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 $R Mod$.

This is the dual notion of a projective module.

Properties

Equivalent characterizations

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

Theorem

(Baer's criterion)

If the axiom of choice holds, then a module $Q \in R Mod$ 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 $i \colon M \hookrightarrow N$ be a monomorphism in $R Mod$, and let $f \colon M \to Q$ be a map. We must extend $f$ to a map $h \colon N \to Q$. Consider the poset whose elements are pairs $(M', f')$ where $M'$ is an intermediate submodule between $M$ and $N$ and $f' \colon M' \to Q$ is an extension of $f$, ordered by $(M', f') \leq (M'', f'')$ if $M''$ contains $M'$ and $f''$ extends $f'$. By an application of Zorn's lemma, this poset has a maximal element, say $(M', f')$. Suppose $M'$ is not all of $N$, and let $x \in N$ be an element not in $M'$; we show that $f'$ extends to a map $M'' = \langle x \rangle + M' \to Q$, a contradiction.

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

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

is well-defined and extends $f'$, 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 $Q = \bigoplus_{j \in J} Q_j$ is also injective.

Proof

By Baer’s criterion, it suffices to show that for any ideal $I$ of $R$, a module homomorphism $f \colon I \to Q$ extends to a map $R \to Q$. Since $R$ is Noetherian, $I$ is finitely generated as an $R$-module, say by elements $x_1, \ldots, x_n$. Let $p_j \colon Q \to Q_j$ 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

$$\prod_{j \in J'} Q_j = \bigoplus_{j \in J'} Q_j \hookrightarrow Q$$

for some finite $J' \subset J$. But a product of injectives is injective, hence $f$ extends to a map $R \to \prod_{j \in J'} Q_j$, 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).

Preservation of injectives

Lemma

Given a pair of additive adjoint functors

$$(L \dashv R) \;\colon\; \mathcal{B} \stackrel{\overset{L}{\longleftarrow}}{\underset{R}{\longrightarrow}} \mathcal{A}$$

between abelian categories such that the left adjoint $L$ is an exact functor, then the right adjoint preserves injective objects.

Proof

Observe that an object in an abelian category is injective precisely if the hom-functor into it sends monomorphisms to epimorphisms (prop.), and that $L$ preserves monomorphisms by assumption of exactness. With this the statement follows via adjunction isomorphism

$$Hom_{\mathcal{A}}(-,R(I))\simeq Hom_{\mathcal{B}}(L(-),I) \,.$$

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. , 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. 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 $\mathbb{Q}$.

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

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.

Lemma

Given a pair of additive adjoint functors

$$(L \dashv R) \;\colon\; \mathcal{B} \stackrel{\overset{L}{\longleftarrow}}{\underset{R}{\longrightarrow}} \mathcal{A}$$

between abelian categories such that the left adjoint $L$ is

1. an exact functor,

2. a faithful functor.

Then if $\mathcal{B}$ has enough injectives, also $\mathcal{A}$ has enough injectives.

Proof

Consider $A \in \mathcal{A}$. By the assumption that $\mathcal{B}$ has enough injectives, there is an injective object $I \in \mathcal{B}$ and a monomorphism $i \colon L(A) \hookrightarrow I$. The adjunct of this is a morphism

$$\tilde i \colon A \longrightarrow R(I)$$

and so it is sufficient to show that

1. $R(I)$ is injective in $\mathcal{A}$;

2. $\tilde i$ is a monomorphism.

The first point is the statement of lemma .

For the second point, consider the kernel of $\tilde i$ as part of the exact sequence

$$ker(\tilde i)\longrightarrow A \stackrel{\tilde i}{\longrightarrow} R(I) \,.$$

By the assumption that $L$ is an exact functor, the image of this sequence under $L$ is still exact

$$L(ker(\tilde i)) \longrightarrow L(A) \stackrel{L(\tilde i)}{\longrightarrow} L(R(I)) \,.$$

Now observe that $L(\tilde i)$ is a monomorphism: this is because its composite $L(A) \stackrel{L(\tilde i)}{\longrightarrow} L(R(I)) \stackrel{\epsilon}{\longrightarrow} I$ with the adjunction unit is (by the formula for adjuncts) the original morphism $i$, which by construction is a monomorphism. Therefore the exactness of the above sequence means that $L(ker(\tilde i)) \to L(A)$ is the zero morphism; and by the assumption that $L$ is a faithful functor this means that already $ker(\tilde i) \to A$ is zero, hence that $ker(\tilde i) = 0$, hence that $\tilde i$ is a monomorphism.

Proposition

As soon as the category Ab of abelian groups has enough injectives, so does the abelian category $R$Mod of modules over some ring $R$.

In particular if the axiom of choice holds, then $R Mod$ has enough injectives.

Proof

Observe that the forgetful functor $U\colon R Mod \to AbGp$ has both a left adjoint $R_!$ (extension of scalars from $\mathbb{Z}$ to $R$) and a right adjoint $R_*$ (coextension of scalars). Since it has a left adjoint, it is exact. Thus the statement follows via lemma from prop. .

Proposition

For $R = k$ a field, hence $R$Mod = $k$Vect, every object is both injective as well as projective.

Examples

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

Let $C = \mathbb{Z} Mod \simeq$ 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 $n G = G$.

Using Baer’s criterion, prop. .

Example

By prop. 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 $\mathbb{R}$ 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 $n \gt 1$.

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 module was introduced in

• R. Baer, Abelian groups that are direct summands of every containing abelian group , Bulletin AMS 46 no. 10 (1940) pp.800-806. (projecteuclid)

(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)

For abelian sheaves over the etale site:

• James Milne, section 8 of Lectures on Étale Cohomology

A study of injective modules in higher algebra:

• Liran Shaul, Injective DG-modules over non-positive DG-rings arxiv