Example 19.2.1. Consider the category of sets. Let $A = \mathbf{N}$ and $B_ n = \{ 1, \ldots , n\} $ be the inductive system indexed by the natural numbers where $B_ n \to B_ m$ for $n \leq m$ is the obvious map. Then $\mathop{\mathrm{colim}}\nolimits B_ n = \mathbf{N}$, so there is a map $A \to \mathop{\mathrm{colim}}\nolimits B_ n$, which does not factor as $A \to B_ m$ for any $m$. Consequently, $\mathop{\mathrm{colim}}\nolimits \mathop{Mor}\nolimits (A, B_ n) \to \mathop{Mor}\nolimits (A, \mathop{\mathrm{colim}}\nolimits B_ n)$ is not surjective.

## 19.2 Baer's argument for modules

There is another, more set-theoretic approach to showing that any $R$-module $M$ can be imbedded in an injective module. This approach constructs the injective module by a transfinite colimit of push-outs. While this method is somewhat abstract and more complicated than the one of More on Algebra, Section 15.54, it is also more general. Apparently this method originates with Baer, and was revisited by Cartan and Eilenberg in [Cartan-Eilenberg] and by Grothendieck in [Tohoku]. There Grothendieck uses it to show that many other abelian categories have enough injectives. We will get back to the general case later (Section 19.11).

We begin with a few set theoretic remarks. Let $\{ B_{\beta }\} _{\beta \in \alpha }$ be an inductive system of objects in some category $\mathcal{C}$, indexed by an ordinal $\alpha $. Assume that $\mathop{\mathrm{colim}}\nolimits _{\beta \in \alpha } B_\beta $ exists in $\mathcal{C}$. If $A$ is an object of $\mathcal{C}$, then there is a natural map

because if one is given a map $A \to B_\beta $ for some $\beta $, one naturally gets a map from $A$ into the colimit by composing with $B_\beta \to \mathop{\mathrm{colim}}\nolimits _{\beta \in \alpha } B_\alpha $. Note that the left colimit is one of sets! In general, (19.2.0.1) is neither injective or surjective.

Example 19.2.2. Next we give an example where the map fails to be injective. Let $B_ n = \mathbf{N}/\{ 1, 2, \ldots , n\} $, that is, the quotient set of $\mathbf{N}$ with the first $n$ elements collapsed to one element. There are natural maps $B_ n \to B_ m$ for $n \leq m$, so the $\{ B_ n\} $ form a system of sets over $\mathbf{N}$. It is easy to see that $\mathop{\mathrm{colim}}\nolimits B_ n = \{ *\} $: it is the one-point set. So it follows that $\mathop{Mor}\nolimits (A, \mathop{\mathrm{colim}}\nolimits B_ n)$ is a one-element set for every set $A$. However, $\mathop{\mathrm{colim}}\nolimits \mathop{Mor}\nolimits (A , B_ n)$ is **not** a one-element set. Consider the family of maps $A \to B_ n$ which are just the natural projections $\mathbf{N} \to \mathbf{N}/\{ 1, 2, \ldots , n\} $ and the family of maps $A \to B_ n$ which map the whole of $A$ to the class of $1$. These two families of maps are distinct at each step and thus are distinct in $\mathop{\mathrm{colim}}\nolimits \mathop{Mor}\nolimits (A, B_ n)$, but they induce the same map $A \to \mathop{\mathrm{colim}}\nolimits B_ n$.

Nonetheless, if we map out of a finite set then (19.2.0.1) is an isomorphism always.

Lemma 19.2.3. Suppose that, in (19.2.0.1), $\mathcal{C}$ is the category of sets and $A$ is a *finite set*, then the map is a bijection.

**Proof.**
Let $f : A \to \mathop{\mathrm{colim}}\nolimits B_\beta $. The range of $f$ is finite, containing say elements $c_1, \ldots , c_ r \in \mathop{\mathrm{colim}}\nolimits B_\beta $. These all come from some elements in $B_\beta $ for $\beta \in \alpha $ large by definition of the colimit. Thus we can define $\widetilde{f} : A \to B_\beta $ lifting $f$ at a finite stage. This proves that (19.2.0.1) is surjective. Next, suppose two maps $f : A \to B_\gamma , f' : A \to B_{\gamma '}$ define the same map $A \to \mathop{\mathrm{colim}}\nolimits B_\beta $. Then each of the finitely many elements of $A$ gets sent to the same point in the colimit. By definition of the colimit for sets, there is $\beta \geq \gamma , \gamma '$ such that the finitely many elements of $A$ get sent to the same points in $B_\beta $ under $f$ and $f'$. This proves that (19.2.0.1) is injective.
$\square$

The most interesting case of the lemma is when $\alpha = \omega $, i.e., when the system $\{ B_\beta \} $ is a system $\{ B_ n\} _{n \in \mathbf{N}}$ over the natural numbers as in Examples 19.2.1 and 19.2.2. The essential idea is that $A$ is “small” relative to the long chain of compositions $B_1 \to B_2 \to \ldots $, so that it has to factor through a finite step. A more general version of this lemma can be found in Sets, Lemma 3.7.1. Next, we generalize this to the category of modules.

Definition 19.2.4. Let $\mathcal{C}$ be a category, let $I \subset \text{Arrows}(\mathcal{C})$, and let $\alpha $ be an ordinal. An object $A$ of $\mathcal{C}$ is said to be *$\alpha $-small with respect to $I$* if whenever $\{ B_\beta \} $ is a system over $\alpha $ with transition maps in $I$, then the map (19.2.0.1) is an isomorphism.

In the rest of this section we shall restrict ourselves to the category of $R$-modules for a fixed commutative ring $R$. We shall also take $I$ to be the collection of injective maps, i.e., the *monomorphisms* in the category of modules over $R$. In this case, for any system $\{ B_\beta \} $ as in the definition each of the maps

is an injection. It follows that the map (19.2.0.1) is an *injection*. We can in fact interpret the $B_\beta $'s as submodules of the module $B = \mathop{\mathrm{colim}}\nolimits _{\beta \in \alpha } B_\beta $, and then we have $B = \bigcup _{\beta \in \alpha } B_\beta $. This is not an abuse of notation if we identify $B_\alpha $ with the image in the colimit. We now want to show that modules are always small for “large” ordinals $\alpha $.

Proposition 19.2.5. Let $R$ be a ring. Let $M$ be an $R$-module. Let $\kappa $ the cardinality of the set of submodules of $M$. If $\alpha $ is an ordinal whose cofinality is bigger than $\kappa $, then $M$ is $\alpha $-small with respect to injections.

**Proof.**
The proof is straightforward, but let us first think about a special case. If $M$ is finite, then the claim is that for any inductive system $\{ B_\beta \} $ with injections between them, parametrized by a limit ordinal, any map $M \to \mathop{\mathrm{colim}}\nolimits B_\beta $ factors through one of the $B_\beta $. And this we proved in Lemma 19.2.3.

Now we start the proof in the general case. We need only show that the map (19.2.0.1) is a surjection. Let $f : M \to \mathop{\mathrm{colim}}\nolimits B_\beta $ be a map. Consider the subobjects $\{ f^{-1}(B_\beta )\} $ of $M$, where $B_\beta $ is considered as a subobject of the colimit $B = \bigcup _\beta B_\beta $. If one of these, say $f^{-1}(B_\beta )$, fills $M$, then the map factors through $B_\beta $.

So suppose to the contrary that all of the $f^{-1}(B_\beta )$ were proper subobjects of $M$. However, we know that

Now there are at most $\kappa $ different subobjects of $M$ that occur among the $f^{-1}(B_\alpha )$, by hypothesis. Thus we can find a subset $S \subset \alpha $ of cardinality at most $\kappa $ such that as $\beta '$ ranges over $S$, the $f^{-1}(B_{\beta '})$ range over *all* the $f^{-1}(B_\alpha )$.

However, $S$ has an upper bound $\widetilde{\alpha } < \alpha $ as $\alpha $ has cofinality bigger than $\kappa $. In particular, all the $f^{-1}(B_{\beta '})$, $\beta ' \in S$ are contained in $f^{-1}(B_{\widetilde{\alpha }})$. It follows that $f^{-1}(B_{\widetilde{\alpha }}) = M$. In particular, the map $f$ factors through $B_{\widetilde{\alpha }}$. $\square$

From this lemma we will be able to deduce the existence of lots of injectives. Let us recall Baer's criterion.

Lemma 19.2.6 (Baer's criterion). Let $R$ be a ring. An $R$-module $Q$ is injective if and only if in every commutative diagram

for $\mathfrak {a} \subset R$ an ideal, the dotted arrow exists.

**Proof.**
This is the equivalence of (1) and (3) in More on Algebra, Lemma 15.54.4; please observe that the proof given there is elementary (and does not use $\text{Ext}$ groups or the existence of injectives or projectives in the category of $R$-modules).
$\square$

If $M$ is an $R$-module, then in general we may have a semi-complete diagram as in Lemma 19.2.6. In it, we can form the *push-out*

Here the vertical map is injective, and the diagram commutes. The point is that we can extend $\mathfrak {a} \to Q$ to $R$ *if* we extend $Q$ to the larger module $R \oplus _{\mathfrak {a}} Q$.

The key point of Baer's argument is to repeat this procedure transfinitely many times. To do this we first define, given an $R$-module $M$ the following (huge) pushout

Here the top horizontal arrow maps the element $a \in \mathfrak a$ in the summand corresponding to $\varphi $ to the element $\varphi (a) \in M$. The left vertical arrow maps $a \in \mathfrak a$ in the summand corresponding to $\varphi $ simply to the element $a \in R$ in the summand corresponding to $\varphi $. The fundamental properties of this construction are formulated in the following lemma.

Lemma 19.2.7. Let $R$ be a ring.

The construction $M \mapsto (M \to \mathbf{M}(M))$ is functorial in $M$.

The map $M \to \mathbf{M}(M)$ is injective.

For any ideal $\mathfrak {a}$ and any $R$-module map $\varphi : \mathfrak a \to M$ there is an $R$-module map $\varphi ' : R \to \mathbf{M}(M)$ such that

\[ \xymatrix{ \mathfrak {a} \ar[d] \ar[r]_\varphi & M \ar[d] \\ R \ar[r]^{\varphi '} & \mathbf{M}(M) } \]commutes.

**Proof.**
Parts (2) and (3) are immediate from the construction. To see (1), let $\chi : M \to N$ be an $R$-module map. We claim there exists a canonical commutative diagram

which induces the desired map $\mathbf{M}(M) \to \mathbf{M}(N)$. The middle east-south-east arrow maps the summand $\mathfrak a$ corresponding to $\varphi $ via $\text{id}_{\mathfrak a}$ to the summand $\mathfrak a$ corresponding to $\psi = \chi \circ \varphi $. Similarly for the lower east-south-east arrow. Details omitted. $\square$

The idea will now be to apply the functor $\mathbf{M}$ a transfinite number of times. We define for each ordinal $\alpha $ a functor $\mathbf{M}_\alpha $ on the category of $R$-modules, together with a natural injection $N \to \mathbf{M}_\alpha (N)$. We do this by transfinite induction. First, $\mathbf{M}_1 = \mathbf{M}$ is the functor defined above. Now, suppose given an ordinal $\alpha $, and suppose $\mathbf{M}_{\alpha '}$ is defined for $\alpha ' < \alpha $. If $\alpha $ has an immediate predecessor $\widetilde{\alpha }$, we let

If not, i.e., if $\alpha $ is a limit ordinal, we let

It is clear (e.g., inductively) that the $\mathbf{M}_{\alpha }(N)$ form an inductive system over ordinals, so this is reasonable.

Theorem 19.2.8. Let $\kappa $ be the cardinality of the set of ideals in $R$, and let $\alpha $ be an ordinal whose cofinality is greater than $\kappa $. Then $\mathbf{M}_\alpha (N)$ is an injective $R$-module, and $N \to \mathbf{M}_\alpha (N)$ is a functorial injective embedding.

**Proof.**
By Baer's criterion Lemma 19.2.6, it suffices to show that if $\mathfrak {a} \subset R$ is an ideal, then any map $f : \mathfrak {a} \to \mathbf{M}_\alpha (N)$ extends to $R \to \mathbf{M}_\alpha (N)$. However, we know since $\alpha $ is a limit ordinal that

so by Proposition 19.2.5, we find that

This means in particular that there is some $\beta ' < \alpha $ such that $f$ factors through the submodule $\mathbf{M}_{\beta '}(N)$, as

However, by the fundamental property of the functor $\mathbf{M}$, see Lemma 19.2.7 part (3), we know that the map $\mathfrak {a} \to \mathbf{M}_{\beta '}(N)$ can be extended to

and the last object imbeds in $\mathbf{M}_{\alpha }(N)$ (as $\beta ' + 1 < \alpha $ since $\alpha $ is a limit ordinal). In particular, $f$ can be extended to $\mathbf{M}_{\alpha }(N)$. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #2351 by Antoine Chambert-Loir on

Comment #2418 by Johan on