注射模

Injective Modules and ${\operatorname{Hom}}_{R}\left( {\_ \_ ,D}\right)$

注射模块和 ${\operatorname{Hom}}_{R}\left( {\_ \_ ,D}\right)$

If $0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ is a short exact sequence of $R$ -modules then,instead of considering maps from an $R$ -module $D$ into $L$ or $N$ and the extent to which these determine maps from $D$ into $M$ ,we can consider the "dual" question of maps from $L$ or $N$ to $D$ . In this case,it is easy to dispose of the situation of a map from $N$ to $D$ : an $R$ -module map from $N$ to $D$ immediately gives a map from $M$ to $D$ simply by composing with $\varphi$ . It is easy to check that this defines an injective homomorphism of abelian groups

如果 $0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ 是一个 $R$ -模的短正合序列，那么，我们不必考虑从 $R$ -模 $D$ 到 $L$ 或 $N$ 的映射，以及这些映射在多大程度上决定了从 $D$ 到 $M$ 的映射，而是可以考虑“对偶”问题，即从 $L$ 或 $N$ 到 $D$ 的映射。在这种情况下，很容易处理从 $N$ 到 $D$ 的映射情况：一个从 $N$ 到 $D$ 的 $R$ -模映射，通过简单地与 $\varphi$ 结合，立即给出了从 $M$ 到 $D$ 的映射。很容易验证这定义了一个阿贝尔群的注入同态。

or, put another way,

或者换句话说，

then $0 \rightarrow {\operatorname{Hom}}_{R}\left( {N,D}\right) \overset{{\varphi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {M,D}\right)$ is exact.

那么 $0 \rightarrow {\operatorname{Hom}}_{R}\left( {N,D}\right) \overset{{\varphi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {M,D}\right)$ 是精确的。

(Note that the associated maps on the homomorphism groups are in the reverse direction from the original maps.)

（注意，同态群上的相关映射与原始映射的方向相反。）

On the other hand,given an $R$ -module homomorphism $f$ from $L$ to $D$ it may not be possible to extend $f$ to a map $F$ from $M$ to $D$ ,i.e.,given $f$ it may not be possible to find a map $F$ making the following diagram commute:

另一方面，给定一个从 $L$ 到 $D$ 的 $R$ -模同态 $f$，可能无法将 $f$ 扩展为从 $M$ 到 $D$ 的映射 $F$，即给定 $f$ 后，可能无法找到一个使以下图表成立的映射 $F$。

For example,consider the exact sequence $0 \rightarrow \mathbb{Z}\overset{\psi }{ \rightarrow }\mathbb{Z}\overset{\varphi }{ \rightarrow }\mathbb{Z}/2\mathbb{Z} \rightarrow 0$ of $\mathbb{Z}$ -modules,where $\psi$ is multiplication by 2 and $\varphi$ is the natural projection. Take $D = \mathbb{Z}/2\mathbb{Z}$ and let $f : \mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}$ be reduction modulo 2 on the first $\mathbb{Z}$ in the sequence. There is only one nonzero homomorphism $F$ from the second $\mathbb{Z}$ in the sequence to $\mathbb{Z}/2\mathbb{Z}$ (namely,reduction modulo 2),but this $F$ does not lift the map $f$ since $F \circ \psi \left( \mathbb{Z}\right) = F\left( {2\mathbb{Z}}\right) = 0$ ,so $F \circ \psi \neq f$ .

例如，考虑确切序列 $0 \rightarrow \mathbb{Z}\overset{\psi }{ \rightarrow }\mathbb{Z}\overset{\varphi }{ \rightarrow }\mathbb{Z}/2\mathbb{Z} \rightarrow 0$ 的 $\mathbb{Z}$ -模，其中 $\psi$ 是乘以2，$\varphi$ 是自然投影。取 $D = \mathbb{Z}/2\mathbb{Z}$ 并让 $f : \mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}$ 是序列中第一个 $\mathbb{Z}$ 的模2约化。序列中的第二个 $\mathbb{Z}$ 到 $\mathbb{Z}/2\mathbb{Z}$ 只存在一个非零同态 $F$（即模2约化），但这个 $F$ 不能提升映射 $f$，因为 $F \circ \psi \left( \mathbb{Z}\right) = F\left( {2\mathbb{Z}}\right) = 0$，所以 $F \circ \psi \neq f$。

Composition with $\psi$ induces an abelian group homomorphism ${\psi }^{\prime }$ from ${\mathrm{{Hom}}}_{R}\left( {M,D}\right)$ to ${\operatorname{Hom}}_{R}\left( {L,D}\right)$ ,and in terms of the map ${\psi }^{\prime }$ ,the homomorphism $f \in {\operatorname{Hom}}_{R}\left( {L,D}\right)$ can be lifted to a homomorphism from $M$ to $D$ if and only if $f$ is in the image of ${\psi }^{\prime }$ . The example above shows that

与 $\psi$ 的复合诱导出一个从 ${\mathrm{{Hom}}}_{R}\left( {M,D}\right)$ 到 ${\operatorname{Hom}}_{R}\left( {L,D}\right)$ 的阿贝尔群同态 ${\psi }^{\prime }$，并且就映射 ${\psi }^{\prime }$ 而言，当且仅当 $f$ 在 ${\psi }^{\prime }$ 的像中时，同态 $f \in {\operatorname{Hom}}_{R}\left( {L,D}\right)$ 可以提升为从 $M$ 到 $D$ 的同态。上面的例子表明

then $\;{\operatorname{Hom}}_{R}\left( {M,D}\right) \overset{{\psi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {L,D}\right) \rightarrow 0$ is not necessarily exact.

那么 $\;{\operatorname{Hom}}_{R}\left( {M,D}\right) \overset{{\psi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {L,D}\right) \rightarrow 0$ 不一定是确切的。

We can summarize these results in the following dual version of Theorem 28:

我们可以将这些结果总结为定理28的对偶版本如下：

Theorem 33. Let $D,L,M$ ,and $N$ be $R$ -modules. If

定理33。设 $D,L,M$，$N$ 是 $R$ -模。如果

then the associated sequence

那么关联的序列

A homomorphism $f : L \rightarrow D$ lifts to a homomorphism $F : M \rightarrow D$ if and only if $f \in {\operatorname{Hom}}_{R}\left( {L,D}\right)$ is in the image of ${\psi }^{\prime }$ . In general ${\psi }^{\prime } : {\operatorname{Hom}}_{R}\left( {M,D}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {L,D}\right)$ need not be surjective; the map ${\psi }^{\prime }$ is surjective if and only if every homomorphism from $L$ to $D$ lifts to a homomorphism from $M$ to $D$ ,in which case the sequence (12) can be extended to a short exact sequence.

一个同态 $f : L \rightarrow D$ 提升为同态 $F : M \rightarrow D$ 当且仅当 $f \in {\operatorname{Hom}}_{R}\left( {L,D}\right)$ 在 ${\psi }^{\prime }$ 的像中。一般来说 ${\psi }^{\prime } : {\operatorname{Hom}}_{R}\left( {M,D}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {L,D}\right)$ 不一定是满射；映射 ${\psi }^{\prime }$ 是满射当且仅当从 $L$ 到 $D$ 的每个同态都可以提升为从 $M$ 到 $D$ 的同态，在这种情况下序列（12）可以扩展为一个短确切序列。

The sequence (12) is exact for all $R$ -modules $D$ if and only if the sequence

序列 (12) 对于所有 $R$ -模块 $D$ 是精确的当且仅当序列

Proof: The only item remaining to be proved in the first statement is the exactness of (12) at ${\operatorname{Hom}}_{R}\left( {M,D}\right)$ . The proof of this statement is very similar to the proof of the corresponding result in Theorem 28 and is left as an exercise. Note also that the injectivity of $\psi$ is not required,which proves the "if" portion of the final statement of the theorem.

证明：在第一个陈述中唯一需要证明的是序列 (12) 在 ${\operatorname{Hom}}_{R}\left( {M,D}\right)$ 处的精确性。这个陈述的证明与定理 28 中相应结果的证明非常相似，留作练习。注意，$\psi$ 的单射性不是必需的，这证明了定理最后陈述的“如果”部分。

Suppose now that the sequence (12) is exact for all $R$ -modules $D$ . We first show that $\varphi : M \rightarrow N$ is a surjection. Take $D = N/\varphi \left( M\right)$ . If ${\pi }_{1} : N \rightarrow N/\varphi \left( M\right)$ is the natural projection homomorphism,then ${\pi }_{1} \circ \varphi \left( M\right) = 0$ by definition of ${\pi }_{1}$ . Since ${\pi }_{1} \circ \varphi = {\varphi }^{\prime }\left( {\pi }_{1}\right)$ ,this means that the element ${\pi }_{1} \in {\mathrm{{Hom}}}_{R}\left( {N,N/\varphi \left( M\right) }\right)$ is mapped to 0 by ${\varphi }^{\prime }$ . Since ${\varphi }^{\prime }$ is assumed to be injective for all modules $D$ ,this means ${\pi }_{1}$ is the zero map,i.e., $N = \varphi \left( M\right)$ and so $\varphi$ is a surjection. We next show that $\varphi \circ \psi = 0$ ,which will imply that image $\psi \subseteq \ker \varphi$ . For this we take $D = N$ and observe that the identity $\operatorname{map}i{d}_{N}$ on $N$ is contained in ${\operatorname{Hom}}_{R}\left( {N,N}\right)$ ,hence ${\varphi }^{\prime }\left( {i{d}_{N}}\right) \in {\operatorname{Hom}}_{R}\left( {M,N}\right)$ . Then the exactness of (12) for $D = N$ implies that ${\varphi }^{\prime }\left( {i{d}_{N}}\right) \in \ker {\psi }^{\prime }$ ,so ${\psi }^{\prime }\left( {{\varphi }^{\prime }\left( {i{d}_{N}}\right) }\right) = 0$ . Then ${\text{id}}_{N} \circ \psi \circ \varphi = 0,$ i.e., $\psi \circ \varphi = 0,$ as claimed. Finally,we show that $\ker \varphi \subseteq \mathrm{{image}}\;\psi .$ Let $D = M/\psi \left( L\right)$ and let ${\pi }_{2} : M \rightarrow M/\psi \left( L\right)$ be the natural projection. Then ${\psi }^{\prime }\left( {\pi }_{2}\right) = 0$ since ${\pi }_{2}\left( {\psi \left( L\right) }\right) = 0$ by definition of ${\pi }_{2}$ . The exactness of (12) for this $D$ then implies that ${\pi }_{2}$ is in the image of ${\varphi }^{\prime }$ ,say ${\pi }_{2} = {\varphi }^{\prime }\left( f\right)$ for some homomorphism $f \in {\operatorname{Hom}}_{R}\left( {N,M/\psi \left( L\right) }\right) ,$ i.e., ${\pi }_{2} = f \circ \varphi .$ If $m \in \ker \varphi$ then ${\pi }_{2}\left( m\right) = f\left( {\varphi \left( m\right) }\right) = 0,$ which means that $m \in \psi \left( L\right)$ since ${\pi }_{2}$ is just the projection from $M$ into the quotient $M/\psi \left( L\right)$ . Hence $\ker \varphi \subseteq$ image $\psi$ ,completing the proof.

假设现在序列 (12) 对所有 $R$ -模块 $D$ 都是精确的。我们首先证明 $\varphi : M \rightarrow N$ 是满射。取 $D = N/\varphi \left( M\right)$ 。如果 ${\pi }_{1} : N \rightarrow N/\varphi \left( M\right)$ 是自然投影同态，那么根据 ${\pi }_{1}$ 的定义，${\pi }_{1} \circ \varphi \left( M\right) = 0$ 。由于 ${\pi }_{1} \circ \varphi = {\varphi }^{\prime }\left( {\pi }_{1}\right)$ ，这意味着元素 ${\pi }_{1} \in {\mathrm{{Hom}}}_{R}\left( {N,N/\varphi \left( M\right) }\right)$ 被 ${\varphi }^{\prime }$ 映射到 0。由于假设 ${\varphi }^{\prime }$ 对所有模块 $D$ 都是单射，这意味着 ${\pi }_{1}$ 是零映射，即 $N = \varphi \left( M\right)$ ，因此 $\varphi$ 是满射。接下来我们证明 $\varphi \circ \psi = 0$ ，这将意味着像 $\psi \subseteq \ker \varphi$ 。为此我们取 $D = N$ 并观察到在 $N$ 上的恒等式 $\operatorname{map}i{d}_{N}$ 包含在 ${\operatorname{Hom}}_{R}\left( {N,N}\right)$ 中，因此 ${\varphi }^{\prime }\left( {i{d}_{N}}\right) \in {\operatorname{Hom}}_{R}\left( {M,N}\right)$ 。那么序列 (12) 对 $D = N$ 的精确性意味着 ${\varphi }^{\prime }\left( {i{d}_{N}}\right) \in \ker {\psi }^{\prime }$ ，所以 ${\psi }^{\prime }\left( {{\varphi }^{\prime }\left( {i{d}_{N}}\right) }\right) = 0$ 。那么 ${\text{id}}_{N} \circ \psi \circ \varphi = 0,$ ，即 $\psi \circ \varphi = 0,$ 如所声称的。最后，我们证明 $\ker \varphi \subseteq \mathrm{{image}}\;\psi .$ 。设 $D = M/\psi \left( L\right)$ 并设 ${\pi }_{2} : M \rightarrow M/\psi \left( L\right)$ 为自然投影。那么 ${\psi }^{\prime }\left( {\pi }_{2}\right) = 0$ ，因为根据 ${\pi }_{2}$ 的定义 ${\pi }_{2}\left( {\psi \left( L\right) }\right) = 0$ 。这个 $D$ 的序列 (12) 的精确性意味着 ${\pi }_{2}$ 在 ${\varphi }^{\prime }$ 的像中，即存在某个同态 $f \in {\operatorname{Hom}}_{R}\left( {N,M/\psi \left( L\right) }\right) ,$ 使得 ${\pi }_{2} = {\varphi }^{\prime }\left( f\right)$ ，即 ${\pi }_{2} = f \circ \varphi .$ 。如果 $m \in \ker \varphi$ ，那么 ${\pi }_{2}\left( m\right) = f\left( {\varphi \left( m\right) }\right) = 0,$ ，这意味着 $m \in \psi \left( L\right)$ ，因为 ${\pi }_{2}$ 只是从 $M$ 到商 $M/\psi \left( L\right)$ 的投影。因此 $\ker \varphi \subseteq$ 像是 $\psi$ ，完成了证明。

By Theorem 33, the sequence

根据定理 33，序列

is in general not a short exact sequence since ${\psi }^{\prime }$ need not be surjective,and the question of whether this sequence is exact precisely measures the extent to which homomorphisms from $M$ to $D$ are uniquely determined by pairs of homomorphisms from $L$ and $N$ to $D$ .

通常来说，这并不是一个短正合序列，因为 ${\psi }^{\prime }$ 不一定是满射，而这个问题是否精确地衡量了从 $M$ 到 $D$ 的同态是否由从 $L$ 和 $N$ 到 $D$ 的同态对唯一确定。

The second statement in Proposition 29 shows that this sequence is exact when the original exact sequence $0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0$ is a split exact sequence. In fact in this case the sequence $0 \rightarrow {\operatorname{Hom}}_{R}\left( {N,D}\right) \overset{{\varphi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {M,D}\right) \overset{{\psi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {L,D}\right) \rightarrow 0$ is also a split exact sequence of abelian groups for every $R$ -module $D$ . Exercise 14 shows that a converse holds: if $0 \rightarrow {\operatorname{Hom}}_{R}\left( {N,D}\right) \overset{{\varphi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {M,D}\right) \overset{{\psi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {L,D}\right) \rightarrow 0$ is exact for every $R$ -module $D$ then $0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ is a split short exact sequence (which then implies that if the Hom sequence is exact for every $D$ ,then in fact it is split exact for every $D$ ).

命题29中的第二个陈述表明，当原始的短正合序列 $0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0$ 是分裂正合序列时，这个序列是正合的。实际上，在这种情况下，序列 $0 \rightarrow {\operatorname{Hom}}_{R}\left( {N,D}\right) \overset{{\varphi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {M,D}\right) \overset{{\psi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {L,D}\right) \rightarrow 0$ 对于每个 $R$ -模 $D$ 也是分裂正合序列。练习14表明一个逆命题成立：如果 $0 \rightarrow {\operatorname{Hom}}_{R}\left( {N,D}\right) \overset{{\varphi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {M,D}\right) \overset{{\psi }^{\prime }}{ \rightarrow }{\operatorname{Hom}}_{R}\left( {L,D}\right) \rightarrow 0$ 对于每个 $R$ -模 $D$ 都是正合的，那么 $0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ 是一个分裂的短正合序列（这进而意味着如果 Hom 序列对于每个 $D$ 是正合的，那么实际上它对于每个 $D$ 也是分裂正合的）。

There is also a dual version of the first three parts of Proposition 30, which describes the $\textit{R-modules }\textit{D}$ having the property that the sequence $\left( \frac{1}{2}\right)$ in Theorem $\frac{3}{2}$ can always be extended to a short exact sequence:

命题30的前三部分也有一个对偶版本，它描述了 $\textit{R-modules }\textit{D}$ 具有这样的性质：定理 $\frac{3}{2}$ 中的序列 $\left( \frac{1}{2}\right)$ 总是可以扩展为一个短正合序列：

Proposition 34. Let $Q$ be an $R$ -module. Then the following are equivalent:

命题34。设 $Q$ 是一个 $R$ -模。那么以下条件是等价的：

(1) For any $R$ -modules $L,M$ ,and $N$ ,if

（1）对于任何 $R$ -模 $L,M$ 和 $N$ ，如果

is a short exact sequence, then

是一个短正合序列，那么

is also a short exact sequence.

也是一个短正合序列。

(2) For any $R$ -modules $L$ and $M$ ,if $0 \rightarrow L\overset{\psi }{ \rightarrow }M$ is exact,then every $R$ -module homomorphism from $L$ into $Q$ lifts to an $R$ -module homomorphism of $M$ into $Q$ ,i.e.,given $f \in {\operatorname{Hom}}_{R}\left( {L,Q}\right)$ there is a lift $F \in {\operatorname{Hom}}_{R}\left( {M,Q}\right)$ making the following diagram commute:

(2) 对于任意的 $R$ -模块 $L$ 和 $M$ ，如果 $0 \rightarrow L\overset{\psi }{ \rightarrow }M$ 是精确的，那么从 $L$ 到 $Q$ 的每个 $R$ -模块同态都可以提升为从 $M$ 到 $Q$ 的 $R$ -模块同态，即给定 $f \in {\operatorname{Hom}}_{R}\left( {L,Q}\right)$ 存在一个提升 $F \in {\operatorname{Hom}}_{R}\left( {M,Q}\right)$ 使得以下图表可交换：

(3) If $Q$ is a submodule of the $R$ -module $M$ then $Q$ is a direct summand of $M$ ,i.e., every short exact sequence $0 \rightarrow Q \rightarrow M \rightarrow N \rightarrow 0$ splits.

(3) 如果 $Q$ 是 $R$ -模块 $M$ 的子模块，那么 $Q$ 是 $M$ 的直和项，即每个短精确序列 $0 \rightarrow Q \rightarrow M \rightarrow N \rightarrow 0$ 都可以分解。

Proof: The equivalence of (1) and (2) is part of Theorem 33. Suppose now that (2) is satisfied and let $0 \rightarrow Q\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ be exact. Taking $L = Q$ and $f$ the identity map from $Q$ to itself,it follows by (2) that there is a homomorphism $F : M \rightarrow Q$ with $F \circ \psi = 1$ ,so $F$ is a splitting homomorphism for the sequence,which proves (3). The proof that (3) implies (2) is outlined in the exercises.

证明： (1) 和 (2) 的等价性是定理33的一部分。现在假设 (2) 成立，并且让 $0 \rightarrow Q\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ 是精确的。取 $L = Q$ 和 $f$ 为从 $Q$ 到其自身的恒等映射，根据 (2) 可知存在一个同态 $F : M \rightarrow Q$ 使得 $F \circ \psi = 1$ ，因此 $F$ 是该序列的分解同态，这证明了 (3)。 (3) 推导出 (2) 的证明在练习中概述。

Definition. An $R$ -module $Q$ is called injective if it satisfies any of the equivalent conditions of Proposition 34.

定义。一个 $R$ -模块 $Q$ 如果满足命题34中的任何等价条件，则称为可注入的。

The third statement in Proposition 34 can be rephrased as saying that any module $M$ into which $Q$ injects has (an isomorphic copy of) $Q$ as a direct summand,which explains the terminology.

命题34中的第三条陈述可以重新表述为：任何 $M$ 模块，如果 $Q$ 可以注入其中，则具有（一个同构副本的）$Q$ 作为直和项，这解释了术语的使用。

If $D$ is fixed,then given any $R$ -module $X$ we have an associated abelian group ${\operatorname{Hom}}_{R}\left( {X,D}\right)$ . Further,an $R$ -module homomorphism $\alpha : X \rightarrow Y$ induces an abelian group homomorphism ${\alpha }^{\prime } : {\operatorname{Hom}}_{R}\left( {Y,D}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {X,D}\right)$ ,defined by ${\alpha }^{\prime }\left( f\right) = f \circ \alpha$ , that “reverses” the direction of the arrow. Put another way,the map ${\operatorname{Hom}}_{R}\left( {D,\underline{\;}}\right)$ is a $\begin{matrix} \text{contravariant} \\ \text{functor} \\ \text{from the category of}\;R\text{-modules} \\ \text{to the category of abelian groups} \end{matrix}$ (cf. Appendix II). Theorem 33 shows that applying this functor to the terms in the exact sequence

如果 $D$ 是固定的，那么对于任意的 $R$ -模 $X$，我们有一个相关的阿贝尔群 ${\operatorname{Hom}}_{R}\left( {X,D}\right)$。进一步地，一个 $R$ -模同态 $\alpha : X \rightarrow Y$ 诱导出一个阿贝尔群同态 ${\alpha }^{\prime } : {\operatorname{Hom}}_{R}\left( {Y,D}\right) \rightarrow {\operatorname{Hom}}_{R}\left( {X,D}\right)$，由 ${\alpha }^{\prime }\left( f\right) = f \circ \alpha$ 定义，它“逆转”了箭头的方向。换句话说，映射 ${\operatorname{Hom}}_{R}\left( {D,\underline{\;}}\right)$ 是一个 $\begin{matrix} \text{contravariant} \\ \text{functor} \\ \text{from the category of}\;R\text{-modules} \\ \text{to the category of abelian groups} \end{matrix}$（参见附录II）。定理33表明，将这个函子应用于精确序列中的项

produces an exact sequence

产生一个精确序列

This is referred to by saying that ${\operatorname{Hom}}_{R}\left( {\_ ,D}\right)$ is a left exact (contravariant) functor. Note that the functor ${\operatorname{Hom}}_{R}\left( {\_ ,D}\right)$ and the functor ${\operatorname{Hom}}_{R}\left( {D,\_ }\right)$ considered earlier are both left exact; the former reverses the directions of the maps in the original short exact sequence, the latter maintains the directions of the maps.

这被称作 ${\operatorname{Hom}}_{R}\left( {\_ ,D}\right)$ 是一个左精确（反对称）函子。注意，函子 ${\operatorname{Hom}}_{R}\left( {\_ ,D}\right)$ 和之前考虑的函子 ${\operatorname{Hom}}_{R}\left( {D,\_ }\right)$ 都是左精确的；前者逆转了原始短精确序列中映射的方向，后者保持了映射的方向。

By Proposition 34,the functor ${\mathrm{{Hom}}}_{R}\left( {\_ \_ ,D}\right)$ is exact,i.e.,always takes short exact sequences to short exact sequences (and hence exact sequences of any length to exact sequences), if and only if $D$ is injective. We summarize this in the following proposition, which is dual to the covariant result of Corollary 32.

根据命题34，函子 ${\mathrm{{Hom}}}_{R}\left( {\_ \_ ,D}\right)$ 是精确的，即总是将短精确序列映射到短精确序列（因此也将任意长度的精确序列映射到精确序列），当且仅当 $D$ 是可注入的。我们在以下命题中总结这一点，该命题是32号推论的协变结果的对偶。

Corollary 35. If $D$ is an $R$ -module,then the functor ${\operatorname{Hom}}_{R}\left( {\_ \_ ,D}\right)$ from the category of $R$ -modules to the category of abelian groups is left exact. It is exact if and only if $D$ is an injective $R$ -module.

推论35。如果 $D$ 是一个 $R$ -模，那么从 $R$ -模范畴到阿贝尔群范畴的函子 ${\operatorname{Hom}}_{R}\left( {\_ \_ ,D}\right)$ 是左精确的。当且仅当 $D$ 是一个可注入的 $R$ -模时，它是精确的。

We have seen that an $R$ -module is projective if and only if it is a direct summand of a free $R$ -module. Providing such a simple characterization of injective $R$ -modules is not so easy. The next result gives a criterion for $Q$ to be an injective $R$ -module (a result due to Baer, who introduced the notion of injective modules around 1940), and using it we can give a characterization of injective modules when $R = \mathbb{Z}$ (or,more generally,when $R$ is a P.I.D.). Recall that a $\mathbb{Z}$ -module $A$ (i.e.,an abelian group,written additively) is said to be divisible if $A = {nA}$ for all nonzero integers $n$ . For example, both $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$ are divisible (cf. Exercises 18 and 19 in Section 2.4 and Exercise 15 in Section 3.1).

我们已经看到，一个 $R$ -模是投射的当且仅当它是自由 $R$ -模的直接和项。给出一个如此简单的注入 $R$ -模的特征并不是那么容易。下一个结果给出了 $Q$ 为注入 $R$ -模的一个准则（Baer 的结果，他在大约1940年引入了可注入模的概念），使用它我们可以给出当 $R = \mathbb{Z}$ （或者更一般地，当 $R$ 是一个P.I.D.时）注入模的特征。回顾一下，一个 $\mathbb{Z}$ -模 $A$（即，一个加法书写的阿贝尔群）被称为可除的，如果 $A = {nA}$ 对于所有非零整数 $n$ 成立。例如，$\mathbb{Q}$ 和 $\mathbb{Q}/\mathbb{Z}$ 都是可除的（参见第2.4节的练习18和19以及第3.1节的练习15）。

Proposition 36. Let $Q$ be an $R$ -module.

命题36。设 $Q$ 是一个 $R$ -模。

(1) (Baer’s Criterion) The module $Q$ is injective if and only if for every left ideal $I$ of $R$ any $R$ -modulehomomorphism $g : I \rightarrow Q$ can be extended to an $R$ -module homomorphism $G : R \rightarrow Q$ .

（1）（Baer准则）模 $Q$ 是可注入的当且仅当对于 $R$ 的每一个左理想 $I$ ，任何 $R$ -模同态 $g : I \rightarrow Q$ 都可以扩展为一个 $R$ -模同态 $G : R \rightarrow Q$ 。

(2) If $R$ is a P.I.D. then $Q$ is injective if and only if ${rQ} = Q$ for every nonzero $r \in R$ . In particular,a $\mathbb{Z}$ -module is injective if and only if it is divisible. When $R$ is a P.I.D.,quotient modules of injective $R$ -modules are again injective.

（2）如果 $R$ 是一个P.I.D.，那么 $Q$ 是可注入的当且仅当对于每一个非零 $r \in R$ 成立 ${rQ} = Q$ 。特别地，一个 $\mathbb{Z}$ -模是可注入的当且仅当它是可除的。当 $R$ 是一个P.I.D.时，可注入 $R$ -模的商模仍然是可注入的。

Proof: If $Q$ is injective and $g : I \rightarrow Q$ is an $R$ -module homomorphism from the nonzero ideal $I$ of $R$ into $Q$ ,then $g$ can be extended to an $R$ -module homomorphism from $R$ into $Q$ by Proposition 34(2) applied to the exact sequence $0 \rightarrow I \rightarrow R$ ,which proves the “only if” portion of (1). Suppose conversely that every homomorphism $g : I \rightarrow Q$ can be lifted to a homomorphism $G : R \rightarrow Q$ . To show that $Q$ is injective we must show that if $0 \rightarrow L \rightarrow M$ is exact and $f : L \rightarrow Q$ is an $R$ - module homomorphism then there is a lift $F : M \rightarrow Q$ extending $f$ . If $\mathcal{S}$ is the collection $\left( {{f}^{\prime },{L}^{\prime }}\right)$ of lifts ${f}^{\prime } : {L}^{\prime } \rightarrow Q$ of $f$ to a submodule ${L}^{\prime }$ of $M$ containing $L$ , then the ordering $\left( {{f}^{\prime },{L}^{\prime }}\right) \leq \left( {{f}^{\prime \prime },{L}^{\prime \prime }}\right)$ if ${L}^{\prime } \subseteq {L}^{\prime \prime }$ and ${f}^{\prime \prime } = {f}^{\prime }$ on ${L}^{\prime }$ partially orders $\mathcal{S}$ . Since $\mathcal{S} \neq \varnothing$ ,by Zorn’s Lemma there is a maximal element $\left( {F,{M}^{\prime }}\right)$ in $\mathcal{S}$ . The map $F : {M}^{\prime } \rightarrow Q$ is a lift of $f$ and it suffices to show that ${M}^{\prime } = M$ . Suppose that there is some element $m \in M$ not contained in ${M}^{\prime }$ and let $I = \left\{ {r \in R \mid {rm} \in {M}^{\prime }}\right\}$ . It is easy to check that $I$ is a left ideal in $R$ ,and the map $g : I \rightarrow Q$ defined by $g\left( x\right) = F\left( {xm}\right)$ is an $R$ -module homomorphism from $I$ to $Q$ . By hypothesis,there is a lift $G : R \rightarrow Q$ of $g$ . Consider the submodule ${M}^{\prime } + {Rm}$ of $M$ ,and define the map ${F}^{\prime } : {M}^{\prime } + {Rm} \rightarrow Q$ by ${F}^{\prime }\left( {{m}^{\prime } + {rm}}\right) = F\left( {m}^{\prime }\right) + G\left( r\right)$ . If ${m}_{1} + {r}_{1}m = {m}_{2} + {r}_{2}m$ then $\left( {{r}_{1} - {r}_{2}}\right) m = {m}_{2} - {m}_{1}$ shows that ${r}_{1} - {r}_{2} \in I$ ,so that

证明：如果 $Q$ 是单射且 $g : I \rightarrow Q$ 是从 $R$ 的非零理想 $I$ 到 $Q$ 的 $R$ -模同态，那么根据命题34(2)应用于精确序列 $0 \rightarrow I \rightarrow R$ ，$g$ 可以扩展为从 $R$ 到 $Q$ 的 $R$ -模同态，这证明了(1)的“仅当”部分。假设反过来，每个同态 $g : I \rightarrow Q$ 都可以提升为同态 $G : R \rightarrow Q$ 。为了证明 $Q$ 是单射，我们必须证明如果 $0 \rightarrow L \rightarrow M$ 是精确的且 $f : L \rightarrow Q$ 是一个 $R$ -模同态，那么存在一个提升 $F : M \rightarrow Q$ 来扩展 $f$ 。如果 $\mathcal{S}$ 是提升 ${f}^{\prime } : {L}^{\prime } \rightarrow Q$ 的集合 $\left( {{f}^{\prime },{L}^{\prime }}\right)$ 到包含 $L$ 的 $M$ 的子模块 ${L}^{\prime }$ ，那么在 ${L}^{\prime }$ 上的排序 $\left( {{f}^{\prime },{L}^{\prime }}\right) \leq \left( {{f}^{\prime \prime },{L}^{\prime \prime }}\right)$ 如果 ${L}^{\prime } \subseteq {L}^{\prime \prime }$ 和 ${f}^{\prime \prime } = {f}^{\prime }$ 部分排序 $\mathcal{S}$ 。由于 $\mathcal{S} \neq \varnothing$ ，根据佐恩引理，在 $\mathcal{S}$ 中存在一个极大元素 $\left( {F,{M}^{\prime }}\right)$ 。映射 $F : {M}^{\prime } \rightarrow Q$ 是 $f$ 的提升，足够证明 ${M}^{\prime } = M$ 。假设存在某个元素 $m \in M$ 不包含在 ${M}^{\prime }$ 中，并令 $I = \left\{ {r \in R \mid {rm} \in {M}^{\prime }}\right\}$ 。容易验证 $I$ 是 $R$ 中的左理想，且由 $g\left( x\right) = F\left( {xm}\right)$ 定义的映射 $g : I \rightarrow Q$ 是从 $I$ 到 $Q$ 的 $R$ -模同态。根据假设，存在 $g$ 的提升 $G : R \rightarrow Q$ 。考虑 $M$ 的子模块 ${M}^{\prime } + {Rm}$ ，并通过 ${F}^{\prime }\left( {{m}^{\prime } + {rm}}\right) = F\left( {m}^{\prime }\right) + G\left( r\right)$ 定义映射 ${F}^{\prime } : {M}^{\prime } + {Rm} \rightarrow Q$ 。如果 ${m}_{1} + {r}_{1}m = {m}_{2} + {r}_{2}m$ ，那么 $\left( {{r}_{1} - {r}_{2}}\right) m = {m}_{2} - {m}_{1}$ 显示 ${r}_{1} - {r}_{2} \in I$ ，因此使得

and so $F\left( {m}_{1}\right) + G\left( {r}_{1}\right) = F\left( {m}_{2}\right) + G\left( {r}_{2}\right)$ . Hence ${F}^{\prime }$ is well defined and it is then immediate that ${F}^{\prime }$ is an $R$ -module homomorphism extending $f$ to ${M}^{\prime } + {Rm}$ . This contradicts the maximality of ${M}^{\prime }$ ,so that ${M}^{\prime } = M$ ,which completes the proof of (1).

因此 $F\left( {m}_{1}\right) + G\left( {r}_{1}\right) = F\left( {m}_{2}\right) + G\left( {r}_{2}\right)$ 。所以 ${F}^{\prime }$ 是良好定义的，并且立即可以得出 ${F}^{\prime }$ 是一个 $R$ -模同态，将 $f$ 扩展到 ${M}^{\prime } + {Rm}$ 。这与 ${M}^{\prime }$ 的极大性相矛盾，因此 ${M}^{\prime } = M$ ，这完成了 (1) 的证明。

To prove (2),suppose $R$ is a P.I.D. Any nonzero ideal $I$ of $R$ is of the form $I = \left( r\right)$ for some nonzero element $r$ of $R$ . An $R$ -module homomorphism $f : I \rightarrow Q$ is completely determined by the image $f\left( r\right) = q$ in $Q$ . This homomorphism can be extended to a homomorphism $F : R \rightarrow Q$ if and only if there is an element ${q}^{\prime }$ in $Q$ with $F\left( 1\right) = {q}^{\prime }$ satisfying $q = f\left( r\right) = F\left( r\right) = r{q}^{\prime }$ . It follows that Baer’s criterion for $Q$ is satisfied if and only if ${rQ} = Q$ ,which proves the first two statements in (2). The final statement follows since a quotient of a module $Q$ with ${rQ} = Q$ for all $r \neq 0$ in $R$ has the same property.

为了证明 (2)，假设 $R$ 是一个 P.I.D.。$R$ 的任何非零理想 $I$ 都可以表示为 $I = \left( r\right)$ 的形式，其中 $r$ 是 $R$ 中的某个非零元素。一个 $R$ -模同态 $f : I \rightarrow Q$ 完全由在 $Q$ 中的像 $f\left( r\right) = q$ 确定。当且仅当 $Q$ 中存在一个元素 ${q}^{\prime }$ 满足 $F\left( 1\right) = {q}^{\prime }$ 使得 $q = f\left( r\right) = F\left( r\right) = r{q}^{\prime }$ 时，这个同态可以扩展为一个同态 $F : R \rightarrow Q$ 。因此，如果且仅当 ${rQ} = Q$ 时，$Q$ 满足 Baer 准则，这证明了 (2) 中的前两个陈述。由于一个模 $Q$ 的商，对于 $R$ 中所有 $r \neq 0$ 都具有 ${rQ} = Q$ 的性质，最后一个陈述随之成立。

Examples

示例

(1) Since $\mathbb{Z}$ is not divisible, $\mathbb{Z}$ is not an injective $\mathbb{Z}$ -module. This also follows from the fact that the exact sequence $0 \rightarrow \mathbb{Z}\overset{2}{ \rightarrow }\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow 0$ corresponding to multiplication by 2 does not split.

(1) 由于 $\mathbb{Z}$ 不可除，$\mathbb{Z}$ 不是一个注入的 $\mathbb{Z}$ -模。这也从对应的乘以 2 的精确序列 $0 \rightarrow \mathbb{Z}\overset{2}{ \rightarrow }\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow 0$ 不分裂的事实得出。

(2) The rational numbers $\mathbb{Q}$ is an injective $\mathbb{Z}$ -module.

(2) 有理数 $\mathbb{Q}$ 是一个注入的 $\mathbb{Z}$ -模。

(3) The quotient $\mathbb{Q}/\mathbb{Z}$ of the injective $\mathbb{Z}$ -module $\mathbb{Q}$ is an injective $\mathbb{Z}$ -module.

(3) 注入 $\mathbb{Z}$ -模 $\mathbb{Q}$ 的商 $\mathbb{Q}/\mathbb{Z}$ 是一个注入的 $\mathbb{Z}$ -模。

(4) It is immediate that a direct sum of divisible $\mathbb{Z}$ -modules is again divisible,hence a direct sum of injective $\mathbb{Z}$ -modules is again injective. For example, $\mathbb{Q} \oplus \mathbb{Q}/\mathbb{Z}$ is an injective $\mathbb{Z}$ -module. (See also Exercise 4).

(4) 显然，可除 $\mathbb{Z}$ -模的直接和仍然是可除的，因此可除 $\mathbb{Z}$ -模的直接和仍然是可注的。例如，$\mathbb{Q} \oplus \mathbb{Q}/\mathbb{Z}$ 是一个可注 $\mathbb{Z}$ -模。（也见练习4）。

(5) We shall see in Chapter 12 that no nonzero finitely generated $\mathbb{Z}$ -module is injective.

(5) 我们将在第12章看到，没有非零有限生成的 $\mathbb{Z}$ -模是可注的。

(6) Suppose that the ring $R$ is an integral domain. An $R$ -module $A$ is said to be a divisible $R$ -module if ${rA} = A$ for every nonzero $r \in R$ . The proof of Proposition 36 shows that in this case an injective $R$ -module is divisible.

(6) 假设环 $R$ 是一个整环。一个 $R$ -模 $A$ 被称为可除 $R$ -模，如果对于每一个非零 $r \in R$ 成立 ${rA} = A$。命题36的证明表明，在这种情况下，一个可注 $R$ -模是可除的。

(7) We shall see in Section 11.1 that if $R = F$ is a field then every $F$ -module is injective.

(7) 我们将在11.1节看到，如果 $R = F$ 是一个域，那么每一个 $F$ -模都是可注的。

(8) We shall see in Part VI that if $F$ is any field and $n \in {\mathbb{Z}}^{ + }$ then the ring $R = {M}_{n}\left( F\right)$ of all $n \times n$ matrices with entries from $F$ has the property that every $R$ -module is injective (and also projective). We shall also see that if $G$ is a finite group of order $n$ and $n \neq 0$ in the field $F$ then the group ring ${FG}$ also has the property that every module is injective (and also projective).

(8) 我们将在第六部分看到，如果 $F$ 是任意域且 $n \in {\mathbb{Z}}^{ + }$ ，那么所有 $n \times n$ 矩阵的环 $R = {M}_{n}\left( F\right)$ ，其元素来自 $F$ ，具有每个 $R$ -模都是可注（并且也是可投射）的性质。我们还将看到，如果 $G$ 是一个有限阶为 $n$ 的群，并且 $n \neq 0$ 在域 $F$ 中，那么群环 ${FG}$ 也具有每个模都是可注（并且也是可投射）的性质。

Corollary 37. Every $\mathbb{Z}$ -module is a submodule of an injective $\mathbb{Z}$ -module.

推论37。每一个 $\mathbb{Z}$ -模都是某个可注 $\mathbb{Z}$ -模的子模。

Proof: Let $M$ be a $\mathbb{Z}$ -module and let $A$ be any set of $\mathbb{Z}$ -module generators of $M$ . Let $\mathcal{F} = F\left( A\right)$ be the free $\mathbb{Z}$ -module on the set $A$ . Then by Theorem 6 there is a surjective $\mathbb{Z}$ -module homomorphism from $\mathcal{F}$ to $M$ and if $\mathcal{K}$ denotes the kernel of this homomorphism then $\mathcal{K}$ is a $\mathbb{Z}$ -submodule of $\mathcal{F}$ and we can identify $M = \mathcal{F}/\mathcal{K}$ . Let $\mathcal{Q}$ be the free $\mathbb{Q}$ -module on the set $A$ . Then $\mathcal{Q}$ is a direct sum of a number of copies of $\mathbb{Q}$ , so is a divisible,hence (by Proposition 36) injective, $\mathbb{Z}$ -module containing $\mathcal{F}$ . Then $\mathcal{K}$ is also a $\mathbb{Z}$ -submodule of $\mathcal{Q}$ ,so the quotient $\mathcal{Q}/\mathcal{K}$ is injective,again by Proposition 36. Since $M = \mathcal{F}/\mathcal{K} \subseteq \mathcal{Q}/\mathcal{K}$ ,it follows that $M$ is contained in an injective $\mathbb{Z}$ -module.

证明：设 $M$ 是一个 $\mathbb{Z}$ -模，设 $A$ 是 $M$ 的一组任意 $\mathbb{Z}$ -模生成元。设 $\mathcal{F} = F\left( A\right)$ 是在集合 $A$ 上的自由 $\mathbb{Z}$ -模。根据定理6，存在从 $\mathcal{F}$ 到 $M$ 的满射 $\mathbb{Z}$ -模同态，如果 $\mathcal{K}$ 表示这个同态的核，那么 $\mathcal{K}$ 是 $\mathcal{F}$ 的一个 $\mathbb{Z}$ -子模，我们可以识别 $M = \mathcal{F}/\mathcal{K}$。设 $\mathcal{Q}$ 是在集合 $A$ 上的自由 $\mathbb{Q}$ -模。那么 $\mathcal{Q}$ 是 $\mathbb{Q}$ 的若干个副本的直接和，因此是可除的（由命题36），所以是包含 $\mathcal{F}$ 的内射 $\mathbb{Z}$ -模。因此 $\mathcal{K}$ 也是 $\mathcal{Q}$ 的一个 $\mathbb{Z}$ -子模，所以商 $\mathcal{Q}/\mathcal{K}$ 是内射的，再次由命题36得出。由于 $M = \mathcal{F}/\mathcal{K} \subseteq \mathcal{Q}/\mathcal{K}$ ，因此 $M$ 包含在一个内射的 $\mathbb{Z}$ -模中。

Corollary 37 can be used to prove the following more general version valid for arbitrary $R$ -modules. This theorem is the injective analogue of the results in Theorem 6 and Corollary 31 showing that every $R$ -module is a quotient of a projective $R$ -module.

推论37可以用来证明以下更一般的版本，对于任意的 $R$ -模都是有效的。这个定理是定理6和推论31的内射类似物，表明每个 $R$ -模都是某个投射 $R$ -模的商。

Theorem 38. Let $R$ be a ring with 1 and let $M$ be an $R$ -module. Then $M$ is contained in an injective $R$ -module.

定理38。设 $R$ 是一个带有单位元的环，设 $M$ 是一个 $R$ -模。那么 $M$ 包含在一个内射的 $R$ -模中。

Proof: A proof is outlined in Exercises 15 to 17.

证明：证明大纲在练习15到17中概述。

It is possible to prove a sharper result than Theorem 38, namely that there is a minimal injective $R$ -module $H$ containing $M$ in the sense that any injective map of $M$ into an injective $R$ -module $Q$ factors through $H$ . More precisely,if $M \subseteq Q$ for an injective $R$ -module $Q$ then there is an injection $\iota : H \hookrightarrow Q$ that restricts to the identity map on $M$ ; using $\iota$ to identify $H$ as a subset of $Q$ we have $M \subseteq H \subseteq Q$ . (cf. Theorem 57.13 in Representation Theory of Finite Groups and Associative Algebras by C. Curtis and I. Reiner,John Wiley &Sons,1966). This module $H$ is called the injective hull or injective envelope of $M$ . The universal property of the injective hull of $M$ with respect to inclusions of $M$ into injective $R$ -modules should be compared to the universal property with respect to homomorphisms of $M$ of the free module $F\left( A\right)$ on a set of generators $A$ for $M$ in Theorem 6. For example,the injective hull of $\mathbb{Z}$ is $\mathbb{Q}$ ,and the injective hull of any field is itself (cf. the exercises).

可以证明一个比定理38更强的结果，即存在一个包含 $R$ -模 $H$ 的最小单射模 $M$ ，在 $M$ 到任何 $R$ -单射模 $Q$ 的单射映射的意义上，它可以通过 $H$ 实现。更准确地说，如果 $M \subseteq Q$ 对于一个 $R$ -单射模 $Q$ ，那么存在一个单射 $\iota : H \hookrightarrow Q$ ，它限制在 $M$ 上是恒等映射；使用 $\iota$ 识别 $H$ 作为 $Q$ 的子集，我们得到 $M \subseteq H \subseteq Q$。（参见C. Curtis和I. Reiner的《有限群和结合代数的表示论》中的定理57.13，John Wiley &Sons, 1966）。这个模 $H$ 被称为 $M$ 的单射 hull 或单射 envelope。关于 $M$ 的单射 hull 的普遍性质，与定理6中关于生成元集合 $A$ 的自由模 $F\left( A\right)$ 的同态的普遍性质相比，应考虑 $M$ 到单射 $R$ -模的包含。例如，$\mathbb{Z}$ 的单射 hull 是 $\mathbb{Q}$ ，任何域的单射 hull 是其本身（参见练习）。

Flat Modules and $D{ \otimes }_{R}$ -

扁平模和 $D{ \otimes }_{R}$ -

We now consider the behavior of extensions $0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ of $R$ -modules with respect to tensor products.

现在我们考虑 $R$ -模的扩张 $0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ 对张量积的行为。

Suppose that $D$ is a right $R$ -module. For any homomorphism $f : X \rightarrow Y$ of left $R$ -modules we obtain a homomorphism $1 \otimes f : D{ \otimes }_{R}X \rightarrow D{ \otimes }_{R}Y$ of abelian groups (Theorem 13). If in addition $D$ is an(S,R)-bimodule (for example,when $S = R$ is commutative and $D$ is given the standard(R,R)-bimodule structure as in Section 4), then $1 \otimes f$ is a homomorphism of left $S$ -modules. Put another way,

假设 $D$ 是一个右 $R$ -模。对于任意左 $R$ -模的同态 $f : X \rightarrow Y$，我们可以得到一个阿贝尔群的同态 $1 \otimes f : D{ \otimes }_{R}X \rightarrow D{ \otimes }_{R}Y$（定理13）。此外，如果 $D$ 是一个(S,R)-双模（例如，当 $S = R$ 是交换的并且 $D$ 具有如第4节中给出的标准(R,R)-双模结构时），那么 $1 \otimes f$ 是一个左 $S$ -模的同态。换句话说，

is a covariant functor from the category of left $R$ -modules to the category of abelian groups (respectively,to the category of left $S$ -modules when $D$ is an(S,R)-bimodule), cf. Appendix II. In a similar way,if $D$ is a left $R$ -module then $\_ { \otimes }_{R}D$ is a covariant functor from the category of right $\textit{R-modules}$ to the category of abelian groups (respectively,to the category of right $S$ -modules when $D$ is an(R,S)-bimodule). Note that, unlike Hom, the tensor product is covariant in both variables, and we shall therefore concentrate on $D{ \otimes }_{R}$ ,leaving as an exercise the minor alterations necessary for _____ ${ \otimes }_{R}D$ .

是从左 $R$ -模的类别到阿贝尔群类别（相应地，当 $D$ 是一个(S,R)-双模时，到左 $S$ -模的类别）的协变函子，参见附录II。类似地，如果 $D$ 是一个左 $R$ -模，那么 $\_ { \otimes }_{R}D$ 是从右 $\textit{R-modules}$ 的类别到阿贝尔群类别（相应地，当 $D$ 是一个(R,S)-双模时，到右 $S$ -模的类别）的协变函子。注意，与 Hom 不同，张量积在两个变量上都是协变的，因此我们将集中研究 $D{ \otimes }_{R}$，将 ${ \otimes }_{R}D$ 的必要小改动留作练习。

We have already seen examples where the map $1 \otimes \psi : D{ \otimes }_{R}L \rightarrow D{ \otimes }_{R}M$ induced by an injective map $\psi : L \hookrightarrow M$ is no longer injective (for example the injection $\mathbb{Z} \hookrightarrow \mathbb{Q}$ of $\mathbb{Z}$ -modules induces the zero map from $\mathbb{Z}/2\mathbb{Z}{ \otimes }_{\mathbb{Z}}\mathbb{Z} = \mathbb{Z}/2\mathbb{Z}$ to $\mathbb{Z}/2\mathbb{Z}{ \otimes }_{\mathbb{Z}}\mathbb{Q} = 0).\;$ On the other hand,suppose that $\varphi : M \rightarrow N$ is a surjective $R$ -module homomorphism. The tensor product $D{ \otimes }_{R}N$ is generated as an abelian group by the simple tensors $d \otimes n$ for $d \in D$ and $n \in N$ . The surjectivity of $\varphi$ implies that $n = \varphi \left( m\right)$ for some $m \in M$ ,and then $1 \otimes \varphi \left( {d \otimes m}\right) = d \otimes \varphi \left( m\right) = d \otimes n$ shows that $1 \otimes \varphi$ is a surjective homomorphism of abelian groups from $D{ \otimes }_{R}M$ to $D{ \otimes }_{R}N$ . This proves most of the following theorem.

我们已经看到了一些例子，其中由单射 $1 \otimes \psi : D{ \otimes }_{R}L \rightarrow D{ \otimes }_{R}M$ 引导的映射 $\psi : L \hookrightarrow M$ 不再是单射（例如，$\mathbb{Z} \hookrightarrow \mathbb{Q}$ 的注入 $\mathbb{Z}$ -模导致从 $\mathbb{Z}/2\mathbb{Z}{ \otimes }_{\mathbb{Z}}\mathbb{Z} = \mathbb{Z}/2\mathbb{Z}$ 到 $\mathbb{Z}/2\mathbb{Z}{ \otimes }_{\mathbb{Z}}\mathbb{Q} = 0).\;$ 的零映射）。另一方面，假设 $\varphi : M \rightarrow N$ 是一个满射 $R$ -模同态。张量积 $D{ \otimes }_{R}N$ 作为一个阿贝尔群由简单张量 $d \otimes n$ 生成，对于 $d \in D$ 和 $n \in N$ 。$\varphi$ 的满射性意味着对于某些 $m \in M$ ，存在 $n = \varphi \left( m\right)$，然后 $1 \otimes \varphi \left( {d \otimes m}\right) = d \otimes \varphi \left( m\right) = d \otimes n$ 显示 $1 \otimes \varphi$ 是从 $D{ \otimes }_{R}M$ 到 $D{ \otimes }_{R}N$ 的阿贝尔群的满射同态。这证明了以下定理的大部分内容。

Theorem 39. Suppose that $D$ is a right $R$ -module and that $L,M$ and $N$ are left $R$ -modules. If

定理39。假设 $D$ 是一个右 $R$ -模，且 $L,M$ 和 $N$ 是左 $R$ -模。如果

then the associated sequence of abelian groups

那么，相关的阿贝尔群序列

If $D$ is an(S,R)-bimodule then (13) is an exact sequence of left $S$ -modules. In particular,if $S = R$ is a commutative ring,then (13) is an exact sequence of $R$ -modules with respect to the standard $R$ -module structures. The map $1 \otimes \varphi$ is not in general injective i.e., the sequence (13) cannot in general be extended to a short exact sequence.

如果 $D$ 是一个 (S,R)-双边模，那么 (13) 是左 $S$ -模的精确序列。特别是，如果 $S = R$ 是一个交换环，那么 (13) 是关于标准 $R$ -模结构的 $R$ -模的精确序列。映射 $1 \otimes \varphi$ 通常不是单射的，即序列 (13) 通常不能扩展为短精确序列。

The sequence (13) is exact for all right $R$ -modules $D$ if and only if

序列 (13) 对于所有右 $R$ -模 $D$ 是精确的，当且仅当

Proof: For the first statement it remains to prove the exactness of (13) at $D{ \otimes }_{R}M$ . Since $\varphi \circ \psi = 0$ ,we have

证明：对于第一个陈述，需要证明 (13) 在 $D{ \otimes }_{R}M$ 处的精确性。由于 $\varphi \circ \psi = 0$ ，我们有

and it follows that image $\left( {1 \otimes \psi }\right) \subseteq \ker \left( {1 \otimes \varphi }\right)$ . In particular,there is a natural projection $\pi : \left( {D{ \otimes }_{R}M}\right) /\;\mathrm{{image}}\left( {1 \otimes \psi }\right) \rightarrow \left( {D{ \otimes }_{R}M}\right) /\ker \left( {1 \otimes \varphi }\right) = D{ \otimes }_{R}N.\;\mathrm{{The}\;{composite}}$ of the two projection homomorphisms

因此图像 $\left( {1 \otimes \psi }\right) \subseteq \ker \left( {1 \otimes \varphi }\right)$ 。特别是，存在两个投影同态的自然投影 $\pi : \left( {D{ \otimes }_{R}M}\right) /\;\mathrm{{image}}\left( {1 \otimes \psi }\right) \rightarrow \left( {D{ \otimes }_{R}M}\right) /\ker \left( {1 \otimes \varphi }\right) = D{ \otimes }_{R}N.\;\mathrm{{The}\;{composite}}$。

is the quotient of $D{ \otimes }_{R}M$ by $\ker \left( {1 \otimes \varphi }\right)$ ,so is just the map $1 \otimes \varphi$ . We shall show that $\pi$ is an isomorphism,which will show that the kernel of $1 \otimes \varphi$ is just the kernel of the first projection above,i.e.,image $\left( {1 \otimes \psi }\right)$ ,giving the exactness of (13) at $D{ \otimes }_{R}M$ . To see that $\pi$ is an isomorphism we define an inverse map. First define ${\pi }^{\prime } : D \times N \rightarrow$ $\left( {D{ \otimes }_{R}M}\right) /\operatorname{image}\left( {1 \otimes \psi }\right)$ by ${\pi }^{\prime }\left( \left( {d,n}\right) \right) = d \otimes m$ for any $m \in M$ with $\varphi \left( m\right) = n.$ Note that this is well defined: any other element ${m}^{\prime } \in M$ mapping to $n$ differs from $m$ by an element in $\ker \varphi = \operatorname{image}\psi$ ,i.e., ${m}^{\prime } = m + \psi \left( l\right)$ for some $l \in L$ ,and $d \otimes \psi \left( l\right) \in \operatorname{image}\left( {1 \otimes \psi }\right)$ . It is easy to check that ${\pi }^{\prime }$ is a balanced map,so induces a homomorphism $\widetilde{\pi } : D \times N \rightarrow \left( {D{ \otimes }_{R}M}\right) /$ image $\left( {1 \otimes \psi }\right)$ with $\widetilde{\pi }\left( {d \otimes n}\right) = d \otimes m.$ Then $\widetilde{\pi } \circ \pi \left( {d \otimes m}\right) = \widetilde{\pi }\left( {d \otimes \varphi \left( m\right) }\right) = d \otimes m$ shows that $\widetilde{\pi } \circ \pi = 1$ . Similarly, $\pi \circ \widetilde{\pi } = 1$ ,so that $\pi$ and $\widetilde{\pi }$ are inverse isomorphisms,completing the proof that (13) is exact. Note also that the injectivity of $\psi$ was not required for the proof.

是 $D{ \otimes }_{R}M$ 除以 $\ker \left( {1 \otimes \varphi }\right)$ 的商，因此仅仅是映射 $1 \otimes \varphi$ 。我们将证明 $\pi$ 是同构，这将表明 $1 \otimes \varphi$ 的核仅仅是上述第一个投影的核，即图像 $\left( {1 \otimes \psi }\right)$ ，从而证明了（13）在 $D{ \otimes }_{R}M$ 处的精确性。为了看出 $\pi$ 是同构，我们定义一个逆映射。首先定义 ${\pi }^{\prime } : D \times N \rightarrow$ $\left( {D{ \otimes }_{R}M}\right) /\operatorname{image}\left( {1 \otimes \psi }\right)$ 通过 ${\pi }^{\prime }\left( \left( {d,n}\right) \right) = d \otimes m$ 对于任何满足 $\varphi \left( m\right) = n.$ 的 $m \in M$ 。注意这是良好定义的：映射到 $n$ 的任何其他元素 ${m}^{\prime } \in M$ 都与 $m$ 相差一个属于 $\ker \varphi = \operatorname{image}\psi$ 的元素，即存在某个 $l \in L$ 使得 ${m}^{\prime } = m + \psi \left( l\right)$ ，并且 $d \otimes \psi \left( l\right) \in \operatorname{image}\left( {1 \otimes \psi }\right)$ 。容易验证 ${\pi }^{\prime }$ 是一个平衡映射，因此诱导出一个同态 $\widetilde{\pi } : D \times N \rightarrow \left( {D{ \otimes }_{R}M}\right) /$ 图像 $\left( {1 \otimes \psi }\right)$ 并满足 $\widetilde{\pi }\left( {d \otimes n}\right) = d \otimes m.$ 。然后 $\widetilde{\pi } \circ \pi \left( {d \otimes m}\right) = \widetilde{\pi }\left( {d \otimes \varphi \left( m\right) }\right) = d \otimes m$ 显示 $\widetilde{\pi } \circ \pi = 1$ 。类似地， $\pi \circ \widetilde{\pi } = 1$ ，因此 $\pi$ 和 $\widetilde{\pi }$ 是互为逆的同构，完成了（13）是精确的证明。还应注意，证明中并未要求 $\psi$ 的单射性。

Finally,suppose (13) is exact for every right $R$ -module $D$ . In general, $R{ \otimes }_{R}X \cong X$ for any left $R$ -module $X$ (Example 1 following Corollary 9). Taking $D = R$ the exactness of the sequence $L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ follows.

最后，假设对于每个右 $R$ -模 $D$ ，（13）都是精确的。一般来说， $R{ \otimes }_{R}X \cong X$ 对于任何左 $R$ -模 $X$ （例1跟随推论9）。取 $D = R$ 序列的精确性随之而来 $L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ 。

By Theorem 39, the sequence

根据定理39，序列

is not in general exact since $1 \otimes \psi$ need not be injective. If $0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ is a split short exact sequence, however, then since tensor products commute with direct sums by Theorem 17, it follows that

通常来说并不精确，因为 $1 \otimes \psi$ 不一定是单射的。然而，如果 $0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ 是一个分裂的短正合序列，由于定理17，张量积与直和可交换，因此可以得出

is also a split short exact sequence.

也是一个分裂的短正合序列。

The following result relating to modules $D$ having the property that (13) can always be extended to a short exact sequence is immediate from Theorem 39:

下面这个关于具有性质（13）总能扩展为短正合序列的模 $D$ 的结果直接来源于定理39：

Proposition 40. Let $A$ be a right $R$ -module. Then the following are equivalent:

命题40。设 $A$ 是一个右 $R$ -模。那么以下条件是等价的：

(1) For any left $R$ -modules $L,M$ ,and $N$ ,if

（1）对于任何左 $R$ -模 $L,M$ 和 $N$，如果

is a short exact sequence, then

是一个短正合序列，那么

is also a short exact sequence.

也是一个短正合序列。

(2) For any left $R$ -modules $L$ and $M$ ,if $0 \rightarrow L\overset{\psi }{ \rightarrow }M$ is an exact sequence of left $R$ -modules (i.e., $\psi : L \rightarrow M$ is injective) then $0 \rightarrow A{ \otimes }_{R}L\overset{1 \otimes \psi }{ \rightarrow }A{ \otimes }_{R}M$ is an exact sequence of abelian groups (i.e., $1 \otimes \psi : A{ \otimes }_{R}L \rightarrow A{ \otimes }_{R}M$ is injective).

（2）对于任何左 $R$ -模 $L$ 和 $M$，如果 $0 \rightarrow L\overset{\psi }{ \rightarrow }M$ 是左 $R$ -模的精确序列（即 $\psi : L \rightarrow M$ 是单射的），那么 $0 \rightarrow A{ \otimes }_{R}L\overset{1 \otimes \psi }{ \rightarrow }A{ \otimes }_{R}M$ 是阿贝尔群的精确序列（即 $1 \otimes \psi : A{ \otimes }_{R}L \rightarrow A{ \otimes }_{R}M$ 是单射的）。

Definition. A right $R$ -module $A$ is called flat if it satisfies either of the two equivalent conditions of Proposition 40.

定义。一个右 $R$ -模 $A$ 如果满足命题40中的两个等价条件之一，则被称为平坦模。

For a fixed right $R$ -module $D$ ,the first part of Theorem 39 is referred to by saying that the functor $D{ \otimes }_{R}$ is right exact.

对于一个固定的右 $R$ -模 $D$，定理39的第一部分指的是说函子 $D{ \otimes }_{R}$ 是右正合的。

Corollary 41. If $D$ is a right $R$ -module,then the functor $D{ \otimes }_{R}$ _____from the category of left $R$ -modules to the category of abelian groups is right exact. If $D$ is an(S,R)- bimodule (for example when $S = R$ is commutative and $D$ is given the standard $R$ -module structure),then $D{ \otimes }_{R}$ _____is a right exact functor from the category of left $R$ -modules to the category of left $S$ -modules. The functor is exact if and only if $D$ is a flat $R$ -module.

推论 41. 如果 $D$ 是一个右 $R$ -模，那么从左 $R$ -模范畴到阿贝尔群范畴的函子 $D{ \otimes }_{R}$ _____ 是右正合的。如果 $D$ 是一个 (S,R) 双模（例如当 $S = R$ 是交换的并且 $D$ 具有标准的 $R$ -模结构时），那么 $D{ \otimes }_{R}$ _____ 是从左 $R$ -模范畴到左 $S$ -模范畴的右正合函子。当且仅当 $D$ 是一个平坦 $R$ -模时，该函子是正合的。

We have already seen some flat modules:

我们已经看到了一些平坦模：

Corollary 42. Free modules are flat; more generally, projective modules are flat.

推论 42. 自由模是平坦的；更一般地，投射模也是平坦的。

Proof: To show that the free $R$ -module $F$ is flat it suffices to show that for any injective map $\psi : L \rightarrow M$ of $R$ -modules $L$ and $M$ the induced map $1 \otimes \psi : F{ \otimes }_{R}L \rightarrow$ $F{ \otimes }_{R}M$ is also injective. Suppose first that $F \cong {R}^{n}$ is a finitely generated free $R$ - module. In this case $F{ \otimes }_{R}L = {R}^{n}{ \otimes }_{R}L \cong {L}^{n}$ since $R{ \otimes }_{R}L \cong L$ and tensor products commute with direct sums. Similarly $F{ \otimes }_{R}M \cong {M}^{n}$ and under these isomorphisms the map $1 \otimes \psi : F{ \otimes }_{R}L \rightarrow F{ \otimes }_{R}M$ is just the natural map of ${L}^{n}$ to ${M}^{n}$ induced by the inclusion $\psi$ in each component. In particular, $1 \otimes \psi$ is injective and it follows that any finitely generated free module is flat. Suppose now that $F$ is an arbitrary free module and that the element $\sum {f}_{i} \otimes {l}_{i} \in F{ \otimes }_{R}L$ is mapped to 0 by $1 \otimes \psi$ . This means that the element $\sum \left( {{f}_{i},\psi \left( {l}_{i}\right) }\right)$ can be written as a sum of generators as in equation (6) in the previous section in the free group on $F \times M$ . Since this sum of elements is finite, all of the first coordinates of the resulting equation lie in some finitely generated free submodule ${F}^{\prime }$ of $F$ . Then this equation implies that $\sum {f}_{i} \otimes {l}_{i} \in {F}^{\prime }{ \otimes }_{R}L$ is mapped to 0 in ${F}^{\prime }{ \otimes }_{R}M$ . Since ${F}^{\prime }$ is a finitely generated free module,the injectivity we proved above shows that $\sum {f}_{i} \otimes {l}_{i}$ is 0 in ${F}^{\prime }{ \otimes }_{R}L$ and so also in $F{ \otimes }_{R}L$ . It follows that $1 \otimes \psi$ is injective and hence that $F$ is flat.

证明：要证明自由的 $R$ -模 $F$ 是平坦的，只需证明对于任何 $R$ -模 $L$ 和 $M$ 之间的单射映射 $\psi : L \rightarrow M$，所诱导的映射 $1 \otimes \psi : F{ \otimes }_{R}L \rightarrow$ $F{ \otimes }_{R}M$ 也是单射的。首先假设 $F \cong {R}^{n}$ 是一个有限生成的自由 $R$ -模。在这种情况下 $F{ \otimes }_{R}L = {R}^{n}{ \otimes }_{R}L \cong {L}^{n}$，由于 $R{ \otimes }_{R}L \cong L$ 且张量积与直和交换。类似地 $F{ \otimes }_{R}M \cong {M}^{n}$，在这些同构下，映射 $1 \otimes \psi : F{ \otimes }_{R}L \rightarrow F{ \otimes }_{R}M$ 只是从 ${L}^{n}$ 到 ${M}^{n}$ 的自然映射，由每个分量中的包含 $\psi$ 诱导。特别是，$1 \otimes \psi$ 是单射的，因此任何有限生成的自由模都是平坦的。现在假设 $F$ 是一个任意的自由模，且元素 $\sum {f}_{i} \otimes {l}_{i} \in F{ \otimes }_{R}L$ 通过 $1 \otimes \psi$ 映射到 0。这意味着元素 $\sum \left( {{f}_{i},\psi \left( {l}_{i}\right) }\right)$ 可以写成如上一节方程（6）中的生成元的和，在 $F \times M$ 的自由群上。由于这个元素的和是有限的，所得方程的所有第一个坐标都位于 $F$ 的某个有限生成的自由子模 ${F}^{\prime }$ 中。那么这个方程意味着 $\sum {f}_{i} \otimes {l}_{i} \in {F}^{\prime }{ \otimes }_{R}L$ 在 ${F}^{\prime }{ \otimes }_{R}M$ 中映射到 0。由于 ${F}^{\prime }$ 是一个有限生成的自由模，我们上面证明的单射性表明 $\sum {f}_{i} \otimes {l}_{i}$ 在 ${F}^{\prime }{ \otimes }_{R}L$ 中是 0，因此在 $F{ \otimes }_{R}L$ 中也是 0。因此 $1 \otimes \psi$ 是单射的，从而 $F$ 是平坦的。

Suppose now that $P$ is a projective module. Then $P$ is a direct summand of a free module $F$ (Proposition 30),say $F = P \oplus {P}^{\prime }$ . If $\psi : L \rightarrow M$ is injective then $1 \otimes \psi : F{ \otimes }_{R}L \rightarrow F{ \otimes }_{R}M$ is also injective by what we have already shown. Since $F = P \oplus {P}^{\prime }$ and tensor products commute with direct sums,this shows that

假设现在 $P$ 是一个射影模。那么 $P$ 是自由模 $F$ 的直和项（命题30），比如说 $F = P \oplus {P}^{\prime }$。如果 $\psi : L \rightarrow M$ 是内射的，那么根据我们已经证明的，$1 \otimes \psi : F{ \otimes }_{R}L \rightarrow F{ \otimes }_{R}M$ 也是内射的。由于 $F = P \oplus {P}^{\prime }$ 和张量积与直和交换，这表明

is injective. Hence $1 \otimes \psi : P{ \otimes }_{R}L \rightarrow P{ \otimes }_{R}M$ is injective,proving that $P$ is flat.

是内射的。因此 $1 \otimes \psi : P{ \otimes }_{R}L \rightarrow P{ \otimes }_{R}M$ 是内射的，证明了 $P$ 是平坦的。

Examples

示例

(1) Since $\mathbb{Z}$ is a projective $\mathbb{Z}$ -module it is flat. The example before Theorem 39 shows that $\mathbb{Z}/2\mathbb{Z}$ not a flat $\mathbb{Z}$ -module.

(1) 由于 $\mathbb{Z}$ 是一个射影 $\mathbb{Z}$ -模，它是平坦的。定理39之前的例子表明 $\mathbb{Z}/2\mathbb{Z}$ 不是平坦的 $\mathbb{Z}$ -模。

(2) The $\mathbb{Z}$ -module $\mathbb{Q}$ is a flat $\mathbb{Z}$ -module,as follows. Suppose $\psi : L \rightarrow M$ is an injective map of $\mathbb{Z}$ -modules. Every element of $\mathbb{Q}{ \otimes }_{\mathbb{Z}}L$ can be written in the form $\left( {1/d}\right) \otimes l$ for some nonzero integer $d$ and some $l \in L$ (Exercise 7 in Section 4). If $\left( {1/d}\right) \otimes l$ is in the kernel of $1 \otimes \psi$ then $\left( {1/d}\right) \otimes \psi \left( l\right)$ is0in $\mathbb{Q}{ \otimes }_{\mathbb{Z}}M$ . By Exercise 8 in Section 4 this means ${c\psi }\left( l\right) = 0$ in $M$ for some nonzero integer $c$ . Then $\psi \left( {c \cdot l}\right) = 0$ ,and the injectivity of $\psi$ implies $c \cdot l = 0$ in $L$ . But this implies that $\left( {1/d}\right) \otimes l = \left( {1/{cd}}\right) \otimes \left( {c \cdot l}\right) = 0$ in $L$ , which shows that $1 \otimes \psi$ is injective.

(2) $\mathbb{Z}$ -模 $\mathbb{Q}$ 是一个平坦 $\mathbb{Z}$ -模，如下所示。假设 $\psi : L \rightarrow M$ 是一个 $\mathbb{Z}$ -模的内射映射。$\mathbb{Q}{ \otimes }_{\mathbb{Z}}L$ 的每个元素都可以写成 $\left( {1/d}\right) \otimes l$ 的形式，对于某些非零整数 $d$ 和某些 $l \in L$（第4节的练习7）。如果 $\left( {1/d}\right) \otimes l$ 在 $1 \otimes \psi$ 的核中，那么 $\left( {1/d}\right) \otimes \psi \left( l\right)$ 在 $\mathbb{Q}{ \otimes }_{\mathbb{Z}}M$ 中为0。根据第4节的练习8，这意味着 ${c\psi }\left( l\right) = 0$ 在 $M$ 中对于某些非零整数 $c$。那么 $\psi \left( {c \cdot l}\right) = 0$，并且 $\psi$ 的内射性意味着 $c \cdot l = 0$ 在 $L$ 中。但这意味着 $\left( {1/d}\right) \otimes l = \left( {1/{cd}}\right) \otimes \left( {c \cdot l}\right) = 0$ 在 $L$ 中，这表明 $1 \otimes \psi$ 是内射的。

(3) The $\mathbb{Z}$ -module $\mathbb{Q}/\mathbb{Z}$ is injective (by Proposition 36),but is not flat: the injective map $\psi \left( z\right) = {2z}$ from $\mathbb{Z}$ to $\mathbb{Z}$ does not remain injective after tensoring with $\mathbb{Q}/\mathbb{Z}$ $(1 \otimes \psi : \mathbb{Q}/\mathbb{Z}{ \otimes }_{\mathbb{Z}}\mathbb{Z} \rightarrow \mathbb{Q}/\mathbb{Z} \otimes \mathbb{Z}$ has the nonzero element $\left( {\frac{1}{2} + \mathbb{Z}}\right) \otimes 1$ in its kernel - identifying $\mathbb{Q}/\mathbb{Z} = \mathbb{Q}/\mathbb{Z}{ \otimes }_{\mathbb{Z}}\mathbb{Z}$ this is the statement that multiplication by 2 has the element $1/2$ in its kernel).

(3) $\mathbb{Z}$ -模 $\mathbb{Q}/\mathbb{Z}$ 是可注入的（由命题36得出），但不是平坦的：从 $\mathbb{Z}$ 到 $\mathbb{Z}$ 的可注入映射 $\psi \left( z\right) = {2z}$ 在与 $\mathbb{Q}/\mathbb{Z}$ 张量积后不再保持可注入性 $(1 \otimes \psi : \mathbb{Q}/\mathbb{Z}{ \otimes }_{\mathbb{Z}}\mathbb{Z} \rightarrow \mathbb{Q}/\mathbb{Z} \otimes \mathbb{Z}$ 在其核中具有非零元素 $\left( {\frac{1}{2} + \mathbb{Z}}\right) \otimes 1$ - 识别 $\mathbb{Q}/\mathbb{Z} = \mathbb{Q}/\mathbb{Z}{ \otimes }_{\mathbb{Z}}\mathbb{Z}$ 这意味着乘以2在核中有元素 $1/2$）。

(4) The direct sum of flat modules is flat (Exercise 5). In particular, $\mathbb{Q} \oplus \mathbb{Z}$ is flat. This module is neither projective nor injective (since $\mathbb{Q}$ is not projective by Exercise 8 and $\mathbb{Z}$ is not injective by Proposition 36 (cf. Exercises 3 and 4).

(4) 扁平模的直和是扁平的（练习5）。特别是，$\mathbb{Q} \oplus \mathbb{Z}$ 是扁平的。这个模既不是投射的也不是可注入的（因为 $\mathbb{Q}$ 不是投射的，由练习8得出，$\mathbb{Z}$ 不是可注入的，由命题36得出（参见练习3和4））。

We close this section with an important relation between Hom and tensor products:

我们以一个Hom和张量积之间的重要关系结束这一节：

Theorem 43. (Adjoint Associativity) Let $R$ and $S$ be rings,let $A$ be a right $R$ -module,let $B$ be an(R,S)-bimodule and let $C$ be a right $S$ -module. Then there is an isomorphism of abelian groups:

定理43.（伴随结合律）设 $R$ 和 $S$ 是环，$A$ 是一个右 $R$ -模，$B$ 是一个(R,S)-双边模，$C$ 是一个右 $S$ -模。那么存在阿贝尔群的同构：

(the homomorphism groups are right module homomorphisms-note that ${\operatorname{Hom}}_{S}\left( {B,C}\right)$ has the structure of a right $R$ -module,cf. the exercises). If $R = S$ is commutative this is an isomorphism of $R$ -modules with the standard $R$ -module structures.

（同态群是右模同态 - 注意 ${\operatorname{Hom}}_{S}\left( {B,C}\right)$ 具有右 $R$ -模的结构，参见练习）。如果 $R = S$ 是交换的，这是一个 $R$ -模的同构，具有标准的 $R$ -模结构。

Proof: Suppose $\varphi : A{ \otimes }_{R}B \rightarrow C$ is a homomorphism. For any fixed $a \in A$ define the map $\Phi \left( a\right)$ from $B$ to $C$ by $\Phi \left( a\right) \left( b\right) = \varphi \left( {a \otimes b}\right)$ . It is easy to check that $\Phi \left( a\right)$ is a homomorphism of right $S$ -modules and that the map $\Phi$ from $A$ to ${\operatorname{Hom}}_{S}\left( {B,C}\right)$ given by mapping $a$ to $\Phi \left( a\right)$ is a homomorphism of right $R$ -modules. Then $f\left( \varphi \right) = \Phi$ defines a group homomorphism from ${\operatorname{Hom}}_{S}\left( {A{ \otimes }_{R}B,C}\right)$ to ${\operatorname{Hom}}_{R}\left( {A,{\operatorname{Hom}}_{S}\left( {B,C}\right) }\right)$ . Conversely,suppose $\Phi : A \rightarrow {\operatorname{Hom}}_{S}\left( {B,C}\right)$ is a homomorphism. The map from $A \times B$ to $C$ defined by mapping(a,b)to $\Phi \left( a\right) \left( c\right)$ is an $R$ -balanced map,so induces a homomorphism $\varphi$ from $A{ \otimes }_{R}B$ to $C$ . Then $g\left( \Phi \right) = \varphi$ defines a group homomorphism inverse to $f$ and gives the isomorphism in the theorem.

证明：假设 $\varphi : A{ \otimes }_{R}B \rightarrow C$ 是一个同态。对于任意固定的 $a \in A$ ，定义从 $B$ 到 $C$ 的映射 $\Phi \left( a\right)$ 通过 $\Phi \left( a\right) \left( b\right) = \varphi \left( {a \otimes b}\right)$ 。容易验证 $\Phi \left( a\right)$ 是一个右 $S$ -模的同态，且由将 $a$ 映射到 $\Phi \left( a\right)$ 的映射 $\Phi$ 从 $A$ 到 ${\operatorname{Hom}}_{S}\left( {B,C}\right)$ 是一个右 $R$ -模的同态。那么 $f\left( \varphi \right) = \Phi$ 定义了一个从 ${\operatorname{Hom}}_{S}\left( {A{ \otimes }_{R}B,C}\right)$ 到 ${\operatorname{Hom}}_{R}\left( {A,{\operatorname{Hom}}_{S}\left( {B,C}\right) }\right)$ 的群同态。反之，假设 $\Phi : A \rightarrow {\operatorname{Hom}}_{S}\left( {B,C}\right)$ 是一个同态。由将 (a,b) 映射到 $\Phi \left( a\right) \left( c\right)$ 定义的从 $A \times B$ 到 $C$ 的映射是一个 $R$ -平衡映射，因此诱导出一个从 $A{ \otimes }_{R}B$ 到 $C$ 的同态 $\varphi$ 。那么 $g\left( \Phi \right) = \varphi$ 定义了一个与 $f$ 逆的群同态，并给出了定理中的同构。

As a first application of Theorem 43 we give an alternate proof of the first result in Theorem 39 that the tensor product is right exact in the case where $S = R$ is a commutative ring. If $0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0$ is exact,then by Theorem 33 the sequence

作为定理 43 的第一个应用，我们给出了定理 39 中第一个结果的另一种证明，即当 $S = R$ 是一个交换环时，张量积在右边是准确的。如果 $0 \rightarrow L \rightarrow M \rightarrow N \rightarrow 0$ 是准确的，那么根据定理 33，序列

is exact for every $R$ -module $E$ . Then by Theorem 28,the sequence

对于每个 $R$ -模 $E$ 是准确的。然后根据定理 28，序列

is exact for all $D$ and all $E$ . By adjoint associativity,this means the sequence

对于所有 $D$ 和所有 $E$ 是准确的。由于伴随结合律，这意味着序列

is exact for any $D$ and all $E$ . Then,by the second part of Theorem 33,it follows that the sequence

对于任意的 $D$ 和所有 $E$ 是准确的。接着，根据定理 33 的第二部分，可以得出序列

is exact for all $D$ ,which is the right exactness of the tensor product.

对于所有 $D$ 是准确的，这就是张量积在右边的准确性。

As a second application of Theorem 43 we prove that the tensor product of two projective modules over a commutative ring $R$ is again projective (see also Exercise 9 for a more direct proof).

作为定理43的第二个应用，我们证明了两个在交换环上的投影模的张量积 $R$ 再次是投影的（也参见练习9以获得更直接的证明）。

Corollary 44. If $R$ is commutative then the tensor product of two projective $R$ -modules is projective.

推论44。如果 $R$ 是交换的，那么两个投影 $R$ -模的张量积是投影的。

Proof: Let ${P}_{1}$ and ${P}_{2}$ be projective modules. Then by Corollary 32, ${\operatorname{Hom}}_{R}\left( {{P}_{2},\_ }\right)$ is an exact functor from the category of $R$ -modules to the category of $R$ -modules. Then the composition ${\operatorname{Hom}}_{R}\left( {{P}_{1},{\operatorname{Hom}}_{R}\left( {{P}_{2},\_ \_ }\right) }\right)$ is an exact functor by the same corollary. By Theorem 43 this means that ${\mathrm{{Hom}}}_{R}\left( {{P}_{1}{ \otimes }_{R}{P}_{2},\_ \_ }\right)$ is an exact functor on $R$ -modules. It follows again from Corollary 32 that ${P}_{1}{ \otimes }_{R}{P}_{2}$ is projective.

证明：设 ${P}_{1}$ 和 ${P}_{2}$ 是投影模。那么根据推论32，${\operatorname{Hom}}_{R}\left( {{P}_{2},\_ }\right)$ 是从 $R$ -模范畴到 $R$ -模范畴的精确函子。因此，复合 ${\operatorname{Hom}}_{R}\left( {{P}_{1},{\operatorname{Hom}}_{R}\left( {{P}_{2},\_ \_ }\right) }\right)$ 也是由同一推论得出的精确函子。根据定理43，这意味着 ${\mathrm{{Hom}}}_{R}\left( {{P}_{1}{ \otimes }_{R}{P}_{2},\_ \_ }\right)$ 是在 $R$ -模上的精确函子。再次根据推论32，得出 ${P}_{1}{ \otimes }_{R}{P}_{2}$ 是投影的。

Summary

总结

Each of the functors ${\operatorname{Hom}}_{R}\left( {A,\_ }\right) ,{\operatorname{Hom}}_{R}\left( {\_ ,A}\right)$ ,and $A{ \otimes }_{R}$ _____,map left $R$ -modules to abelian groups; the functor $\_ { \otimes }_{R}A$ maps right $R$ -modules to abelian groups. When $R$ is commutative all four functors map $R$ -modules to $R$ -modules.

(1) Let $A$ be a left $R$ -module. The functor ${\operatorname{Hom}}_{R}\left( {A,\_ }\right)$ is covariant and left exact; the module $A$ is projective if and only if ${\operatorname{Hom}}_{R}\left( {A,\_ }\right)$ is exact (i.e.,is also right exact).

(2) Let $A$ be a left $R$ -module. The functor ${\operatorname{Hom}}_{R}\left( {\_ ,A}\right)$ is contravariant and left exact; the module $A$ is injective if and only if ${\operatorname{Hom}}_{R}\left( {\_ ,A}\right)$ is exact.

(2) 设 $A$ 是一个左 $R$ -模。函子 ${\operatorname{Hom}}_{R}\left( {\_ ,A}\right)$ 是反变和左精确的；模块 $A$ 是可内射的当且仅当 ${\operatorname{Hom}}_{R}\left( {\_ ,A}\right)$ 是精确的。

(3) Let $A$ be a right $R$ -module. The functor $A{ \otimes }_{R}$ is covariant and right exact; the module $A$ is flat if and only if $A{ \otimes }_{R}$ _____is exact (i.e.,is also left exact).

(3) 设 $A$ 是一个右 $R$ -模。函子 $A{ \otimes }_{R}$ 是协变和右精确的；模块 $A$ 是平坦的当且仅当 $A{ \otimes }_{R}$ _____是精确的（即，也是左精确的）。

(4) Let $A$ be a left $R$ -module. The functor $\_ { \otimes }_{R}A$ is covariant and right exact; the module $A$ is flat if and only if $\_ { \otimes }_{R}A$ is exact.

(4) 设 $A$ 是一个左 $R$ -模。函子 $\_ { \otimes }_{R}A$ 是协变和右精确的；模块 $A$ 是平坦的当且仅当 $\_ { \otimes }_{R}A$ 是精确的。

(5) Projective modules are flat. The $\mathbb{Z}$ -module $\mathbb{Q}/\mathbb{Z}$ is injective but not flat. The $\mathbb{Z}$ -module $\mathbb{Z} \oplus \mathbb{Q}$ is flat but neither projective nor injective.

(5) 投影模是平坦的。 $\mathbb{Z}$ -模 $\mathbb{Q}/\mathbb{Z}$ 是可内射的但不是平坦的。 $\mathbb{Z}$ -模 $\mathbb{Z} \oplus \mathbb{Q}$ 是平坦的但既不投影也不可内射。

EXERCISES

练习题

Let $R$ be a ring with 1 . 1. Suppose that

设 $R$ 是一个带有单位元的环。1. 假设

is a commutative diagram of groups and that the rows are exact. Prove that

是一个群交换图，且行是精确的。证明

(a) if $\varphi$ and $\alpha$ are surjective,and $\beta$ is injective then $\gamma$ is injective. [If $c \in \ker \gamma$ ,show there is a $b \in B$ with $\varphi \left( b\right) = c$ . Show that ${\varphi }^{\prime }\left( {\beta \left( b\right) }\right) = 0$ and deduce that $\beta \left( b\right) = {\psi }^{\prime }\left( {a}^{\prime }\right)$ for some ${a}^{\prime } \in {A}^{\prime }$ . Show there is an $a \in A$ with $\alpha \left( a\right) = {a}^{\prime }$ and that $\beta \left( {\psi \left( a\right) }\right) = \beta \left( b\right)$ . Conclude that $b = \psi \left( a\right)$ and hence $c = \varphi \left( b\right) = 0$ .]

(a) 如果 $\varphi$ 和 $\alpha$ 是满射，且 $\beta$ 是单射，那么 $\gamma$ 是单射。 [如果 $c \in \ker \gamma$ ，则存在一个 $b \in B$ 使得 $\varphi \left( b\right) = c$ 。证明 ${\varphi }^{\prime }\left( {\beta \left( b\right) }\right) = 0$ 并推导出 $\beta \left( b\right) = {\psi }^{\prime }\left( {a}^{\prime }\right)$ 对于某个 ${a}^{\prime } \in {A}^{\prime }$ 成立。证明存在一个 $a \in A$ 使得 $\alpha \left( a\right) = {a}^{\prime }$ 并且 $\beta \left( {\psi \left( a\right) }\right) = \beta \left( b\right)$ 。得出 $b = \psi \left( a\right)$ ，因此 $c = \varphi \left( b\right) = 0$ 。]

(b) if ${\psi }^{\prime },\alpha$ ,and $\gamma$ are injective,then $\beta$ is injective,

(b) 如果 ${\psi }^{\prime },\alpha$ 和 $\gamma$ 是单射，那么 $\beta$ 是单射，

(c) if $\varphi ,\alpha$ ,and $\gamma$ are surjective,then $\beta$ is surjective,

(c) 如果 $\varphi ,\alpha$ 和 $\gamma$ 是满射，那么 $\beta$ 是满射，

(d) if $\beta$ is injective, $\alpha$ and $\gamma$ are surjective,then $\gamma$ is injective,

(d) 如果 $\beta$ 是单射， $\alpha$ 和 $\gamma$ 是满射，那么 $\gamma$ 是单射，

(e) if $\beta$ is surjective, $\gamma$ and ${\psi }^{\prime }$ are injective,then $\alpha$ is surjective.

(e) 如果 $\beta$ 是满射， $\gamma$ 和 ${\psi }^{\prime }$ 是单射，那么 $\alpha$ 是满射。

2. Suppose that

2. 假设

is a commutative diagram of groups, and that the rows are exact. Prove that

是一个群交换图，且行是精确的。证明

(a) if $\alpha$ is surjective,and $\beta ,\delta$ are injective,then $\gamma$ is injective.

(a) 如果 $\alpha$ 是满射，且 $\beta ,\delta$ 是单射，那么 $\gamma$ 是单射。

(b) if $\delta$ is injective,and $\alpha ,\gamma$ are surjective,then $\beta$ is surjective.

(b) 如果 $\delta$ 是单射，且 $\alpha ,\gamma$ 是满射，那么 $\beta$ 是满射。

3. Let ${P}_{1}$ and ${P}_{2}$ be $R$ -modules. Prove that ${P}_{1} \oplus {P}_{2}$ is a projective $R$ -module if and only if both ${P}_{1}$ and ${P}_{2}$ are projective.

4. 设 ${P}_{1}$ 和 ${P}_{2}$ 是 $R$ -模。证明 ${P}_{1} \oplus {P}_{2}$ 是一个投射 $R$ -模当且仅当 ${P}_{1}$ 和 ${P}_{2}$ 都是投射的。

5. Let ${Q}_{1}$ and ${Q}_{2}$ be $R$ -modules. Prove that ${Q}_{1} \oplus {Q}_{2}$ is an injective $R$ -module if and only if both ${Q}_{1}$ and ${Q}_{2}$ are injective.

6. 设 ${Q}_{1}$ 和 ${Q}_{2}$ 是 $R$ -模。证明 ${Q}_{1} \oplus {Q}_{2}$ 是一个可注入 $R$ -模当且仅当 ${Q}_{1}$ 和 ${Q}_{2}$ 都是不可注入的。

7. Let ${A}_{1}$ and ${A}_{2}$ be $R$ -modules. Prove that ${A}_{1} \oplus {A}_{2}$ is a flat $R$ -module if and only if both ${A}_{1}$ and ${A}_{2}$ are flat. More generally,prove that an arbitrary direct sum $\sum {A}_{i}$ of $R$ -modules is flat if and only if each ${A}_{i}$ is flat. [Use the fact that tensor product commutes with arbitrary direct sums.]

8. 设 ${A}_{1}$ 和 ${A}_{2}$ 是 $R$ -模。证明 ${A}_{1} \oplus {A}_{2}$ 是一个平坦 $R$ -模当且仅当 ${A}_{1}$ 和 ${A}_{2}$ 都是平坦的。更一般地，证明任意直和 $\sum {A}_{i}$ 的 $R$ -模是平坦的当且仅当每个 ${A}_{i}$ 都是平坦的。[使用张量积与任意直和交换的事实。]

9. Prove that the following are equivalent for a ring $R$ :

10. 证明对于一个环 $R$ 以下条件是等价的：

(i) Every $R$ -module is projective.

(i) 每个 $R$ -模都是投射的。

(ii) Every $R$ -module is injective.

(ii) 每个 $R$ -模都是可注入的。

7. Let $A$ be a nonzero finite abelian group.

8. 设 $A$ 是一个非零有限阿贝尔群。

(a) Prove that $A$ is not a projective $\mathbb{Z}$ -module.

(a) 证明 $A$ 不是一个投射 $\mathbb{Z}$ -模。

(b) Prove that $A$ is not an injective $\mathbb{Z}$ -module.

(b) 证明 $A$ 不是一个可注入 $\mathbb{Z}$ -模。

8. Let $Q$ be a nonzero divisible $\mathbb{Z}$ -module. Prove that $Q$ is not a projective $\mathbb{Z}$ -module. Deduce that the rational numbers $\mathbb{Q}$ is not a projective $\mathbb{Z}$ -module. [Show first that if $F$ is any free module then ${ \cap }_{n = 1}^{\infty }{nF} = 0$ (use a basis of $F$ to prove this). Now suppose to the contrary that $Q$ is projective and derive a contradiction from Proposition 30(4).]

9. 设 $Q$ 是一个非零的可除 $\mathbb{Z}$ -模。证明 $Q$ 不是一个投射 $\mathbb{Z}$ -模。推导出有理数 $\mathbb{Q}$ 不是一个投射 $\mathbb{Z}$ -模。（首先证明如果 $F$ 是任意自由模，那么 ${ \cap }_{n = 1}^{\infty }{nF} = 0$（使用 $F$ 的基来证明这一点）。现在假设 $Q$ 是投射的，并从命题 30(4) 推导出矛盾。]

10. Assume $R$ is commutative with 1 .

11. 假设 $R$ 是带有单位元 1 的交换环。

(a) Prove that the tensor product of two free $R$ -modules is free. [Use the fact that tensor products commute with direct sums.]

(a) 证明两个自由 $R$ -模的张量积是自由的。[使用张量积与直和可交换的事实。]

(b) Use (a) to prove that the tensor product of two projective $R$ -modules is projective.

(b) 利用 (a) 证明两个投射 $R$ -模的张量积是投射的。

10. Let $R$ and $S$ be rings with 1 and let $M$ and $N$ be left $R$ -modules. Assume also that $M$ is an(R,S)-bimodule.

11. 设 $R$ 和 $S$ 是带有单位元的环，并且设 $M$ 和 $N$ 是左 $R$ -模。还假设 $M$ 是一个 (R,S)-双模。

(a) For $s \in S$ and for $\varphi \in {\operatorname{Hom}}_{R}\left( {M,N}\right)$ define $\left( {s\varphi }\right) : M \rightarrow N$ by $\left( {s\varphi }\right) \left( m\right) = \varphi \left( {ms}\right)$ . Prove that ${s\varphi }$ is a homomorphism of left $R$ -modules,and that this action of $S$ on ${\operatorname{Hom}}_{R}\left( {M,N}\right)$ makes it into a left $S$ -module.

(a) 对于 $s \in S$ 和 $\varphi \in {\operatorname{Hom}}_{R}\left( {M,N}\right)$，通过 $\left( {s\varphi }\right) \left( m\right) = \varphi \left( {ms}\right)$ 定义 $\left( {s\varphi }\right) : M \rightarrow N$。证明 ${s\varphi }$ 是左 $R$ -模的同态，并且 $S$ 对 ${\operatorname{Hom}}_{R}\left( {M,N}\right)$ 的这种作用使其成为一个左 $S$ -模。

(b) Let $S = R$ and let $M = R$ (considered as an(R,R)-bimodule by left and right ring multiplication on itself). For each $n \in N$ define ${\varphi }_{n} : R \rightarrow N$ by ${\varphi }_{n}\left( r\right) = {rn}$ , i.e., ${\varphi }_{n}$ is the unique $R$ -module homomorphism mapping ${1}_{R}$ to $n$ . Show that ${\varphi }_{n} \in$ ${\operatorname{Hom}}_{R}\left( {R,N}\right)$ . Use part (a) to show that the map $n \mapsto {\varphi }_{n}$ is an isomorphism of left $R$ -modules: $N \cong {\operatorname{Hom}}_{R}\left( {R,N}\right)$ .

(b) 设 $S = R$ 并且设 $M = R$（将其视为通过自身上的左和右环乘法作为 (R,R)-双模）。对于每个 $n \in N$，定义 ${\varphi }_{n} : R \rightarrow N$ 为 ${\varphi }_{n}\left( r\right) = {rn}$，即 ${\varphi }_{n}$ 是唯一的 $R$ -模同态，将 ${1}_{R}$ 映射到 $n$。证明 ${\varphi }_{n} \in$ ${\operatorname{Hom}}_{R}\left( {R,N}\right)$。使用部分 (a) 来证明映射 $n \mapsto {\varphi }_{n}$ 是左 $R$ -模的同构：$N \cong {\operatorname{Hom}}_{R}\left( {R,N}\right)$。

(c) Deduce that if $N$ is a free (respectively,projective,injective,fiat) left $R$ -module,then ${\operatorname{Hom}}_{R}\left( {R,N}\right)$ is also a free (respectively,projective,injective,flat) left $R$ -module.

(c) 推导出如果 $N$ 是一个自由的（分别地，投射的、内射的、平坦的）左 $R$ -模，那么 ${\operatorname{Hom}}_{R}\left( {R,N}\right)$ 也是一个自由的（分别地，投射的、内射的、平坦的）左 $R$ -模。

11. Let $R$ and $S$ be rings with 1 and let $M$ and $N$ be left $R$ -modules. Assume also that $N$ is an (R,S)-bimodule.

12. 设 $R$ 和 $S$ 是带有单位元的环，并且设 $M$ 和 $N$ 是左 $R$ -模。还假设 $N$ 是一个 (R,S)-双模。

(a) For $s \in S$ and for $\varphi \in {\operatorname{Hom}}_{R}\left( {M,N}\right)$ define $\left( {\varphi s}\right) : M \rightarrow N$ by $\left( {\varphi s}\right) \left( m\right) = \varphi \left( m\right) s$ . Prove that ${\varphi s}$ is a homomorphism of left $R$ -modules,and that this action of $S$ on ${\operatorname{Hom}}_{R}\left( {M,N}\right)$ makes it into a right $S$ -module. Deduce that ${\operatorname{Hom}}_{R}\left( {M,R}\right)$ is a right $R$ -module,for any $R$ -module $M$ -called the dual module to $M.$ . .

(a) 对于 $s \in S$ 和对于 $\varphi \in {\operatorname{Hom}}_{R}\left( {M,N}\right)$，定义 $\left( {\varphi s}\right) : M \rightarrow N$ 为 $\left( {\varphi s}\right) \left( m\right) = \varphi \left( m\right) s$。证明 ${\varphi s}$ 是左 $R$ -模的同态，并且这个 $S$ 对 ${\operatorname{Hom}}_{R}\left( {M,N}\right)$ 的作用使其成为一个右 $S$ -模。推导出 ${\operatorname{Hom}}_{R}\left( {M,R}\right)$ 是一个右 $R$ -模，对于任何 $R$ -模 $M$ -称为 $M.$ 的对偶模。

(b) Let $N = R$ be considered as an(R,R)-bimodule as usual. Under the action defined in part (a) show that the map $r \mapsto {\varphi }_{r}$ is an isomorphism of right $R$ -modules: ${\operatorname{Hom}}_{R}\left( {R,R}\right) \cong R$ ,where ${\varphi }_{r}$ is the homomorphism that maps ${1}_{R}$ to $r$ . Deduce that if $M$ is a finitely generated free left $R$ -module,then ${\operatorname{Hom}}_{R}\left( {M,R}\right)$ is a free right $R$ -module of the same rank. (cf. also Exercise 13.)

(b) 将 $N = R$ 视为一个 (R,R)-双模，如通常情况。在部分 (a) 定义的作用下，证明映射 $r \mapsto {\varphi }_{r}$ 是一个右 $R$ -模的同构：${\operatorname{Hom}}_{R}\left( {R,R}\right) \cong R$，其中 ${\varphi }_{r}$ 是同态，将 ${1}_{R}$ 映射到 $r$。推导出如果 $M$ 是一个有限生成的自由左 $R$ -模，那么 ${\operatorname{Hom}}_{R}\left( {M,R}\right)$ 是一个同阶的自由右 $R$ -模。（参见练习 13。）

(c) Show that if $M$ is a finitely generated projective $R$ -module then its dual module ${\operatorname{Hom}}_{R}\left( {M,R}\right)$ is also projective.

(c) 证明如果 $M$ 是一个有限生成的投射 $R$ -模，那么它的对偶模 ${\operatorname{Hom}}_{R}\left( {M,R}\right)$ 也是投射的。

12. Let $A$ be an $R$ -module,let $I$ be any nonempty index set and for each $i \in I$ let ${B}_{i}$ be an $R$ -module. Prove the following isomorphisms of abelian groups; when $R$ is commutative prove also that these are $R$ -module isomorphisms. (Arbitrary direct sums and direct products of modules are introduced in Exercise 20 of Section 3.)

13. 设 $A$ 是一个 $R$ -模，$I$ 是任意非空指标集，对于每个 $i \in I$，设 ${B}_{i}$ 是一个 $R$ -模。证明以下阿贝尔群的同构；当 $R$ 是交换的时候，也证明这些是 $R$ -模的同构。（模块的任意直和与直积在第3节的练习20中引入。）

(a) ${\operatorname{Hom}}_{R}\left( {{\bigoplus }_{i \in I}{B}_{i},A}\right) \cong \mathop{\prod }\limits_{{i \in I}}{\operatorname{Hom}}_{R}\left( {{B}_{i},A}\right)$

(a) ${\operatorname{Hom}}_{R}\left( {{\bigoplus }_{i \in I}{B}_{i},A}\right) \cong \mathop{\prod }\limits_{{i \in I}}{\operatorname{Hom}}_{R}\left( {{B}_{i},A}\right)$

(b) ${\operatorname{Hom}}_{R}\left( {A,\mathop{\prod }\limits_{{i \in I}}{B}_{i}}\right) \cong \mathop{\prod }\limits_{{i \in I}}{\operatorname{Hom}}_{R}\left( {A,{B}_{i}}\right)$ .

(b) ${\operatorname{Hom}}_{R}\left( {A,\mathop{\prod }\limits_{{i \in I}}{B}_{i}}\right) \cong \mathop{\prod }\limits_{{i \in I}}{\operatorname{Hom}}_{R}\left( {A,{B}_{i}}\right)$。

13. (a) Show that the dual of the free $\mathbb{Z}$ -module with countable basis is not free. [Use the preceding exercise and Exercise 24, Section 3.] (See also Exercise 5 in Section 11.3.)

14. (a) 证明具有可数基的自由 $\mathbb{Z}$ -模的对偶不是自由的。[使用前一个练习和第3节的练习24。]（也参见第11.3节的练习5。）

(b) Show that the dual of the free $\mathbb{Z}$ -module with countable basis is also not projective. [You may use the fact that any submodule of a free $\mathbb{Z}$ -module is free.]

(b) 证明具有可数基的自由 $\mathbb{Z}$ -模的对偶也不是投射的。[你可以使用这样一个事实：任何自由 $\mathbb{Z}$ -模的子模都是自由的。]

14. Let $0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ be a sequence of $R$ -modules.

15. 设 $0 \rightarrow L\overset{\psi }{ \rightarrow }M\overset{\varphi }{ \rightarrow }N \rightarrow 0$ 是一个 $R$ -模的序列。

(a) Prove that the associated sequence

is a short exact sequence of abelian groups for all $R$ -modules $D$ if and only if the original sequence is a split short exact sequence. [To show the sequence splits, take $D = N$ and show the lift of the identity map in ${\operatorname{Hom}}_{R}\left( {N,N}\right)$ to ${\operatorname{Hom}}_{R}\left( {N,M}\right)$ is a splitting homomorphism for $\varphi$ .]

(b) Prove that the associated sequence

is a short exact sequence of abelian groups for all $R$ -modules $D$ if and only if the original sequence is a split short exact sequence.

15. Let $M$ be a left $R$ -module where $R$ is a ring with 1 .

(a) Show that ${\operatorname{Hom}}_{\mathbb{Z}}\left( {R,M}\right)$ is a left $R$ -module under the action $\left( {r\varphi }\right) \left( {r}^{\prime }\right) = \varphi \left( {{r}^{\prime }r}\right)$ (see Exercise 10).

(b) Suppose that $0 \rightarrow A\overset{\psi }{ \rightarrow }B$ is an exact sequence of $R$ -modules. Prove that if every homomorphism $f$ from $A$ to $M$ lifts to a homomorphism $F$ from $B$ to $M$ with $f =$ $F \circ \psi ,\mathrm{{then}}\;\mathrm{{every}}\;\mathrm{{homomorphism}}\;{f}^{\prime }\;\mathrm{{from}}\;A\;\mathrm{{to}}\;{\mathrm{{Hom}}}_{\mathbb{Z}}\left( {R,M}\right) \;\mathrm{{lifts}}\;\mathrm{{to}}\;\mathrm{a}\;\mathrm{{homomorphism}}$ ${F}^{\prime }\operatorname{from}B\operatorname{to}{\operatorname{Hom}}_{\mathbb{Z}}\left( {R,M}\right)$ with ${f}^{\prime } = {F}^{\prime } \circ \psi .$ [Given ${f}^{\prime },$ show that $f\left( a\right) = {f}^{\prime }\left( a\right) \left( {1}_{R}\right)$ defines a homomorphism of $A$ to $M$ . If $F$ is the associated lift of $f$ to $B$ ,show that ${F}^{\prime }\left( b\right) \left( r\right) = F\left( {rb}\right)$ defines a homomorphism from $B$ to ${\operatorname{Hom}}_{\mathbb{Z}}\left( {R,M}\right)$ that lifts ${f}^{\prime }$ .]

(c) Prove that if $Q$ is an injective $R$ -module then ${\operatorname{Hom}}_{\mathbb{Z}}\left( {R,Q}\right)$ is also an injective $R$ - module.

16. This exercise proves Theorem 38 that every left $R$ -module $M$ is contained in an injective left $R$ -module.

(a) Show that $M$ is contained in an injective $\mathbb{Z}$ -module $Q$ . [ $M$ is a $\mathbb{Z}$ -module-use Corollary 37.]

(b) Show that ${\operatorname{Hom}}_{R}\left( {R,M}\right) \subseteq {\operatorname{Hom}}_{\mathbb{Z}}\left( {R,M}\right) \subseteq {\operatorname{Hom}}_{\mathbb{Z}}\left( {R,Q}\right)$ .

(c) Use the $R$ -module isomorphism $M \cong {\operatorname{Hom}}_{R}\left( {R,M}\right)$ (Exercise 10) and the previous exercise to conclude that $M$ is contained in an injective module.

17. This exercise completes the proof of Proposition 34. Suppose that $Q$ is an $R$ -module with the property that every short exact sequence $0 \rightarrow Q \rightarrow {M}_{1} \rightarrow N \rightarrow 0$ splits and suppose that the sequence $0 \rightarrow L\overset{\psi }{ \rightarrow }M$ is exact. Prove that every $R$ -module homomorphism $f$ from $L$ to $Q$ can be lifted to an $R$ -module homomorphism $F$ from $M$ to $Q$ with $f = F \circ \psi$ . [By the previous exercise, $Q$ is contained in an injective $R$ -module. Use the splitting property together with Exercise 4 (noting that Exercise 4 can be proved using (2) in Proposition 34 as the definition of an injective module).]

18. Prove that the injective hull of the $\mathbb{Z}$ -module $\mathbb{Z}$ is $\mathbb{Q}$ . [Let $H$ be the injective hull of $\mathbf{Z}$ and argue that $\mathbb{Q}$ contains an isomorphic copy of $H$ . Use the divisibility of $H$ to show $1/n \in H$ for all nonzero integers $n$ ,and deduce that $H = \mathbb{Q}$ .]

19. If $F$ is a field,prove that the injective hull of $F$ is $F$ .

20. Prove that the polynomial ring $R\left\lbrack x\right\rbrack$ in the indeterminate $x$ over the commutative ring $\mathbf{R}$ is a flat $R$ -module.

21. Let $R$ and $S$ be rings with 1 and suppose $M$ is a right $R$ -module,and $N$ is an(R,S)- bimodule. If $M$ is flat over $R$ and $N$ is flat as an $S$ -module prove that $M{ \otimes }_{R}N$ is flat as $\mathbf{a}$ right $S$ -module.

22. Suppose that $R$ is a commutative ring and that $M$ and $N$ are flat $R$ -modules. Prove that $M{ \otimes }_{R}N$ is a flat $R$ -module. [Use the previous exercise.]

23. 假设 $R$ 是一个交换环，并且 $M$ 和 $N$ 是平坦的 $R$ -模。证明 $M{ \otimes }_{R}N$ 是一个平坦的 $R$ -模。 [使用前一个练习。]

24. Prove that the (right) module $M{ \otimes }_{R}S$ obtained by changing the base from the ring $R$ to the ring $S$ (by some homomorphism $f : R \rightarrow S$ with $f\left( {1}_{R}\right) = {1}_{S}$ ,cf. Example 6 following Corollary 12 in Section 4) of the flat (right) $R$ -module $M$ is a flat $S$ -module.

25. 证明通过将基从环 $R$ 变换到环 $S$（通过某个同态 $f : R \rightarrow S$ 使得 $f\left( {1}_{R}\right) = {1}_{S}$ ，参见第4节中例6后的推论12）得到的（右）模 $M{ \otimes }_{R}S$ ，其中 $M$ 是平坦的（右）$R$ -模，是一个平坦的 $S$ -模。

26. Prove that $A$ is a flat $R$ -module if and only if for any left $R$ -modules $L$ and $M$ where $L$ is finitely generated,then $\psi : L \rightarrow M$ injective implies that also $1 \otimes \psi : A{ \otimes }_{R}L \rightarrow A{ \otimes }_{R}M$ is injective. [Use the techniques in the proof of Corollary 42.]

27. 证明 $A$ 是一个平坦的 $R$ -模当且仅当对于任何左 $R$ -模 $L$ 和 $M$ ，其中 $L$ 是有限生成的，如果 $\psi : L \rightarrow M$ 内射则 $1 \otimes \psi : A{ \otimes }_{R}L \rightarrow A{ \otimes }_{R}M$ 也是内射。 [使用推论42证明中的技术。]

28. (A Flatness Criterion) Parts (a)-(c) of this exercise prove that $A$ is a flat $R$ -module if and only if for every finitely generated ideal $I$ of $R$ ,the map from $A{ \otimes }_{R}I \rightarrow A{ \otimes }_{R}R \cong A$ induced by the inclusion $\;I \subseteq R\;$ is again injective (or,equivalently, $A{ \otimes }_{R}I \cong {AI} \subseteq A$ ).

29. （平坦性准则）本练习的(a)-(c)部分证明了如果对于每个 $R$ 的有限生成理想 $I$ ，由包含 $\;I \subseteq R\;$ 引发的映射 $A{ \otimes }_{R}I \rightarrow A{ \otimes }_{R}R \cong A$ 再次是内射的（或者等价地 $A{ \otimes }_{R}I \cong {AI} \subseteq A$ ），那么 $A$ 是一个平坦的 $R$ -模。

(a) Prove that if $A$ is flat then $A{ \otimes }_{R}I \rightarrow A{ \otimes }_{R}R$ is injective.

(a) 证明如果 $A$ 是平坦的，那么 $A{ \otimes }_{R}I \rightarrow A{ \otimes }_{R}R$ 是内射的。

(b) If $A{ \otimes }_{R}I \rightarrow A{ \otimes }_{R}R$ is injective for every finitely generated ideal $I$ ,prove that $A{ \otimes }_{R}I \rightarrow A{ \otimes }_{R}R$ is injective for every ideal $I$ . Show that if $K$ is any submodule of a finitely generated free module $F$ then $A{ \otimes }_{R}K \rightarrow A{ \otimes }_{R}F$ is injective. Show that the same is true for any free module $F$ . [Cf. the proof of Corollary 42.]

(b) 如果 $A{ \otimes }_{R}I \rightarrow A{ \otimes }_{R}R$ 对每个有限生成理想 $I$ 都是单射，证明 $A{ \otimes }_{R}I \rightarrow A{ \otimes }_{R}R$ 对每个理想 $I$ 也是单射。证明如果 $K$ 是有限生成自由模 $F$ 的任意子模，那么 $A{ \otimes }_{R}K \rightarrow A{ \otimes }_{R}F$ 是单射。证明对于任意自由模 $F$ 也是如此。 [参见命题42的证明。]

(c) Under the assumption in (b),suppose $L$ and $M$ are $R$ -modules and $L\overset{\psi }{ \rightarrow }M$ is injective. Prove that $A{ \otimes }_{R}L\overset{1 \otimes \psi }{ \rightarrow }A{ \otimes }_{R}M$ is injective and conclude that $A$ is flat. [Write $M$ as a quotient of the free module $F$ ,giving a short exact sequence

(c) 在 (b) 的假设下，假设 $L$ 和 $M$ 是 $R$ -模且 $L\overset{\psi }{ \rightarrow }M$ 是单射。证明 $A{ \otimes }_{R}L\overset{1 \otimes \psi }{ \rightarrow }A{ \otimes }_{R}M$ 是单射，并得出 $A$ 是平坦的。 [将 $M$ 写作自由模 $F$ 的商模，给出一个短正合序列]

Show that if $J = {f}^{-1}\left( {\psi \left( L\right) }\right)$ and $\iota : J \rightarrow F$ is the natural injection,then the diagram

证明如果 $J = {f}^{-1}\left( {\psi \left( L\right) }\right)$ 和 $\iota : J \rightarrow F$ 是自然单射，那么该图

is commutative with exact rows. Show that the induced diagram $A{ \otimes }_{R}K \rightarrow A{ \otimes }_{R}F \rightarrow A{ \otimes }_{R}M \rightarrow 0$

是交换的，且行是正合的。证明诱导的图 $A{ \otimes }_{R}K \rightarrow A{ \otimes }_{R}F \rightarrow A{ \otimes }_{R}M \rightarrow 0$

is commutative with exact rows. Use (b) to show that $1 \otimes \iota$ is injective,then use Exercise 1 to conclude that $1 \otimes \psi$ is injective.]

是交换的，且行是正合的。使用 (b) 证明 $1 \otimes \iota$ 是单射，然后使用练习1得出 $1 \otimes \psi$ 是单射。]

(d) ( $A$ Flatness Criterion for quotients) Suppose $A = F/K$ where $F$ is flat (e.g.,if $F$ is free) and $K$ is an $R$ -submodule of $F$ . Prove that $A$ is flat if and only if ${FI} \cap K = {KI}$ for every finitely generated ideal $I$ of $R.$ [Use (a) to prove $F{ \otimes }_{R}I \cong {FI}$ and observe the image of $K{ \otimes }_{R}I$ is ${KI}$ ; tensor the exact sequence $0 \rightarrow K \rightarrow F \rightarrow A \rightarrow 0$ with $I$ to prove that $A{ \otimes }_{R}I \cong {FI}/{KI}$ ,and apply the flatness criterion.]

(d) (商模的平坦性准则) 假设 $A$ 其中 $F$ 是平坦的（例如，如果 $F$ 是自由的）且 $K$ 是 $R$ -子模 $F$ 。证明 $A$ 是平坦的当且仅当 ${FI} \cap K = {KI}$ 对每个 $R.$ 的有限生成理想 $I$ 成立 [使用 (a) 证明 $F{ \otimes }_{R}I \cong {FI}$ 并观察 $K{ \otimes }_{R}I$ 的像为 ${KI}$ ；将正合序列 $0 \rightarrow K \rightarrow F \rightarrow A \rightarrow 0$ 张量 $I$ 以证明 $A{ \otimes }_{R}I \cong {FI}/{KI}$ ，并应用平坦性准则。]

26. Suppose $R$ is a P.I.D. This exercise proves that $A$ is a flat $R$ -module if and only if $A$ is torsion free $R$ -module (i.e.,if $a \in A$ is nonzero and $r \in R$ ,then ${ra} = 0$ implies $r = 0$ ).

27. 假设 $R$ 是一个P.I.D.。这个练习证明了 $A$ 是一个平坦的 $R$ -模当且仅当 $A$ 是一个无扭的 $R$ -模（即，如果 $a \in A$ 是非零的且 $r \in R$ ，那么 ${ra} = 0$ 蕴含 $r = 0$ ）。

(a) Suppose that $A$ is flat and for fixed $r \in R$ consider the map ${\psi }_{r} : R \rightarrow R$ defined by multiplication by $r : {\psi }_{r}\left( x\right) = {rx}$ . If $r$ is nonzero show that ${\psi }_{r}$ is an injection. Conclude from the flatness of $A$ that the map from $A$ to $A$ defined by mapping $a$ to ${ra}$ is injective and that $A$ is torsion free.

(a) 假设 $A$ 是平坦的，对于固定的 $r \in R$ ，考虑由乘以 $r : {\psi }_{r}\left( x\right) = {rx}$ 定义的映射 ${\psi }_{r} : R \rightarrow R$。如果 $r$ 是非零的，证明 ${\psi }_{r}$ 是单射。从 $A$ 的平坦性得出，由将 $a$ 映射到 ${ra}$ 定义的从 $A$ 到 $A$ 的映射是单射，且 $A$ 是无扭的。

(b) Suppose that $A$ is torsion free. If $I$ is a nonzero ideal of $R$ ,then $I = {rR}$ for some nonzero $r \in R$ . Show that the map ${\psi }_{r}$ in (a) induces an isomorphism $R \cong I$ of

(b) 假设 $A$ 是无扭的。如果 $I$ 是 $R$ 的一个非零理想，那么对于某个非零的 $r \in R$ ，有 $I = {rR}$。证明(a)中的映射 ${\psi }_{r}$ 引导出一个 $A$ -模的同构 $R \cong I$，并且 $R$ 与包含映射 $I = {rR}$ 的复合是乘以 $r \in R$。证明复合 ${\psi }_{r}$ 对应于在识别 下的映射 $R \cong I$，并且由于 是无扭的，这个复合是单射。证明 是同构，并推断 是单射。使用前一个练习来得出 是平坦的结论。

$R$ -modules and that the composite $R\overset{\psi }{ \rightarrow }I\overset{\iota }{ \rightarrow }R$ of ${\psi }_{r}$ with the inclusion $\iota : I \subseteq R$ is multiplication by $r$ . Prove that the composite $A{ \otimes }_{R}R\overset{\cup \otimes {\psi }_{r}}{ \rightarrow }A{ \otimes }_{R}I\overset{1 \otimes \iota }{ \rightarrow }A{ \otimes }_{R}R$ corresponds to the map $a \mapsto {ra}$ under the identification $A{ \otimes }_{R}R = A$ and that this composite is injective since $A$ is torsion free. Show that $1 \otimes {\psi }_{r}$ is an isomorphism and deduce that $1 \otimes \iota$ is injective. Use the previous exercise to conclude that $A$ is flat.

$R$ -模和 $R\overset{\psi }{ \rightarrow }I\overset{\iota }{ \rightarrow }R$ 是 ${\psi }_{r}$ -模。

27. Let $M,A$ and $B$ be $R$ -modules.

28. 设 $M,A$ 和 $B$ 是 $R$ -模。

(a) Suppose $f : A \rightarrow M$ and $g : B \rightarrow M$ are $R$ -module homomorphisms. Prove that $X = \{ \left( {a,b}\right) \mid a \in A,b \in B$ with $f\left( a\right) = g\left( b\right) \}$ is an $R$ -submodule of the direct sum $A \oplus B$ (called the pullback or fiber product of $f$ and $g$ ) and that there is a commutative diagram

(a) 假设 $f : A \rightarrow M$ 和 $g : B \rightarrow M$ 是 $R$ -模同态。证明 $X = \{ \left( {a,b}\right) \mid a \in A,b \in B$ 与 $f\left( a\right) = g\left( b\right) \}$ 一起构成 $R$ -子模，该子模是直和 $A \oplus B$（称为 $f$ 和 $g$ 的拉回或纤维积）的一个子模，并且存在一个交换图

where ${\pi }_{1}$ and ${\pi }_{2}$ are the natural projections onto the first and second components.

其中 ${\pi }_{1}$ 和 ${\pi }_{2}$ 是到第一和第二分量的自然投影。

(b) Suppose ${f}^{\prime } : M \rightarrow A$ and ${g}^{\prime } : M \rightarrow B$ are $R$ -module homomorphisms. Prove that the quotient $Y$ of $A \oplus B$ by $\left\{ {\left( {{f}^{\prime }\left( m\right) , - {g}^{\prime }\left( m\right) }\right) \mid m \in M}\right\}$ is an $R$ -module (called the pushout or fiber sum of ${f}^{\prime }$ and ${g}^{\prime }$ ) and that there is a commutative diagram

(b) 假设 ${f}^{\prime } : M \rightarrow A$ 和 ${g}^{\prime } : M \rightarrow B$ 是 $R$ -模同态。证明 $A \oplus B$ 被 $\left\{ {\left( {{f}^{\prime }\left( m\right) , - {g}^{\prime }\left( m\right) }\right) \mid m \in M}\right\}$ 的商 $Y$ 是一个 $R$ -模（称为 ${f}^{\prime }$ 和 ${g}^{\prime }$ 的推出或纤维和），并且存在一个交换图

where ${\pi }_{1}^{\prime }$ and ${\pi }_{2}^{\prime }$ are the natural maps to the quotient induced by the maps into the first and second components.

其中 ${\pi }_{1}^{\prime }$ 和 ${\pi }_{2}^{\prime }$ 是由到第一和第二分量的映射诱导出的到商的天然映射。

28. (a) (Schanuel’s Lemma) If $0 \rightarrow K \rightarrow P\overset{\varphi }{ \rightarrow }M \rightarrow 0$ and $0 \rightarrow {K}^{\prime } \rightarrow {P}^{\prime }\overset{{\varphi }^{\prime }}{ \rightarrow }M \rightarrow 0$ are exact sequences of $R$ -modules where $P$ and ${P}^{\prime }$ are projective,prove $P \oplus {K}^{\prime } \cong {P}^{\prime } \oplus K$ as $R$ -modules. [Show that there is an exact sequence $0 \rightarrow \ker \pi \rightarrow X\overset{\pi }{ \rightarrow }P \rightarrow 0$ with $\ker \pi \cong {K}^{\prime }$ ,where $X$ is the fiber product of $\varphi$ and ${\varphi }^{\prime }$ as in the previous exercise. Deduce that $X \cong P \oplus {K}^{\prime }$ . Show similarly that $X \cong {P}^{\prime } \oplus K$ .]

29. (a) （Schanuel引理）如果 $0 \rightarrow K \rightarrow P\overset{\varphi }{ \rightarrow }M \rightarrow 0$ 和 $0 \rightarrow {K}^{\prime } \rightarrow {P}^{\prime }\overset{{\varphi }^{\prime }}{ \rightarrow }M \rightarrow 0$ 是 $R$ -模的精确序列，其中 $P$ 和 ${P}^{\prime }$ 是投射的，证明 $P \oplus {K}^{\prime } \cong {P}^{\prime } \oplus K$ 作为 $R$ -模。[证明存在一个精确序列 $0 \rightarrow \ker \pi \rightarrow X\overset{\pi }{ \rightarrow }P \rightarrow 0$ ，其中 $\ker \pi \cong {K}^{\prime }$ ，$X$ 是如上练习中的 $\varphi$ 和 ${\varphi }^{\prime }$ 的纤维积。推出 $X \cong P \oplus {K}^{\prime }$ 。类似地证明 $X \cong {P}^{\prime } \oplus K$ 。]

(b) If $0 \rightarrow M \rightarrow Q\overset{\psi }{ \rightarrow }L \rightarrow 0$ and $0 \rightarrow M \rightarrow {Q}^{\prime }\overset{{\psi }^{\prime }}{ \rightarrow }{L}^{\prime } \rightarrow 0$ are exact sequences of $R$ -modules where $Q$ and ${Q}^{\prime }$ are injective,prove $Q \oplus {L}^{\prime } \cong {Q}^{\prime } \oplus L$ as $R$ -modules.

(b) 如果 $0 \rightarrow M \rightarrow Q\overset{\psi }{ \rightarrow }L \rightarrow 0$ 和 $0 \rightarrow M \rightarrow {Q}^{\prime }\overset{{\psi }^{\prime }}{ \rightarrow }{L}^{\prime } \rightarrow 0$ 是 $R$ -模的精确序列，其中 $Q$ 和 ${Q}^{\prime }$ 是内射的，证明 $Q \oplus {L}^{\prime } \cong {Q}^{\prime } \oplus L$ 作为 $R$ -模。

The $R$ -modules $M$ and $N$ are said to be projectively equivalent if $M \oplus P \cong N \oplus {P}^{\prime }$ for some projective modules $P,{P}^{\prime }$ . Similarly, $M$ and $N$ are injectively equivalent if $M \oplus Q \cong N \oplus {Q}^{\prime }$ for some injective modules $Q,{Q}^{\prime }$ . The previous exercise shows $K$ and ${K}^{\prime }$ are projectively equivalent and $L$ and ${L}^{\prime }$ are injectively equivalent.

设 $R$ -模 $M$ 和 $N$ 若存在 $M \oplus P \cong N \oplus {P}^{\prime }$ 对于某些投射模 $P,{P}^{\prime }$ ，则称它们是投射等价的。类似地，$M$ 和 $N$ 若存在 $M \oplus Q \cong N \oplus {Q}^{\prime }$ 对于某些内射模 $Q,{Q}^{\prime }$ ，则称它们是内射等价的。之前的练习表明 $K$ 和 ${K}^{\prime }$ 是投射等价的，而 $L$ 和 ${L}^{\prime }$ 是内射等价的。