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injective module

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Definition

For RR a ring, let RRMod be the category of RR-modules.

An injective module over RR is an injective object in RModR Mod.

This is the dual notion of a projective module.

Properties

Equivalent characterizations

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

Theorem

(Baer's criterion)

If the axiom of choice holds, then a module QRModQ \in R Mod is an injective module precisely if for II any left RR-ideal regarded as an RR-module, any homomorphism g:IQg : I \to Q in CC can be extended to all of RR along the inclusion IRI \hookrightarrow R.

Sketch of proof

Let i:MNi \colon M \hookrightarrow N be a monomorphism in RModR Mod, and let f:MQf \colon M \to Q be a map. We must extend ff to a map h:NQh \colon N \to Q. Consider the poset whose elements are pairs (M,f)(M', f') where MM' is an intermediate submodule between MM and NN and f:MQf' \colon M' \to Q is an extension of ff, ordered by (M,f)(M,f)(M', f') \leq (M'', f'') if MM'' contains MM' and ff'' extends ff'. By an application of Zorn's lemma, this poset has a maximal element, say (M,f)(M', f'). Suppose MM' is not all of NN, and let xNx \in N be an element not in MM'; we show that ff' extends to a map M=x+MQM'' = \langle x \rangle + M' \to Q, a contradiction.

The set {rR:rxM}\{r \in R: r x \in M'\} is an ideal II of RR, and we have a module homomorphism g:IQg \colon I \to Q defined by g(r)=f(rx)g(r) = f'(r x). By hypothesis, we may extend gg to a module map k:RQk \colon R \to Q. Writing a general element of MM'' as rx+yr x + y where yMy \in M', 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 ff', as desired.

Corollary

Assume that the axiom of choice holds.

Let RR be a Noetherian ring, and let {Q j} jJ\{Q_j\}_{j \in J} be a collection of injective modules over RR. Then the direct sum Q= jJQ jQ = \bigoplus_{j \in J} Q_j is also injective.

Proof

By Baer’s criterion, it suffices to show that for any ideal II of RR, a module homomorphism f:IQf \colon I \to Q extends to a map RQR \to Q. Since RR is Noetherian, II is finitely generated as an RR-module, say by elements x 1,,x nx_1, \ldots, x_n. Let p j:QQ jp_j \colon Q \to Q_j be the projection, and put f j=p jff_j = p_j \circ f. Then for each x ix_i, f j(x i)f_j(x_i) is nonzero for only finitely many summands. Taking all of these summands together over all ii, we see that ff factors through

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

for some finite JJJ' \subset J. But a product of injectives is injective, hence ff extends to a map R jJQ jR \to \prod_{j \in J'} Q_j, which completes the proof.

Proposition

Conversely, RR is a Noetherian ring if direct sums of injective RR-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

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

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

Proof

Observe that an object is injective precisely if the hom-functor into it sends monomorphisms to epimorphisms, and that LL preserves monomorphisms by assumption of exactness. With this the statement follows via adjunction isomorphism

Hom 𝒜(,R(I))Hom (L(),I). 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 RRMod has enough injectives in that every module is a submodule of an injective one. We first consider this for R=R = \mathbb{Z}. We do assume prop. 5, 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. 5 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 AA receives an epimorphism ( sS)A(\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 A˜\tilde A of a direct sum of copies of \mathbb{Q}.

ker = ker ( sS) ( sS) A A˜ \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 A˜\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 AA into a divisible abelian group, hence into an injective \mathbb{Z}-module.

Lemma

Given a pair of additive adjoint functors

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

between abelian categories such that the left adjoint LL is

  1. an exact functor,

  2. a faithful functor.

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

Proof

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

i˜:AR(I) \tilde i \colon A \longrightarrow R(I)

and so it is sufficient to show that

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

  2. i˜\tilde i is a monomorphism.

The first point is the statement of lemma 1.

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

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

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

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

Now observe that L(i˜)L(\tilde i) is a monomorphism: this is because its composite L(A)L(i˜)L(R(I))ϵIL(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 ii, which by construction is a monomorphism. Therefore the exactness of the above sequence means that L(ker(i˜))L(A)L(ker(\tilde i)) \to L(A) is the zero morphism; and by the assumption that LL is a faithful functor this means that already ker(i˜)Aker(\tilde i) \to A is zero, hence that ker(i˜)=0ker(\tilde i) = 0, hence that i˜\tilde i is a monomorphism.

Proposition

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

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

Proof

Observe that the forgetful functor U:RModAbGpU\colon R Mod \to AbGp has both a left adjoint R !R_! (extension of scalars from \mathbb{Z} to RR) and a right adjoint R *R_* (coextension of scalars). Since it has a left adjoint, it is exact. Thus the statement follows via lemma 2 from prop. 2.

Proposition

For R=kR = k a field, hence RRMod = kkVect, every object is both injective as well as projective.

Examples

Injective \mathbb{Z}-modules / abelian groups

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

Proposition

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

Using Baer’s criterion, prop. 1.

Example

By prop. 5 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 /n\mathbb{Z}/n\mathbb{Z} for n>1n \gt 1.

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

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

For abelian sheaves over the etale site:

Revised on April 29, 2016 07:42:50 by Urs Schreiber (131.220.184.222)