to Hom_{Ab}(U(R), U(N))

given on $n\in N$ by

$$j(n):r\mapsto rn.$$j(n) : r \mapsto r n \,.

Proof

of prop. 3

Let $N\in R\mathrm{Mod}$. We need to find a monomorphism $N\to \tilde{N}$ such that $\tilde{N}$ is an injective $R$-module.

By prop. 2 there exists a monomorphism

$$i:U(N)\hookrightarrow D$$i \colon U(N) \hookrightarrow D

of the underlying abelian group into an injective abelian group $D$.

Now consider the adjunct $N\to R_{*}(D)$ of $i$, hence the composite

$$N\stackrel{{\eta }_N}{\to }{R}_{*}(U(N))\stackrel{R_{*}(i)}{\to }{R}_{*}(D)$$N \stackrel{\eta_N}{\to} R_*(U(N)) \stackrel{R_*(i)}{\to} R_*(D)

with $R_{*}$ and ${\eta }_N$ from lemma 1. On the underlying abelian groups this is

$$U(N)\stackrel{U({\eta }_N)}{\to }{\mathrm{Hom}}_{\mathrm{Ab}}(U(R),U(N))\stackrel{{\mathrm{Hom}}_{\mathrm{Ab}}(U(R),i)}{\to }{\mathrm{Hom}}_{\mathrm{Ab}}(U(R),U(D)).$$U(N) \stackrel{U(\eta_N)}{\to} Hom_{Ab}(U(R), U(N)) \stackrel{Hom_{Ab}(U(R),i)}{\to} Hom_{Ab}(U(R),U(D)) \,.

Once checks on components that this is a monomorphism. Therefore it is now sufficient to see that ${\mathrm{Hom}}_{\mathrm{Ab}}(U(R),U(D))$ is an injective $R$-module.

This follows from the existence of the adjunction isomorphism given by lemma 1

$${\mathrm{Hom}}_{\mathrm{Ab}}(U(K),U(D))\simeq {\mathrm{Hom}}_{R\mathrm{Mod}}(K,{\mathrm{Hom}}_{\mathrm{Ab}}(U(R),U(D)))$$Hom_{Ab}(U(K),U(D)) \simeq Hom_{R Mod}(K, Hom_{Ab}(U(R), U(D)))

natural in $K\in R\mathrm{Mod}$ and from the injectivity of $D\in \mathrm{Ab}$.

$$\begin{array}{ccc}U(K)& \to & D\\ \downarrow & \nearrow \\ U(L)\end{array}\leftrightarrow \begin{array}{ccc}K& \to & R_{*}D\\ \downarrow & \nearrow \\ L\end{array}.$$\array{ U(K) &\to& D \\ \downarrow & \nearrow \\ U(L) } \;\;\;\;\; \leftrightarrow \;\;\;\;\; \array{ K &\to& R_*D \\ \downarrow & \nearrow \\ L } \,.

Examples

Injective $\mathbb{Z}$-modules / abelian groups

Let $C=\mathbb{Z}\mathrm{Mod}\simeq$ Ab be the abelian category of abelian groups.

Proposition

An abelian group $A$ is injective as a $\mathbb{Z}$-module precisely if it is a divisible group, in that for all integers $n\in \mathbb{N}$ we have $nG=G$.

Using Baer’s criterion, prop. 1.

Example

By prop. 4 the following abelian groups are injective in Ab.

The group of rational numbers $\mathbb{Q}$ is injective in Ab, as is the additive group of real numbers $ℝ$ and generally that underlying any field. The additive group underlying any vector space is injective. The quotient of any injective group by any other group is injective.

Example

Not injective in Ab is for instance the cyclic group $\mathbb{Z}/n\mathbb{Z}$ for $n>1$.

Related concepts

• projective object, projective presentation, projective cover, projective resolution

  • projective module

• injective object, injective presentation, injective envelope, injective resolution

  • injective module

• free object, free resolution

  • free module

• flat object, flat resolution

  • flat module

References

The notion of injective modules was introduced in

• R. Baer (1940)

(The dual notion of projective modules was considered explicitly only much later.)

A general discussion can be found in

• Masaki Kashiwara, Pierre Schapira, Categories and Sheaves

The general notion of injective objects is in section 9.5, the case of injective complexes in section 14.1.

Baer’s criterion is discussed in many texts, for example

• N. Jacobsen, Basic Algebra II, W.H. Freeman and Company, 1980.

See also

• T.-Y. Lam, Lectures on modules and rings, Graduate Texts in Mathematics 189, Springer Verlag (1999).

Section 4.2 of

• Peter May, Notes on Tor and Ext (pdf)