Injective Module

Injective 是projective 的对偶, 但是在问题的处理上却复杂得多. 核心原因是我们在处理projective module 时是去寻找一个打到它的满射, 而injective module 则是需要寻找一个从它出发的单射. Zorn引理在这里具有很强的效果. Baer 判别法也是一个很有效的工具.

Injective Module

Injective is the dual of projective.

Definition 1. An $R$ -module $I$ is injective if for any exact sequence $0\to A\xrightarrow{i} B$ and $f:A\to I$ , there exists an homomorphism $g:B\to I$ s.t. $f=g\circ i$ .

We have similar results as projective module.

Theorem 1. The functor $\mathrm{Hom}(-,I)$ is exact if and only if $I$ is injective.

Proof

Proof. Similar to the projective. Omitted. ◻

Lemma 1. For any family of injective modules $\{M_{i}\}_{i\in I}$ , $\prod\limits_{i\in I}M_{i}$ is also injective.

Proof

Proof. For exact sequence $0\to A\xrightarrow{i}B$ and $f:A\to \prod\limits_{i\in I} M_{i}$ , we will have $\pi_{i}\circ f:A\to M_{i}$ the homomorphisms. Using the universal property of product, we have $g_{i}:B\to M_{i}$ with $g_{i}\circ i=\pi_{i}\circ f$ . ◻

Lemma 2. If $M\oplus N$ is injective, the so are $M$ and $N$ .

Proof

Proof. Consider the diagram

with the bottom row exact, we have

Here $j\circ f:A \to M\oplus N$ , and by injectivity of $M\oplus N$ , we have $g:B\to M\oplus N$ with $g\circ i=j\circ f$ , so $(\pi\circ g)\circ i=\pi\circ (j\circ f)=f$ . Thus $M$ is injective. Similar for $N$ . ◻

Now we give a very useful criterion for injective module:

Theorem 2. An $R$ -module $E$ is injective if and only if every homomorphism $f:I\to E$ from an ideal $I$ of $R$ can be extended to the whole ring, i.e. $\exists g:R\to E$ , for $g|_{I}=f$ .

Proof

Proof. Assume $E$ injective is clear.

Conversely, for any diagram

with the bottom row exact, we can regard $M$ as the submodule of $N$ , consider the set

\Sigma=\{(A,g)|M\subseteq A\subseteq N, g:A\to E, f \text{ can be extend to g}\}

$\Sigma$ is obviously a nonempty partially ordered set. For any chain $C\subset \Sigma$ , choose $(\bigcup\limits_{(A_{i},g_{i})\in C}A_{i}, \bigcup\limits_{(A_{i},g_{i})\in C}g_{i})$ , here $\bigcup\limits_{(A_{i},g_{i})\in C}g_{i}$ means $\forall a\in A_{i}$ , $a\mapsto g_{i}(a)$ . Obviously, $(\bigcup\limits_{(A_{i},g_{i})\in C}A_{i}, \bigcup\limits_{(A_{i},g_{i})\in C}g_{i})$ is an upper bound of $C$ . Thus $\Sigma$ has a maximal element $(A,g)$ .

Now we show $A=N$ . Suppose not $\exists n\in N$ , $x\notin A$ , take an ideal $I=\{r\in R|nr\in A\}$ and $h:I\to E$ with $h:r\to g(rn)$ . By assumption, we can extend $h$ into $k:R\to E$ , with $k|_{I}=E$ . Consider $\varphi:A+Rn\to E$ with $\varphi:a+rn\mapsto g(a)+g(rn)=g(a)+h(r)$ , then $(A+Rn,\varphi)$ is a larger element than $(A,g)$ in $\Sigma$ . Contradiction! Thus $N=A$ . ◻

Theorem 3. Let $R$ be a Noetherian ring. If $\{M_{j}\}_{j\in J}$ is a family of injective modules, then $\bigoplus\limits_{j\in J}M_{j}$ is injective.

Proof

Proof. Only need to show for any ideal $I\subset R$ , and $f:I\to \bigoplus\limits_{j\in J}M_{j}$ , it can be extended to $R$ . Since $R$ is Noetherian, $I$ is finite generated. Assume $I=(a_{1},\dots, a_{n})$ , for $i=1,\dots, n$ , $f(a_{i})=(b_{i,j})_{j\in J}$ , where $b_{i,j}=0$ for all but finite $j$ . Then $K=\{j\in J|f(a_{i})_{j}\neq 0 \text{ for some }i=1,\dots, n\}$ is finite. $f(I)\subset \bigoplus\limits_{j\in K}M_{j}$ , $\bigoplus\limits_{j\in K}M_{j}$ and $\prod\limits_{j\in K}M_{j}$ agree with each other. So we have $\bar{g}:R\to \bigoplus\limits_{j\in K}M_{j}$ , and thus $g:R\to \bigoplus\limits_{j\in J}M_{j}$ . ◻

However, for non Noetherian ring, the direct sum of injective module may not be injective. The theorem is actually if and only if. For any non Noetherian ring $R$ , there is a strictly increasing chain of ideals: $I_{1}\subsetneq I_{2}\subsetneq\cdots$ . Let $E_{i}$ be the injective envelope of $R /I_{i}$ , i.e. $E_{i}$ is injective, $R /I_{i}\subset E_{i}$ and for any submodule $N$ of $E_{i}$ , $R /I_{i}\cap N\neq \varnothing$ . Take $E=\bigoplus\limits_{i=1}^{n}E_{i}$ , $E$ is not injective.

Divisible Module

Definition 2. An $R$ -module $D$ is divisible if for every $r\in R$ and $d\in D$ , $\exists b\in D$ s.t. $rb=d$ .

For any integral domain $R$ , the injective module $E$ is always divisible: For any $r\in R$ and $d\in D$ , take $f:(r)\to E$ with $f:xr\mapsto xd$ . Then extend $f$ to $g:R\to E$ , thus $d=f(r)=rf(1)$ , $E$ is divisible.

The converse is not true in general. For the integral domain $R[x,y]$ and $Q$ is the field of fractions, then the $R$ -module $Q /R$ is divisible but not injective.

Theorem 4. Let $R$ be a PID.

1. An $R$ -module $E$ is injective if and only if it is divisible.

2. Quotient module of injective module are injective.

Proof

Proof.

Only need to show divisible modules are injective. Let $E$ be a divisible module, consider any ideal $I\neq (0)$ with a map $f:I\to E$ , then $I=(a)$ for some $a\in R$ . Since $E$ is divisible, $\exists e\in E$ s.t. $f(a)=ae$ . Now consider $g:R\to E$ with $g:r\mapsto re$ , $g$ extend $f$ to $R$ .
It’s easy to show for the quotient module of divisible module is
divisible. ◻

Lemma 3. For any injective abelian group $D$ , $\mathrm{Hom}_{\mathbb{Z}}(R,D)$ is injective $R$ -module.

Proof

Proof. Let $E=\mathrm{Hom}_{\mathbb{Z}}(R,D)$ , $E$ is obviously an $R$ -module. Only need to show $\mathrm{Hom}_{\mathbb{Z}}(-,E)$ is exact. $\mathrm{Hom}_{\mathbb{Z}}(-,E)\cong \mathrm{Hom}_{\mathbb{Z}}(-\otimes R,D)$ by adjoint functors. Since for $R$ -module $M$ , $M\otimes R\cong R$ , and $\mathrm{Hom}_{\mathbb{Z}}(-,D)$ is exact, thus $\mathrm{Hom}_{\mathbb{Z}}(-,E)$ is exact. ◻

Theorem 5. Every $R$ -module $M$ is a submodule of some injective $R$ -module.

Proof

Proof. There’s a free abelian group $F$ and $\pi:F\to M$ surjective. Since $F\cong \bigoplus\limits_{i\in I}\mathbb{Z}$ , $F$ can be embedded into $D\cong \bigoplus\limits_{i\in I}\mathbb{Q}$ , $D$ is divisible group.

Take $f:D\to M$ be the surjective group homomorphism and $\bar{f}:\mathrm{Hom}_{\mathbb{Z}}(R,M)\to \mathrm{Hom}_{\mathbb{Z}}(R,D)$ . Since every $R$ -module homomorphism is group homomorphism, $\mathrm{Hom}_{R}(R,M)\subset \mathrm{Hom}_{\mathbb{Z}}(R,A)$ . Consider the homomorphism $g:R\to \mathrm{Hom}_{R}(R,M)$ with $g(a):r\mapsto ra$ , then $g$ is injective. Then $R\xrightarrow{g}\mathrm{Hom}_{R}(R,M)\to \mathrm{Hom}_{\mathbb{Z}}(R,M)$ is a way to embed $R$ into divisible $R$ -module $\mathrm{Hom}_{\mathbb{Z}}(R,M)$ . ◻

Finally, we build the criterion of injective via splitting exact sequence.

Theorem 6. An $R$ -module $I$ is injective if and only if every short exact sequence

$$0\to I\to B\to C\to 0$$

splits.

Proof

Proof. Assume $I$ is an injective module, consider any short exact sequence, for the diagram

by injectivity, $\exists g:B\to I$ s.t. $g\circ i=id_{I}$ .

Conversely, suppose every short exact sequence $0\to I\to B\to C\to 0$ splits and consider the diagram

$I$ can be embedded into some injective $R$ -module $E$ via $j: I\to E$ .

0\to I\xrightarrow{j}E\to \mathrm{Coker}j\to 0

splits so $\exists q:E\to I$ s.t. $q\circ i=id_{I}$ .

The there exists $l:B\to E$ s.t. $l\circ i=j\circ f$

Then $q\circ l\circ i=q\circ j\circ f=f$ . Thus $I$ is injective. ◻