(also nonabelian homological algebra)
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.
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).
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 $I \hookrightarrow R$.
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
is well-defined and extends $f'$, as desired.
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.
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
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.
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).
Given a pair of additive adjoint functors
between abelian categories such that the left adjoint $L$ is an exact functor, then the right adjoint preserves injective objects.
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
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. 5, which may be proven using Baer's criterion.
Assuming the axiom of choice, the category $\mathbb{Z}$Mod $\simeq$ Ab has enough injectives.
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 $A$ 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 $\mathbb{Q}$.
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.
Given a pair of additive adjoint functors
between abelian categories such that the left adjoint $L$ is
an exact functor,
Then if $\mathcal{B}$ has enough injectives, also $\mathcal{A}$ has enough injectives.
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
and so it is sufficient to show that
$R(I)$ is injective in $\mathcal{A}$;
$\tilde i$ is a monomorphism.
The first point is the statement of lemma 1.
For the second point, consider the kernel of $\tilde i$ as part of the exact sequence
By the assumption that $L$ is an exact functor, the image of this sequence under $L$ is still exact
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.
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.
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 2 from prop. 2.
For $R = k$ a field, hence $R$Mod = $k$Vect, every object is both injective as well as projective.
Let $C = \mathbb{Z} Mod \simeq$ Ab be the abelian category of abelian groups.
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. 1.
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.
Not injective in Ab is for instance the cyclic group $\mathbb{Z}/n\mathbb{Z}$ for $n \gt 1$.
projective object, projective presentation, projective cover, projective resolution
injective object, injective presentation, injective envelope, injective resolution
flat object, flat resolution
The notion of injective module was introduced in
(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
See also
Section 4.2 of
For abelian sheaves over the etale site:
A study of injective modules in higher algebra:
Last revised on September 22, 2017 at 04:44:14. See the history of this page for a list of all contributions to it.