12.1 THE BASIC THEORY

12.1 THE BASIC THEORY

12.1 基础理论

We first describe some general finiteness conditions. Let $R$ be a ring and let $M$ be a left $R$ -module.

我们首先描述一些一般有限性条件。设 $R$ 是一个环，$M$ 是一个左 $R$ -模。

Definition.

定义。

(1) The left $R$ -module $M$ is said to be a Noetherian $R$ -module or to satisfy the ascending chain condition on submodules (or A.C.C. on submodules) if there are no infinite increasing chains of submodules, i.e., whenever

（1）如果左 $R$ -模 $M$ 是一个诺特 $R$ -模或满足子模的升链条件（或子模的A.C.C.），则不存在无限的子模升链，即每当

is an increasing chain of submodules of $M$ ,then there is a positive integer $m$ such that for all $k \geq m,{M}_{k} = {M}_{m}$ (so the chain becomes stationary at stage $m$ : ${M}_{m} = {M}_{m + 1} = {M}_{m + 2} = \ldots ).$

是一个$M$的递增子模块链，那么存在一个正整数$m$，使得对于所有的$k \geq m,{M}_{k} = {M}_{m}$（因此在阶段$m$上链变为稳定：${M}_{m} = {M}_{m + 1} = {M}_{m + 2} = \ldots ).$。

(2) The ring $R$ is said to be Noetherian if it is Noetherian as a left module over itself,i.e.,if there are no infinite increasing chains of left ideals in $R$ .

(2) 如果环$R$作为自身的左模块是Noetherian的，即$R$中不存在无限递增的左理想链，则称该环为Noetherian环。

One can formulate analogous notions of A.C.C. on right and on two-sided ideals in a (possibly noncommutative) ring $R$ . For noncommutative rings these properties need not be related.

可以在（可能是非交换的）环$R$中对于右理想和双边理想制定A.C.C.的类似概念。对于非交换环，这些属性不一定相关。

Theorem 1. Let $R$ be a ring and let $M$ be a left $R$ -module. Then the following are equivalent:

定理1。设$R$为一个环，$M$为一个左$R$模块。那么以下条件是等价的：

(1) $M$ is a Noetherian $R$ -module.

(1) $M$是一个Noetherian$R$模块。

(2) Every nonempty set of submodules of $M$ contains a maximal element under inclusion.

(2) $M$的每一个非空子模块集合都包含一个包含关系下的极大元素。

(3) Every submodule of $M$ is finitely generated.

(3) $M$的每一个子模块都是有限生成的。

Proof: [(1) implies (2)] Assume $M$ is Noetherian and let $\sum$ be any nonempty collection of submodules of $M$ . Choose any ${M}_{1} \in \sum$ . If ${M}_{1}$ is a maximal element of $\sum$ ,(2) holds,so assume ${M}_{1}$ is not maximal. Then there is some ${M}_{2} \in \sum$ such that ${M}_{1} \subset {M}_{2}$ . If ${M}_{2}$ is maximal in $\sum$ ,(2) holds,so we may assume there is an ${M}_{3} \in \sum$ properly containing ${M}_{2}$ . Proceeding in this way one sees that if (2) fails we can produce by the Axiom of Choice an infinite strictly increasing chain of elements of $\sum$ ,contrary to (1).

证明：[(1)蕴含(2)] 假设$M$是Noetherian的，并且$\sum$是$M$的任意非空子模块集合。选择任意的${M}_{1} \in \sum$。如果${M}_{1}$是$\sum$的一个极大元素，那么(2)成立，因此假设${M}_{1}$不是极大元素。那么存在某个${M}_{2} \in \sum$，使得${M}_{1} \subset {M}_{2}$。如果${M}_{2}$在$\sum$中是极大的，那么(2)成立，所以我们假设存在一个${M}_{3} \in \sum$包含${M}_{2}$。这样继续下去，我们可以看到如果(2)不成立，我们可以通过选择公理产生一个$\sum$中元素的无限严格递增链，这与(1)相矛盾。

[(2) implies (3)] Assume (2) holds and let $N$ be any submodule of $M$ . Let $\sum$ be the collection of all finitely generated submodules of $N$ . Since $\{ 0\} \in \sum$ ,this collection is nonempty. By (2) $\sum$ contains a maximal element ${N}^{\prime }$ . If ${N}^{\prime } \neq N$ ,let $x \in N - {N}^{\prime }$ . Since ${N}^{\prime } \in \sum$ ,the submodule ${N}^{\prime }$ is finitely generated by assumption,hence also the submodule generated by ${N}^{\prime }$ and $x$ is finitely generated. This contradicts the maximality of ${N}^{\prime }$ ,so $N = {N}^{\prime }$ is finitely generated.

[(2) 蕴含 (3)] 假设 (2) 成立，并让 $N$ 是 $M$ 的任意子模。令 $\sum$ 为 $N$ 所有有限生成子模的集合。由于 $\{ 0\} \in \sum$ ，这个集合非空。由 (2) $\sum$ 包含一个极大元素 ${N}^{\prime }$ 。如果 ${N}^{\prime } \neq N$ ，则让 $x \in N - {N}^{\prime }$ 。由于 ${N}^{\prime } \in \sum$ ，子模 ${N}^{\prime }$ 是有限生成的（按假设），因此由 ${N}^{\prime }$ 和 $x$ 生成的子模也是有限生成的。这违背了 ${N}^{\prime }$ 的极大性，所以 $N = {N}^{\prime }$ 是有限生成的。

[(3) implies (1)] Assume (3) holds and let ${M}_{1} \subseteq {M}_{2} \subseteq {M}_{3}\ldots$ be a chain of

[(3) 蕴含 (1)] 假设 (3) 成立，并让 ${M}_{1} \subseteq {M}_{2} \subseteq {M}_{3}\ldots$ 是

submodules of $M$ . Let

$M$ 的子模链。令

and note that $N$ is a submodule. By (3) $N$ is finitely generated by,say, ${a}_{1},{a}_{2},\ldots ,{a}_{n}$ . Since ${a}_{i} \in N$ for all $i$ ,each ${a}_{i}$ lies in one of the submodules in the chain,say ${M}_{{j}_{i}}$ . Let $m = \max \left\{ {{j}_{1},{j}_{2},\ldots ,{j}_{n}}\right\}$ . Then ${a}_{i} \in {M}_{m}$ for all $i$ so the module they generate is contained in ${M}_{m}$ ,i.e., $N \subseteq {M}_{m}$ . This implies ${M}_{m} = N = {M}_{k}$ for all $k \geq m$ ,which proves (1).

并注意到 $N$ 是一个子模。由 (3) $N$ 是有限生成的，比如说，由 ${a}_{1},{a}_{2},\ldots ,{a}_{n}$ 生成。由于 ${a}_{i} \in N$ 对于所有 $i$ ，每个 ${a}_{i}$ 都位于链中的某个子模中，比如说 ${M}_{{j}_{i}}$ 。令 $m = \max \left\{ {{j}_{1},{j}_{2},\ldots ,{j}_{n}}\right\}$ 。那么 ${a}_{i} \in {M}_{m}$ 对于所有 $i$ ，因此它们生成的模包含在 ${M}_{m}$ 中，即 $N \subseteq {M}_{m}$ 。这意味着 ${M}_{m} = N = {M}_{k}$ 对于所有 $k \geq m$ ，这证明了 (1)。

Corollary 2. If $R$ is a P.I.D. then every nonempty set of ideals of $R$ has a maximal element and $R$ is a Noetherian ring.

推论 2。如果 $R$ 是一个 P.I.D.，那么 $R$ 的每个非空理想集都有一个极大元素，且 $R$ 是一个诺特环。

Proof: The P.I.D. $R$ satisfies condition (3) in the theorem with $M = R$ .

证明：P.I.D. $R$ 满足定理中的条件 (3)，其中 $M = R$ 。

Recall that even if $M$ itself is a finitely generated $R$ -module,submodules of $M$ need not be finitely generated,so the condition that $M$ be a Noetherian $R$ -module is in general stronger than the condition that $M$ be a finitely generated $R$ -module.

请记住，即使 $M$ 本身是一个有限生成的 $R$ -模，$M$ 的子模也不一定是有限生成的，因此 $M$ 是一个诺特 $R$ -模的条件通常比 $M$ 是一个有限生成的 $R$ -模的条件更强。

We require a result on "linear dependence" before turning to the main results of this chapter.

在转向本章的主要结果之前，我们需要一个关于“线性相关”的结果。

Proposition 3. Let $R$ be an integral domain and let $M$ be a free $R$ -module of rank $n < \infty$ . Then any $n + 1$ elements of $M$ are $R$ -linearly dependent,i.e.,for any ${y}_{1},{y}_{2},\ldots ,{y}_{n + 1} \in M$ there are elements ${r}_{1},{r}_{2},\ldots ,{r}_{n + 1} \in R$ ,not all zero,such that

命题 3. 设 $R$ 是一个整环，并且 $M$ 是一个秩为 $n < \infty$ 的自由 $R$ -模。那么，$M$ 的任何 $n + 1$ 个元素都是 $R$ -线性相关的，即对于任何 ${y}_{1},{y}_{2},\ldots ,{y}_{n + 1} \in M$ ，都存在元素 ${r}_{1},{r}_{2},\ldots ,{r}_{n + 1} \in R$ ，不全为零，使得

Proof: The quickest way of proving this is to embed $R$ in its quotient field $F$ (since $R$ is an integral domain) and observe that since $M \cong R \oplus R \oplus \cdots \oplus R$ ( $n$ times) we obtain $M \subseteq F \oplus F \oplus \cdots \oplus F$ . The latter is an $n$ -dimensional vector space over $F$ so any $n + 1$ elements of $M$ are $F$ -linearly dependent. By clearing the denominators of the scalars (by multiplying through by the product of all the denominators,for example), we obtain an $R$ -linear dependence relation among the $n + 1$ elements of $M$ .

证明：证明这个结论最快的方法是将 $R$ 嵌入到它的商域 $F$ 中（因为 $R$ 是一个整环），并观察到由于 $M \cong R \oplus R \oplus \cdots \oplus R$（$n$ 次），我们得到 $M \subseteq F \oplus F \oplus \cdots \oplus F$ 。后者是一个 $n$ 维的 $F$ 上的向量空间，因此 $M$ 的任何 $n + 1$ 个元素都是 $F$ -线性相关的。通过清除系数的分母（例如，通过乘以所有分母的乘积），我们得到 $M$ 的 $n + 1$ 个元素之间的一个 $R$ -线性相关关系。

Alternatively,let ${e}_{1},\ldots ,{e}_{n}$ be a basis of the free $R$ -module $M$ and let ${y}_{1},\ldots ,{y}_{n + 1}$ be any $n + 1$ elements of $M$ . For $1 \leq i \leq n + 1$ write ${y}_{i} = {a}_{1i}{e}_{i} + {a}_{2i}{e}_{2} + \ldots + {a}_{ni}{e}_{i}$ in terms of the basis ${e}_{1},{e}_{2},\ldots ,{e}_{n}$ . Let $A$ be the $\left( {n + 1}\right) \times \left( {n + 1}\right)$ matrix whose $i,j$ entry is ${a}_{ij},1 \leq i \leq n,1 \leq j \leq n + 1$ and whose last row is zero,so certainly $\det A = 0$ . Since $R$ is an integral domain,Corollary 27 of Section 11.4 shows that the columns of $A$ are $R$ -linearly dependent. Any dependence relation on the columns of $A$ gives a dependence relation on the ${y}_{i}$ ’s,completing the proof.

或者，让 ${e}_{1},\ldots ,{e}_{n}$ 成为自由 $R$ -模 $M$ 的一个基，并且让 ${y}_{1},\ldots ,{y}_{n + 1}$ 是 $M$ 中的任意 $n + 1$ 元素。对于 $1 \leq i \leq n + 1$，将 ${y}_{i} = {a}_{1i}{e}_{i} + {a}_{2i}{e}_{2} + \ldots + {a}_{ni}{e}_{i}$ 用基 ${e}_{1},{e}_{2},\ldots ,{e}_{n}$ 表示。让 $A$ 是一个 $\left( {n + 1}\right) \times \left( {n + 1}\right)$ 矩阵，其 $i,j$ 条目是 ${a}_{ij},1 \leq i \leq n,1 \leq j \leq n + 1$，最后一行是零，因此必然 $\det A = 0$。由于 $R$ 是一个整环，第11.4节的推论27表明 $A$ 的列线性依赖于 $R$。$A$ 的列上的任何依赖关系都会在 ${y}_{i}$ 上产生依赖关系，从而完成证明。

If $R$ is any integral domain and $M$ is any $R$ -module recall that

如果 $R$ 是任意的整环，并且 $M$ 是任意的 $R$ -模，请记住

is a submodule of $M$ (called the torsion submodule of $M$ ) and if $N$ is any submodule of $\operatorname{Tor}\left( M\right) ,N$ is called $a$ torsion submodule of $M$ (so the torsion submodule of $M$ is the union of all torsion submodules of $M$ ,i.e.,is the maximal torsion submodule of $M$ ). If $\operatorname{Tor}\left( M\right) = 0$ ,the module $M$ is said to be torsion free.

是 $M$ 的一个子模（称为 $M$ 的挠子模），如果 $N$ 是 $\operatorname{Tor}\left( M\right) ,N$ 的一个子模，则称为 $a$ 的挠子模 $M$（因此 $M$ 的挠子模是所有 $M$ 的挠子模的并集，即 $M$ 的最大挠子模）。如果 $\operatorname{Tor}\left( M\right) = 0$，则称模 $M$ 为挠自由的。

For any submodule $N$ of $M$ ,the annihilator of $N$ is the ideal of $R$ defined by

对于 $M$ 的任一子模 $N$，$N$ 的消去子是 $R$ 的理想，由

Note that if $N$ is not a torsion submodule of $M$ then $\operatorname{Ann}\left( N\right) = \left( 0\right)$ . It is easy to see that if $N,L$ are submodules of $M$ with $N \subseteq L$ ,then $\operatorname{Ann}\left( L\right) \subseteq \operatorname{Ann}\left( N\right)$ . If $R$ is a P.I.D. and $N \subseteq L \subseteq M$ with $\operatorname{Ann}\left( N\right) = \left( a\right)$ and $\operatorname{Ann}\left( L\right) = \left( b\right)$ ,then $a \mid b$ . In particular, the annihilator of any element $x$ of $M$ divides the annihilator of $M$ (this is implied by Lagrange’s Theorem when $R = \mathbb{Z}$ ).

注意，如果 $N$ 不是 $M$ 的挠子模，那么 $\operatorname{Ann}\left( N\right) = \left( 0\right)$ 。容易看出，如果 $N,L$ 是 $M$ 的子模，并且 $N \subseteq L$ ，那么 $\operatorname{Ann}\left( L\right) \subseteq \operatorname{Ann}\left( N\right)$ 。如果 $R$ 是一个P.I.D.且 $N \subseteq L \subseteq M$ ，其中 $\operatorname{Ann}\left( N\right) = \left( a\right)$ 和 $\operatorname{Ann}\left( L\right) = \left( b\right)$ ，那么 $a \mid b$ 。特别是，任何元素 $x$ 的消去子模 $M$ 都整除 $M$ 的消去子模（这在 $R = \mathbb{Z}$ 时由拉格朗日定理隐含）。

Definition. For any integral domain $R$ the rank of an $R$ -module $M$ is the maximum number of $R$ -linearly independent elements of $M$ .

定义。对于任何整环 $R$ ，一个 $R$ -模 $M$ 的秩是 $M$ 中 $R$ -线性无关元素的最大数量。

The preceding proposition states that for a free $R$ -module $M$ over an integral domain the rank of a submodule is bounded by the rank of $M$ . This notion of rank agrees with previous uses of the same term. If the ring $R = F$ is a field,then the rank of an $R$ -module $M$ is the dimension of $M$ as a vector space over $F$ and any maximal set of $F$ -linearly independent elements is a basis for $M$ . For a general integral domain, however,an $R$ -module $M$ of rank $n$ need not have a "basis," i.e.,need not be a free $R$ -module even if $M$ is torsion free,so some care is necessary with the notion of rank, particularly with respect to the torsion elements of $M$ . Exercises 1 to 6 and 20 give an alternate characterization of the rank and provide some examples of (torsion free) $R$ -modules (of rank 1) that are not free.

前一个命题表明，对于一个在整环上的自由 $R$ -模 $M$ ，子模的秩受限于 $M$ 的秩。这个秩的概念与之前使用该术语时一致。如果环 $R = F$ 是一个域，那么 $R$ -模 $M$ 的秩就是 $M$ 作为 $F$ 上的向量空间的维数，任何 $F$ -线性无关的元素的最大集合都是 $M$ 的基。然而，对于一般的整环，一个秩为 $n$ 的 $R$ -模 $M$ 不一定有“基”，即即使 $M$ 是挠自由的，也不一定是自由 $R$ -模，因此在秩的概念上需要格外小心，特别是关于 $M$ 的挠元素。练习 1 到 6 和 20 提供了秩的另一种表征，并给出了一些（挠自由）$R$ -模（秩为 1）不是自由的例子。

The next important result shows that if $N$ is a submodule of a free module of finite rank over a P.I.D. then $N$ is again a free module of finite rank and furthermore it is possible to choose generators for the two modules which are related in a simple way.

下一个重要结果表明，如果 $N$ 是一个主理想整环上的有限秩自由模的子模，那么 $N$ 本身也是一个有限秩的自由模，并且可以选择两个模的生成元，它们之间有简单的关系。

Theorem 4. Let $R$ be a Principal Ideal Domain,let $M$ be a free $R$ -module of finite rank $n$ and let $N$ be a submodule of $M$ . Then

定理 4。设 $R$ 是一个主理想整环，$M$ 是一个秩为 $n$ 的自由 $R$ -模，$N$ 是 $M$ 的子模。

(1) $N$ is free of rank $m,m \leq n$ and

（1）$N$ 是秩为 $m,m \leq n$ 的自由模，并且

(2) there exists a basis ${y}_{1},{y}_{2},\ldots ,{y}_{n}$ of $M$ so that ${a}_{1}{y}_{1},{a}_{2}{y}_{2},\ldots ,{a}_{m}{y}_{m}$ is a basis of $N$ where ${a}_{1},{a}_{2},\ldots ,{a}_{m}$ are nonzero elements of $R$ with the divisibility relations

存在一个基 ${y}_{1},{y}_{2},\ldots ,{y}_{n}$ 的 $M$ ，使得 ${a}_{1}{y}_{1},{a}_{2}{y}_{2},\ldots ,{a}_{m}{y}_{m}$ 是 $N$ 的基，其中 ${a}_{1},{a}_{2},\ldots ,{a}_{m}$ 是 $R$ 的非零元素，具有可整除性关系

Proof: The theorem is trivial for $N = \{ 0\}$ ,so assume $N \neq \{ 0\}$ . For each $R$ -module homomorphism $\varphi$ of $M$ into $R$ ,the image $\varphi \left( N\right)$ of $N$ is a submodule of $R$ ,i.e.,an ideal in $R$ . Since $R$ is a P.I.D. this ideal must be principal,say $\varphi \left( N\right) = \left( {a}_{\varphi }\right)$ ,for some ${a}_{\varphi } \in R$ . Let

证明：当 $N = \{ 0\}$ 时定理是平凡的，所以假设 $N \neq \{ 0\}$ 。对于每个 $R$ -模同态 $\varphi$ 从 $M$ 到 $R$ ，$\varphi \left( N\right)$ 的像 $N$ 是 $R$ 的子模，即 $R$ 的一个理想。由于 $R$ 是一个P.I.D.，这个理想必须是主理想，比如说 $\varphi \left( N\right) = \left( {a}_{\varphi }\right)$ ，对于某个 ${a}_{\varphi } \in R$ 。设

be the collection of the principal ideals in $R$ obtained in this way from the $R$ -module homomorphisms of $M$ into $R$ . The collection $\sum$ is certainly nonempty since taking $\varphi$ to be the trivial homomorphism shows that $\left( 0\right) \in \sum$ . By Corollary 2, $\sum$ has at least one maximal element i.e.,there is at least one homomorphism $v$ of $M$ to $R$ so that the principal ideal $v\left( N\right) = \left( {a}_{v}\right)$ is not properly contained in any other element of $\sum$ . Let ${a}_{1} = {a}_{v}$ for this maximal element and let $y \in N$ be an element mapping to the generator ${a}_{1}$ under the homomorphism $v : v\left( y\right) = {a}_{1}$ .

是通过这种方式从 $R$ -模同态从 $M$ 到 $R$ 得到的 $R$ 的主理想的集合。这个集合 $\sum$ 显然是非空的，因为取 $\varphi$ 为平凡同态就表明 $\left( 0\right) \in \sum$ 。根据推论2，$\sum$ 至少有一个极大元素，即至少有一个同态 $v$ 从 $M$ 到 $R$ ，使得主理想 $v\left( N\right) = \left( {a}_{v}\right)$ 不被 $\sum$ 的任何其他元素真包含。设 ${a}_{1} = {a}_{v}$ 为这个极大元素，并设 $y \in N$ 是在同态 $v : v\left( y\right) = {a}_{1}$ 下映射到生成元 ${a}_{1}$ 的元素。

We now show the element ${a}_{1}$ is nonzero. Let ${x}_{1},{x}_{2},\ldots ,{x}_{n}$ be any basis of the free module $M$ and let ${\pi }_{i} \in {\operatorname{Hom}}_{R}\left( {M,R}\right)$ be the natural projection homomorphism onto the ${i}^{\text{th }}$ coordinate with respect to this basis. Since $N \neq \{ 0\}$ ,there exists an $i$ such that ${\pi }_{i}\left( N\right) \neq 0$ ,which in particular shows that $\sum$ contains more than just the trivial ideal (0). Since $\left( {a}_{1}\right)$ is a maximal element of $\sum$ it follows that ${a}_{1} \neq 0$ .

现在我们展示元素 ${a}_{1}$ 是非零的。设 ${x}_{1},{x}_{2},\ldots ,{x}_{n}$ 是自由模块 $M$ 的任意基，${\pi }_{i} \in {\operatorname{Hom}}_{R}\left( {M,R}\right)$ 是到该基的 ${i}^{\text{th }}$ 坐标上的自然投影同态。由于 $N \neq \{ 0\}$ ，存在一个 $i$ 使得 ${\pi }_{i}\left( N\right) \neq 0$ ，这特别表明 $\sum$ 包含的不仅仅是平凡理想 (0)。由于 $\left( {a}_{1}\right)$ 是 $\sum$ 的一个极大元，因此 ${a}_{1} \neq 0$ 。

We next show that this element ${a}_{1}$ divides $\varphi \left( y\right)$ for every $\varphi \in {\operatorname{Hom}}_{R}\left( {M,R}\right)$ . To see this let $d$ be a generator for the principal ideal generated by ${a}_{1}$ and $\varphi \left( y\right)$ . Then $d$ is a divisor of both ${a}_{1}$ and $\varphi \left( y\right)$ in $R$ and $d = {r}_{1}{a}_{1} + {r}_{2}\varphi \left( y\right)$ for some ${r}_{1},{r}_{2} \in R$ . Consider the homomorphism $\psi = {r}_{1}v + {r}_{2}\varphi$ from $M$ to $R$ . Then $\psi \left( y\right) = \left( {{r}_{1}v + {r}_{2}\varphi }\right) \left( y\right) =$ ${r}_{1}{a}_{1} + {r}_{2}\varphi \left( y\right) = d$ so that $d \in \psi \left( N\right)$ ,hence also $\left( d\right) \subseteq \psi \left( N\right)$ . But $d$ is a divisor of ${a}_{1}$ so we also have $\left( {a}_{1}\right) \subseteq \left( d\right)$ . Then $\left( {a}_{1}\right) \subseteq \left( d\right) \subseteq \psi \left( N\right)$ and by the maximality of $\left( {a}_{1}\right)$ we must have equality: $\left( {a}_{1}\right) = \left( d\right) = \psi \left( N\right)$ . In particular $\left( {a}_{1}\right) = \left( d\right)$ shows that ${a}_{1} \mid \varphi \left( y\right)$ since $d$ divides $\varphi \left( y\right)$ .

接下来我们展示这个元素 ${a}_{1}$ 能整除 $\varphi \left( y\right)$ 对于每一个 $\varphi \in {\operatorname{Hom}}_{R}\left( {M,R}\right)$ 。为了看到这一点，设 $d$ 是由 ${a}_{1}$ 和 $\varphi \left( y\right)$ 生成的主理想的生成元。那么 $d$ 是 $R$ 中 ${a}_{1}$ 和 $\varphi \left( y\right)$ 的一个除数，并且 $d = {r}_{1}{a}_{1} + {r}_{2}\varphi \left( y\right)$ 对于某个 ${r}_{1},{r}_{2} \in R$ 。考虑从 $M$ 到 $R$ 的同态 $\psi = {r}_{1}v + {r}_{2}\varphi$ 。那么 $\psi \left( y\right) = \left( {{r}_{1}v + {r}_{2}\varphi }\right) \left( y\right) =$ ${r}_{1}{a}_{1} + {r}_{2}\varphi \left( y\right) = d$ 因此 $d \in \psi \left( N\right)$ ，因此也有 $\left( d\right) \subseteq \psi \left( N\right)$ 。但是 $d$ 是 ${a}_{1}$ 的一个除数，所以我们也有 $\left( {a}_{1}\right) \subseteq \left( d\right)$ 。那么 $\left( {a}_{1}\right) \subseteq \left( d\right) \subseteq \psi \left( N\right)$ 并且由于 $\left( {a}_{1}\right)$ 的极大性，我们必须有等式：$\left( {a}_{1}\right) = \left( d\right) = \psi \left( N\right)$ 。特别地 $\left( {a}_{1}\right) = \left( d\right)$ 表明 ${a}_{1} \mid \varphi \left( y\right)$ ，因为 $d$ 能整除 $\varphi \left( y\right)$ 。

If we apply this to the projection homomorphisms ${\pi }_{i}$ we see that ${a}_{1}$ divides ${\pi }_{i}\left( y\right)$ for all $i$ . Write ${\pi }_{i}\left( y\right) = {a}_{1}{b}_{i}$ for some ${b}_{i} \in R,1 \leq i \leq n$ and define

如果我们将这应用于投影同态 ${\pi }_{i}$ ，我们会看到 ${a}_{1}$ 能整除 ${\pi }_{i}\left( y\right)$ 对于所有的 $i$ 。对于某个 ${b}_{i} \in R,1 \leq i \leq n$ 写 ${\pi }_{i}\left( y\right) = {a}_{1}{b}_{i}$ 并定义

Note that ${a}_{1}{y}_{1} = y$ . Since ${a}_{1} = v\left( y\right) = v\left( {{a}_{1}{y}_{1}}\right) = {a}_{1}v\left( {y}_{1}\right)$ and ${a}_{1}$ is a nonzero element of the integral domain $R$ this shows

注意 ${a}_{1}{y}_{1} = y$ 。由于 ${a}_{1} = v\left( y\right) = v\left( {{a}_{1}{y}_{1}}\right) = {a}_{1}v\left( {y}_{1}\right)$ 并且 ${a}_{1}$ 是整环 $R$ 的一个非零元素，这表明

We now verify that this element ${y}_{1}$ can be taken as one element in a basis for $M$ and that ${a}_{1}{y}_{1}$ can be taken as one element in a basis for $N$ ,namely that we have (a) $M = R{y}_{1} \oplus \ker v$ ,and

我们现在验证这个元素 ${y}_{1}$ 可以作为一个基 $M$ 的一个元素，并且 ${a}_{1}{y}_{1}$ 可以作为基 $N$ 的一个元素，即我们具有 (a) $M = R{y}_{1} \oplus \ker v$，并且

(b) $N = R{a}_{1}{y}_{1} \oplus \left( {N \cap \ker v}\right)$ .

(b) $N = R{a}_{1}{y}_{1} \oplus \left( {N \cap \ker v}\right)$。

To see (a) let $x$ be an arbitrary element in $M$ and write $x = v\left( x\right) {y}_{1} + \left( {x - v\left( x\right) {y}_{1}}\right)$ . Since

为了看到 (a)，设 $x$ 是 $M$ 中的一个任意元素，并写成 $x = v\left( x\right) {y}_{1} + \left( {x - v\left( x\right) {y}_{1}}\right)$。由于

we see that $x - v\left( x\right) {y}_{1}$ is an element in the kernel of $v$ . This shows that $x$ can be written as the sum of an element in $R{y}_{1}$ and an element in the kernel of $v$ ,so $M = R{y}_{1} + \ker v$ . To see that the sum is direct,suppose $r{y}_{1}$ is also an element in the kernel of $v$ . Then $0 = v\left( {r{y}_{1}}\right) = {rv}\left( {y}_{1}\right) = r$ shows that this element is indeed 0 .

我们看到 $x - v\left( x\right) {y}_{1}$ 是 $v$ 的核中的一个元素。这表明 $x$ 可以写成 $R{y}_{1}$ 中的一个元素与 $v$ 的核中的一个元素的和，所以 $M = R{y}_{1} + \ker v$。为了看到这个和是直接的，假设 $r{y}_{1}$ 也是 $v$ 的核中的一个元素。那么 $0 = v\left( {r{y}_{1}}\right) = {rv}\left( {y}_{1}\right) = r$ 显示这个元素确实是0。

For (b) observe that $v\left( {x}^{\prime }\right)$ is divisible by ${a}_{1}$ for every ${x}^{\prime } \in N$ by the definition of ${a}_{1}$ as a generator for $v\left( N\right)$ . If we write $v\left( {x}^{\prime }\right) = b{a}_{1}$ where $b \in R$ then the decomposition we used in (a) above is ${x}^{\prime } = v\left( {x}^{\prime }\right) {y}_{1} + \left( {{x}^{\prime } - v\left( {x}^{\prime }\right) {y}_{1}}\right) = b{a}_{1}{y}_{1} + \left( {{x}^{\prime } - b{a}_{1}{y}_{1}}\right)$ where the second summand is in the kernel of $v$ and is an element of $N$ . This shows that $N = R{a}_{1}{y}_{1} + \left( {N \cap \ker v}\right)$ . The fact that the sum in (b) is direct is a special case of the directness of the sum in (a).

对于 (b)，观察到根据 ${a}_{1}$ 作为 $v\left( N\right)$ 的生成元的定义，$v\left( {x}^{\prime }\right)$ 可以被 ${a}_{1}$ 整除对每个 ${x}^{\prime } \in N$。如果我们写成 $v\left( {x}^{\prime }\right) = b{a}_{1}$，其中 $b \in R$，那么我们在 (a) 中使用的分解是 ${x}^{\prime } = v\left( {x}^{\prime }\right) {y}_{1} + \left( {{x}^{\prime } - v\left( {x}^{\prime }\right) {y}_{1}}\right) = b{a}_{1}{y}_{1} + \left( {{x}^{\prime } - b{a}_{1}{y}_{1}}\right)$，其中第二个加项在 $v$ 的核中，并且是 $N$ 的一个元素。这表明 $N = R{a}_{1}{y}_{1} + \left( {N \cap \ker v}\right)$。(b) 中的和是直接的是一个 ${a}_{1}$ 中和的直接性的特殊情况。

We now prove part (1) of the theorem by induction on the rank, $m$ ,of $N$ . If $m = 0$ , then $N$ is a torsion module,hence $N = 0$ since a free module is torsion free,so (1) holds trivially. Assume then that $m > 0$ . Since the sum in (b) above is direct we see easily that $N \cap \ker v$ has rank $m - 1$ (cf. Exercise 3). By induction $N \cap \ker v$ is then a free $R$ -module of rank $m - 1$ . Again by the directness of the sum in (b) we see that adjoining ${a}_{1}{y}_{1}$ to any basis of $N \cap \ker v$ gives a basis of $N$ ,so $N$ is also free (of rank $m$ ),which proves (1).

我们现在通过归纳法证明定理的第一部分，对$m$的秩$N$进行归纳。如果$m = 0$，那么$N$是一个挠模，因此$N = 0$，因为自由模是无挠的，所以(1)显然成立。假设$m > 0$。由于上面(b)中的和是直和，我们可以容易地看出$N \cap \ker v$的秩为$m - 1$（参见练习3）。通过归纳，$N \cap \ker v$然后是一个秩为$m - 1$的自由$R$模。再次由于(b)中的和是直和，我们看出将${a}_{1}{y}_{1}$添加到$N \cap \ker v$的任何基中都会得到$N$的一个基，所以$N$也是自由的（秩为$m$），这证明了(1)。

Finally,we prove (2) by induction on $n$ ,the rank of $M$ . Applying (1) to the submodule ker $\nu$ shows that this submodule is free and because the sum in (a) is direct it is free of rank $n - 1$ . By the induction assumption applied to the module ker $v$ (which plays the role of $M$ ) and its submodule $\ker v \cap N$ (which plays the role of $N$ ),we see that there is a basis ${y}_{2},{y}_{3},\ldots ,{y}_{n}$ of $\ker v$ such that ${a}_{2}{y}_{2},{a}_{3}{y}_{3},\ldots ,{a}_{m}{y}_{m}$ is a basis of $N \cap \ker v$ for some elements ${a}_{2},{a}_{3},\ldots ,{a}_{m}$ of $R$ with ${a}_{2}\left| {a}_{3}\right| \cdots \mid {a}_{m}$ . Since the sums (a) and (b) are direct, ${y}_{1},{y}_{2},\ldots ,{y}_{n}$ is a basis of $M$ and ${a}_{1}{y}_{1},{a}_{2}{y}_{2},\ldots ,{a}_{m}{y}_{m}$ is a basis of $N$ . To complete the induction it remains to show that ${a}_{1}$ divides ${a}_{2}$ . Define a homomorphism $\varphi$ from $M$ to $R$ by defining $\varphi \left( {y}_{1}\right) = \varphi \left( {y}_{2}\right) = 1$ and $\varphi \left( {y}_{i}\right) = 0$ ,for all $i > 2$ ,on the basis for $M$ . Then for this homomorphism $\varphi$ we have ${a}_{1} = \varphi \left( {{a}_{1}{y}_{1}}\right)$ so ${a}_{1} \in \varphi \left( N\right)$ hence also $\left( {a}_{1}\right) \subseteq \varphi \left( N\right)$ . By the maximality of $\left( {a}_{1}\right)$ in $\sum$ it follows that $\left( {a}_{1}\right) = \varphi \left( N\right)$ . Since ${a}_{2} = \varphi \left( {{a}_{2}{y}_{2}}\right) \in \varphi \left( N\right)$ we then have ${a}_{2} \in \left( {a}_{1}\right)$ i.e., ${a}_{1} \mid {a}_{2}$ . This completes the proof of the theorem.

最后，我们通过对 $n$ 进行归纳，即对 $M$ 的秩进行归纳，来证明 (2)。将 (1) 应用于子模块 ker $\nu$，表明这个子模块是自由的，并且因为 (a) 中的和是直接的，所以它的秩为 $n - 1$ 的自由。通过对模块 ker $v$（它扮演 $M$ 的角色）及其子模块 $\ker v \cap N$（它扮演 $N$ 的角色）应用归纳假设，我们发现存在 ${y}_{2},{y}_{3},\ldots ,{y}_{n}$ 的一个基 $\ker v$，使得 ${a}_{2}{y}_{2},{a}_{3}{y}_{3},\ldots ,{a}_{m}{y}_{m}$ 是 $N \cap \ker v$ 对于某些 ${a}_{2},{a}_{3},\ldots ,{a}_{m}$ 的元素 $R$ 的基，且 ${a}_{2}\left| {a}_{3}\right| \cdots \mid {a}_{m}$ 。由于和 (a) 和 (b) 是直接的，${y}_{1},{y}_{2},\ldots ,{y}_{n}$ 是 $M$ 的基，${a}_{1}{y}_{1},{a}_{2}{y}_{2},\ldots ,{a}_{m}{y}_{m}$ 是 $N$ 的基。为了完成归纳，还需要证明 ${a}_{1}$ 除 ${a}_{2}$ 。定义一个从 $M$ 到 $R$ 的同态 $\varphi$，通过在 $M$ 的基上定义 $\varphi \left( {y}_{1}\right) = \varphi \left( {y}_{2}\right) = 1$ 和 $\varphi \left( {y}_{i}\right) = 0$ ，对于所有的 $i > 2$ 。那么对于这个同态 $\varphi$，我们有 ${a}_{1} = \varphi \left( {{a}_{1}{y}_{1}}\right)$，因此 ${a}_{1} \in \varphi \left( N\right)$，进而也有 $\left( {a}_{1}\right) \subseteq \varphi \left( N\right)$。由于 $\left( {a}_{1}\right)$ 在 $\sum$ 中是最大的，因此得出 $\left( {a}_{1}\right) = \varphi \left( N\right)$。由于 ${a}_{2} = \varphi \left( {{a}_{2}{y}_{2}}\right) \in \varphi \left( N\right)$，然后我们有 ${a}_{2} \in \left( {a}_{1}\right)$，即 ${a}_{1} \mid {a}_{2}$。这完成了定理的证明。

Recall that the left $R$ -module $C$ is a cyclic $R$ -module (for any ring $R$ ,not necessarily commutative nor with 1) if there is an element $x \in C$ such that $C = {Rx}$ . We can then define an $R$ -module homomorphism

回顾一下，左 $R$ -模 $C$ 是一个循环 $R$ -模（对于任何环 $R$ ，不一定是交换的也不一定带有单位元），如果存在一个元素 $x \in C$ 使得 $C = {Rx}$ 。然后我们可以定义一个 $R$ -模同态

by $\pi \left( r\right) = {rx}$ ,which will be surjective by the assumption $C = {Rx}$ . The First Isomorphism Theorem gives an isomorphism of (left) $R$ -modules

通过 $\pi \left( r\right) = {rx}$ ，在假设 $C = {Rx}$ 下这将是一个满射。第一同构定理给出了（左）$R$ -模的同构

If $R$ is a P.I.D.,ker $\pi$ is a principal ideal,(a),so we see that the cyclic $R$ -modules $C$ are of the form $R/\left( a\right)$ where $\left( a\right) = \operatorname{Ann}\left( C\right)$ .

如果 $R$ 是一个P.I.D.，那么ker $\pi$ 是一个主理想，（a），所以我们看到循环 $R$ -模 $C$ 的形式为 $R/\left( a\right)$ ，其中 $\left( a\right) = \operatorname{Ann}\left( C\right)$ 。

The cyclic modules are the simplest modules (since they require only one generator). The existence portion of the Fundamental Theorem states that any finitely generated module over a P.I.D. is isomorphic to the direct sum of finitely many cyclic modules.

循环模是最简单的模（因为它们只需要一个生成元）。基本定理的存在部分指出，任何在P.I.D.上的有限生成模都与有限多个循环模的直接和同构。

Theorem 5. (Fundamental Theorem,Existence: Invariant Factor Form) Let $R$ be a P.I.D. and let $M$ be a finitely generated $R$ -module.

定理5.（基本定理，存在性：不变因子形式）设 $R$ 是一个P.I.D.，$M$ 是一个有限生成的 $R$ -模。

(1) Then $M$ is isomorphic to the direct sum of finitely many cyclic modules. More precisely,

（1）那么 $M$ 与有限多个循环模的直接和同构。更准确地说，

for some integer $r \geq 0$ and nonzero elements ${a}_{1},{a}_{2},\ldots ,{a}_{m}$ of $R$ which are not units in $R$ and which satisfy the divisibility relations

对于某些整数 $r \geq 0$ 和 $R$ 中的非零元素 ${a}_{1},{a}_{2},\ldots ,{a}_{m}$ ，这些元素不是 $R$ 中的单位元，并且满足可除性关系

(2) $M$ is torsion free if and only if $M$ is free.

（2）$M$ 是扭自由的当且仅当 $M$ 是自由的。

(3) In the decomposition in (1),

（3）在（1）中的分解中，

In particular $M$ is a torsion module if and only if $r = 0$ and in this case the annihilator of $M$ is the ideal $\left( {a}_{m}\right)$ .

特别地，$M$ 是一个扭模当且仅当 $r = 0$ ，在这种情况下，$M$ 的消灭理想是理想 $\left( {a}_{m}\right)$ 。

Proof: The module $M$ can be generated by a finite set of elements by assumption so let ${x}_{1},{x}_{2},\ldots ,{x}_{n}$ be a set of generators of $M$ of minimal cardinality. Let ${R}^{n}$ be the free $R$ -module of rank $n$ with basis ${b}_{1},{b}_{2},\ldots ,{b}_{n}$ and define the homomorphism $\pi : {R}^{n} \rightarrow M$ by defining $\pi \left( {b}_{i}\right) = {x}_{i}$ for all $i$ ,which is automatically surjective since ${x}_{1},\ldots ,{x}_{n}$ generate $M$ . By the First Isomorphism Theorem for modules we have ${R}^{n}/\ker \pi \cong M$ . Now,by Theorem 4 applied to ${R}^{n}$ and the submodule ker $\pi$ we can choose another basis ${y}_{1},{y}_{2},\ldots ,{y}_{n}$ of ${R}^{n}$ so that ${a}_{1}{y}_{1},{a}_{2}{y}_{2},\ldots ,{a}_{m}{y}_{m}$ is a basis of $\ker \pi$ for some elements ${a}_{1},{a}_{2},\ldots ,{a}_{m}$ of $R$ with ${a}_{1}\left| {a}_{2}\right| \cdots \mid {a}_{m}$ . This implies

证明：根据假设，模块 $M$ 可以由有限个元素生成，设 ${x}_{1},{x}_{2},\ldots ,{x}_{n}$ 为 $M$ 的生成元集合，其基数最小。令 ${R}^{n}$ 为自由 $R$ -模块，秩为 $n$，基为 ${b}_{1},{b}_{2},\ldots ,{b}_{n}$，并定义同态 $\pi : {R}^{n} \rightarrow M$，通过定义 $\pi \left( {b}_{i}\right) = {x}_{i}$ 对于所有 $i$，这自然是满射，因为 ${x}_{1},\ldots ,{x}_{n}$ 生成 $M$。根据模的第一同构定理，我们有 ${R}^{n}/\ker \pi \cong M$。现在，将定理4应用于 ${R}^{n}$ 和子模 ker $\pi$，我们可以选择 ${R}^{n}$ 的另一个基 ${y}_{1},{y}_{2},\ldots ,{y}_{n}$，使得 ${a}_{1}{y}_{1},{a}_{2}{y}_{2},\ldots ,{a}_{m}{y}_{m}$ 对于某些 $R$ 中的元素 ${a}_{1},{a}_{2},\ldots ,{a}_{m}$ 和 ${a}_{1}\left| {a}_{2}\right| \cdots \mid {a}_{m}$ 是 $\ker \pi$ 的基。这意味着

To identify the quotient on the right hand side we use the natural surjective $R$ -module homomorphism

为了确定右侧商，我们使用自然的满射 $R$ -模块同态

that maps $\left( {{\alpha }_{1}{y}_{1},\ldots ,{\alpha }_{n}{y}_{n}}\right)$ to $\left( {{\alpha }_{1}{\;\operatorname{mod}\;\left( {a}_{1}\right) },\ldots ,{\alpha }_{m}{\;\operatorname{mod}\;\left( {a}_{m}\right) },{\alpha }_{m + 1},\ldots ,{\alpha }_{n}}\right)$ . The kernel of this map is clearly the set of elements where ${a}_{i}$ divides ${\alpha }_{i},i = 1,2,\ldots ,m$ , i.e., $R{a}_{1}{y}_{1} \oplus R{a}_{2}{y}_{2} \oplus \cdots \oplus R{a}_{m}{y}_{m}$ (cf. Exercise 7). Hence we obtain

它将 $\left( {{\alpha }_{1}{y}_{1},\ldots ,{\alpha }_{n}{y}_{n}}\right)$ 映射到 $\left( {{\alpha }_{1}{\;\operatorname{mod}\;\left( {a}_{1}\right) },\ldots ,{\alpha }_{m}{\;\operatorname{mod}\;\left( {a}_{m}\right) },{\alpha }_{m + 1},\ldots ,{\alpha }_{n}}\right)$。这个映射的核显然是元素集合，其中 ${a}_{i}$ 除 ${\alpha }_{i},i = 1,2,\ldots ,m$，即 $R{a}_{1}{y}_{1} \oplus R{a}_{2}{y}_{2} \oplus \cdots \oplus R{a}_{m}{y}_{m}$（参见练习7）。因此我们得到

If $a$ is a unit in $R$ then $R/\left( a\right) = 0$ ,so in this direct sum we may remove any of the initial ${a}_{i}$ which are units. This gives the decomposition in (1) (with $r = n - m$ ).

如果 $a$ 是 $R$ 中的单位元，那么 $R/\left( a\right) = 0$，因此在这个直和中，我们可以移除任何初始的 ${a}_{i}$ 单位元。这给出了（1）中的分解（带有 $r = n - m$）。

Since $R/\left( a\right)$ is a torsion $R$ -module for any nonzero element $a$ of $R,\left( 1\right)$ immediately implies $M$ is a torsion free module if and only if $M \cong {R}^{r}$ ,which is (2). Part (3) is immediate from the definitions since the annihilator of $R/\left( a\right)$ is evidently the ideal(a).

由于 $R/\left( a\right)$ 是扭 $R$ -模块，对于任何非零元素 $a$ 的 $R,\left( 1\right)$，立即意味着 $M$ 是一个无扭模块当且仅当 $M \cong {R}^{r}$，这是（2）。部分（3）根据定义立即得出，因为 $R/\left( a\right)$ 的消去子显然是理想(a)。

We shall shortly prove the uniqueness of the decomposition in Theorem 5, namely that if we have

We shall shortly prove the uniqueness of the decomposition in Theorem 5, namely that if we have

for some integer ${r}^{\prime } \geq 0$ and nonzero elements ${b}_{1},{b}_{2},\ldots ,{b}_{{m}^{\prime }}$ of $R$ which are not units with

for some integer ${r}^{\prime } \geq 0$ and nonzero elements ${b}_{1},{b}_{2},\ldots ,{b}_{{m}^{\prime }}$ of $R$ which are not units with

then $r = {r}^{\prime },m = {m}^{\prime }$ and $\left( {a}_{i}\right) = \left( {b}_{i}\right)$ (so ${a}_{i} = {b}_{i}$ up to units) for all $i$ . It is precisely the divisibility condition ${a}_{1}\left| {a}_{2}\right| \cdots \mid {a}_{m}$ which gives this uniqueness.

then $r = {r}^{\prime },m = {m}^{\prime }$ and $\left( {a}_{i}\right) = \left( {b}_{i}\right)$ (so ${a}_{i} = {b}_{i}$ up to units) for all $i$ . It is precisely the divisibility condition ${a}_{1}\left| {a}_{2}\right| \cdots \mid {a}_{m}$ which gives this uniqueness.

Definition. The integer $r$ in Theorem 5 is called the free rank or the Betti number of $M$ and the elements ${a}_{1},{a}_{2},\ldots ,{a}_{m} \in R$ (defined up to multiplication by units in $R$ ) are called the invariant factors of $M$ .

Definition. The integer $r$ in Theorem 5 is called the free rank or the Betti number of $M$ and the elements ${a}_{1},{a}_{2},\ldots ,{a}_{m} \in R$ (defined up to multiplication by units in $R$ ) are called the invariant factors of $M$ .

Note that until we have proved that the invariant factors of $M$ are unique we should properly refer to $a$ set of invariant factors for $M$ (and similarly for the free rank),by which we mean any elements giving a decomposition for $M$ as in (1) of the theorem above.

Note that until we have proved that the invariant factors of $M$ are unique we should properly refer to $a$ set of invariant factors for $M$ (and similarly for the free rank),by which we mean any elements giving a decomposition for $M$ as in (1) of the theorem above.

Using the Chinese Remainder Theorem it is possible to decompose the cyclic modules in Theorem 5 further so that $M$ is the direct sum of cyclic modules whose annihilators are as simple as possible (namely (0) or generated by powers of primes in $R$ ). This gives an alternate decomposition which we shall also see is unique and which we now describe.

Using the Chinese Remainder Theorem it is possible to decompose the cyclic modules in Theorem 5 further so that $M$ is the direct sum of cyclic modules whose annihilators are as simple as possible (namely (0) or generated by powers of primes in $R$ ). This gives an alternate decomposition which we shall also see is unique and which we now describe.

Suppose $a$ is a nonzero element of the Principal Ideal Domain $R$ . Then since $R$ is also a Unique Factorization Domain we can write

Suppose $a$ is a nonzero element of the Principal Ideal Domain $R$ . Then since $R$ is also a Unique Factorization Domain we can write

where the ${p}_{i}$ are distinct primes in $R$ and $u$ is a unit. This factorization is unique up to units,so the ideals $\left( {p}_{i}^{{\alpha }_{i}}\right) ,i = 1,\ldots ,s$ are uniquely defined. For $i \neq j$ we have $\left( {p}_{i}^{{\alpha }_{i}}\right) + \left( {p}_{j}^{{\alpha }_{j}}\right) = R$ since the sum of these two ideals is generated by a greatest common divisor,which is 1 for distinct primes ${p}_{i},{p}_{j}$ . Put another way,the ideals $\left( {p}_{i}^{{\alpha }_{i}}\right) ,i = 1,\ldots ,s$ ,are comaximal in pairs. The intersection of all these ideals is the ideal (a) since $a$ is the least common multiple of ${p}_{1}^{{\alpha }_{1}},{p}_{2}^{{\alpha }_{2}},\ldots ,{p}_{s}^{{\alpha }_{s}}$ . Then the Chinese Remainder Theorem (Theorem 7.17) shows that

其中 ${p}_{i}$ 是 $R$ 中的不同素数，而 $u$ 是单位。这种分解在单位下是唯一的，因此理想 $\left( {p}_{i}^{{\alpha }_{i}}\right) ,i = 1,\ldots ,s$ 是唯一确定的。对于 $i \neq j$ ，我们有 $\left( {p}_{i}^{{\alpha }_{i}}\right) + \left( {p}_{j}^{{\alpha }_{j}}\right) = R$ ，因为这两个理想的和是由最大公约数生成的，而对于不同的素数 ${p}_{i},{p}_{j}$ ，这个最大公约数是 1。换句话说，理想 $\left( {p}_{i}^{{\alpha }_{i}}\right) ,i = 1,\ldots ,s$ 在一对中是共极大的。所有这些理想的交集是理想 (a)，因为 $a$ 是 ${p}_{1}^{{\alpha }_{1}},{p}_{2}^{{\alpha }_{2}},\ldots ,{p}_{s}^{{\alpha }_{s}}$ 的最小公倍数。然后中国剩余定理（定理 7.17）表明

as rings and also as $R$ -modules.

作为环，也作为 $R$ -模。

Applying this to the modules in Theorem 5 allows us to write each of the direct summands $R/\left( {a}_{i}\right)$ for the invariant factor ${a}_{i}$ of $M$ as a direct sum of cyclic modules whose annihilators are the prime power divisors of ${a}_{i}$ . This proves:

将其应用于定理 5 中的模，允许我们将每个直和项 $R/\left( {a}_{i}\right)$ 对于不变因子 ${a}_{i}$ 的 $M$ 写成循环模的直和，其消去子是 ${a}_{i}$ 的素数幂除数。这证明了：

Theorem 6. (Fundamental Theorem,Existence: Elementary Divisor Form) Let $R$ be a P.I.D. and let $M$ be a finitely generated $R$ -module. Then $M$ is the direct sum of a finite number of cyclic modules whose annihilators are either (0) or generated by powers of primes in $R$ ,i.e.,

定理 6。（基本定理，存在性：初等除数形式）设 $R$ 是一个 P.I.D.，并设 $M$ 是一个有限生成的 $R$ -模。那么 $M$ 是有限个循环模的直和，其消去子要么是 (0)，要么由 $R$ 中素数的幂生成，即，

where $r \geq 0$ is an integer and ${p}_{1}^{{\alpha }_{1}},\ldots ,{p}_{t}^{{\alpha }_{t}}$ are positive powers of (not necessarily distinct) primes in $R$ .

其中 $r \geq 0$ 是一个整数，${p}_{1}^{{\alpha }_{1}},\ldots ,{p}_{t}^{{\alpha }_{t}}$ 是（不一定不同）素数的正幂 $R$。

We proved Theorem 6 by using the prime power factors of the invariant factors for $M$ . In fact we shall see that the decomposition of $M$ into a direct sum of cyclic modules whose annihilators are (0) or prime powers as in Theorem 6 is unique, i.e., the integer $r$ and the ideals $\left( {p}_{1}^{{\alpha }_{1}}\right) ,\ldots ,\left( {p}_{t}^{{\alpha }_{t}}\right)$ are uniquely defined for $M$ . These prime powers are given a name:

我们通过使用不变因子的素数次幂证明了定理6 $M$ 。实际上我们将看到，$M$ 分解为循环模的直接和，其消元子为 (0) 或定理6中的素数次幂是唯一的，即整数 $r$ 和理想 $\left( {p}_{1}^{{\alpha }_{1}}\right) ,\ldots ,\left( {p}_{t}^{{\alpha }_{t}}\right)$ 对于 $M$ 是唯一确定的。这些素数次幂被赋予了一个名称：

Definition. Let $R$ be a P.I.D. and let $M$ be a finitely generated $R$ -module as in Theorem 6 . The prime powers ${p}_{1}^{{\alpha }_{1}},\ldots ,{p}_{t}^{{\alpha }_{t}}$ (defined up to multiplication by units in $R$ ) are called the elementary divisors of $M$ .

定义。设 $R$ 是一个P.I.D.，并且设 $M$ 是定理6中的有限生成 $R$ -模。素数次幂 ${p}_{1}^{{\alpha }_{1}},\ldots ,{p}_{t}^{{\alpha }_{t}}$（乘以 $R$ 中的单位元）被称为 $M$ 的基本除数。

Suppose $M$ is a finitely generated torsion module over the Principal Ideal Domain $R$ . If for the distinct primes ${p}_{1},{p}_{2},\ldots ,{p}_{n}$ occurring in the decomposition in Theorem 6 we group together all the cyclic factors corresponding to the same prime ${p}_{i}$ we see in particular that $M$ can be written as a direct sum

假设 $M$ 是主理想整环 $R$ 上的有限生成挠模。如果在定理6中的分解中出现的不同素数 ${p}_{1},{p}_{2},\ldots ,{p}_{n}$ 将所有对应于同一素数 ${p}_{i}$ 的循环因子分组在一起，我们特别看到 $M$ 可以写为直接和

where ${N}_{i}$ consists of all the elements of $M$ which are annihilated by some power of the prime ${p}_{i}$ . This result holds also for modules over $R$ which may not be finitely generated:

其中 ${N}_{i}$ 包含 $M$ 中所有被某个素数 ${p}_{i}$ 的幂消元的元素。这个结果也适用于可能不是有限生成的 $R$ 上的模：

Theorem 7. (The Primary Decomposition Theorem) Let $R$ be a P.I.D. and let $M$ be a $\frac{\text{nonzero torsion}}{R - \text{module (not necessarily finitely generated) with nonzero annihilator}}$ $a$ . Suppose the factorization of $a$ into distinct prime powers in $R$ is

定理7.（素数分解定理）设 $R$ 是一个P.I.D.，并且设 $M$ 是 $\frac{\text{nonzero torsion}}{R - \text{module (not necessarily finitely generated) with nonzero annihilator}}$ $a$ 。假设 $a$ 在 $R$ 中分解为不同素数次幂的因式分解是

and let ${N}_{i} = \left\{ {x \in M \mid {p}_{i}^{{\alpha }_{i}}x = 0}\right\} ,1 \leq i \leq n$ . Then ${N}_{i}$ is a submodule of $M$ with annihilator ${p}_{i}^{{\alpha }_{i}}$ and is the submodule of $M$ of all elements annihilated by some power of ${p}_{i}$ . We have

并且让 ${N}_{i} = \left\{ {x \in M \mid {p}_{i}^{{\alpha }_{i}}x = 0}\right\} ,1 \leq i \leq n$ 成立。那么 ${N}_{i}$ 是 $M$ 的一个子模，其消去子为 ${p}_{i}^{{\alpha }_{i}}$，并且是所有被 ${p}_{i}$ 的某次幂消去的元素构成的 $M$ 的子模。我们有

If $M$ is finitely generated then each ${N}_{i}$ is the direct sum of finitely many cyclic modules whose annihilators are divisors of ${p}_{i}^{{\alpha }_{i}}$ .

如果 $M$ 是有限生成的，那么每个 ${N}_{i}$ 是有限多个循环模的直接和，其消去子是 ${p}_{i}^{{\alpha }_{i}}$ 的除数。

Proof: We have already proved these results in the case where $M$ is finitely generated over $R$ . In the general case it is clear that ${N}_{i}$ is a submodule of $M$ with annihilator dividing ${p}_{i}^{{\alpha }_{i}}$ . Since $R$ is a P.I.D. the ideals $\left( {p}_{i}^{{\alpha }_{i}}\right)$ and $\left( {p}_{j}^{{\alpha }_{j}}\right)$ are comaximal for $i \neq j$ ,so the direct sum decomposition of $M$ can be proved easily by modifying the argument in the proof of the Chinese Remainder Theorem to apply it to modules. Using this direct sum decomposition it is easy to see that the annihilator of ${N}_{i}$ is precisely ${p}_{i}^{{\alpha }_{i}}$ .

证明：我们已经证明了当 $M$ 在 $R$ 上是有限生成的情况下的这些结果。在一般情况下，显然 ${N}_{i}$ 是 $M$ 的一个子模，其消去子整除 ${p}_{i}^{{\alpha }_{i}}$。由于 $R$ 是一个P.I.D.，理想 $\left( {p}_{i}^{{\alpha }_{i}}\right)$ 和 $\left( {p}_{j}^{{\alpha }_{j}}\right)$ 对于 $i \neq j$ 是极大理想的，因此可以通过修改中国剩余定理证明中的论据来容易地证明 $M$ 的直接和分解。使用这个直接和分解，可以容易地看出 ${N}_{i}$ 的消去子恰好是 ${p}_{i}^{{\alpha }_{i}}$。

Definition. The submodule ${N}_{i}$ in the previous theorem is called the ${p}_{i}$ -primary component of $M$ .

定义。在前面定理中的子模 ${N}_{i}$ 被称为 ${p}_{i}$ -主分量。

Notice that with this terminology the elementary divisors of a finitely generated module $M$ are just the invariant factors of the primary components of $\operatorname{Tor}\left( M\right)$ .

注意，使用这个术语，有限生成模 $M$ 的基本除数只是 $\operatorname{Tor}\left( M\right)$ 的主分量的不变因子。

We now prove the uniqueness statements regarding the decompositions in the Fundamental Theorem.

我们现在证明关于基本定理中分解的唯一性陈述。

Note that if $M$ is any module over a commutative ring $R$ and $a$ is an element of $R$ then ${aM} = \{ {am} \mid m \in M\}$ is a submodule of $M$ . Recall also that in a Principal Ideal Domain $R$ the nonzero prime ideals are maximal,hence the quotient of $R$ by a nonzero prime ideal is a field.

注意，如果 $M$ 是一个交换环 $R$ 上的模，并且 $a$ 是 $R$ 的一个元素，那么 ${aM} = \{ {am} \mid m \in M\}$ 是 $M$ 的一个子模。同时也要记住，在主理想整环 $R$ 中，非零素理想是极大理想，因此 $R$ 除以一个非零素理想的商是一个域。

Lemma 8. Let $R$ be a P.I.D. and let $p$ be a prime in $R$ . Let $F$ denote the field $R/\left( p\right)$ .

引理 8。设 $R$ 是一个主理想整环，$p$ 是 $R$ 中的一个素数。令 $F$ 表示域 $R/\left( p\right)$ 。

(1) Let $M = {R}^{r}$ . Then $M/{pM} \cong {F}^{r}$ .

(1) 设 $M = {R}^{r}$ 。那么 $M/{pM} \cong {F}^{r}$ 。

(2) Let $M = R/\left( a\right)$ where $a$ is a nonzero element of $R$ . Then

(2) 设 $M = R/\left( a\right)$ ，其中 $a$ 是 $R$ 中的一个非零元素。那么

(3) Let $M = R/\left( {a}_{1}\right) \oplus R/\left( {a}_{2}\right) \oplus \cdots \oplus R/\left( {a}_{k}\right)$ where each ${a}_{i}$ is divisible by $p$ . Then $M/{pM} \cong {F}^{k}$ .

(3) 设 $M = R/\left( {a}_{1}\right) \oplus R/\left( {a}_{2}\right) \oplus \cdots \oplus R/\left( {a}_{k}\right)$ ，其中每个 ${a}_{i}$ 都能被 $p$ 整除。那么 $M/{pM} \cong {F}^{k}$ 。

Proof: (1) There is a natural map from ${R}^{r}$ to ${\left( R/\left( p\right) \right) }^{r}$ defined by mapping $\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right)$ to $\left( {{\alpha }_{1}{\;\operatorname{mod}\;\left( p\right) },\ldots ,{\alpha }_{r}{\;\operatorname{mod}\;\left( p\right) }}\right)$ . This is clearly a surjective $R$ -module homomorphism with kernel consisting of the $r$ -tuples all of whose coordinates are divisible by $p$ ,i.e., $p{R}^{r}$ ,so ${R}^{r}/p{R}^{r} \cong {\left( R/\left( p\right) \right) }^{r}$ ,which is (1).

证明：(1) 存在一个从 ${R}^{r}$ 到 ${\left( R/\left( p\right) \right) }^{r}$ 的自然映射，定义为将 $\left( {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right)$ 映射到 $\left( {{\alpha }_{1}{\;\operatorname{mod}\;\left( p\right) },\ldots ,{\alpha }_{r}{\;\operatorname{mod}\;\left( p\right) }}\right)$ 。这显然是一个满射的 $R$ -模同态，其核由所有坐标都能被 $p$ 整除的 $r$ -元组组成，即 $p{R}^{r}$ ，因此 ${R}^{r}/p{R}^{r} \cong {\left( R/\left( p\right) \right) }^{r}$ ，这就是 (1)。

(2) This follows from the Isomorphism Theorems: note first that $p\left( {R/\left( a\right) }\right)$ is the image of the ideal(p)in the quotient $R/\left( a\right)$ ,hence is $\left( p\right) + \left( a\right) /\left( a\right)$ . The ideal $\left( p\right) + \left( a\right)$ is generated by a greatest common divisor of $p$ and $a$ ,hence is(p)if $p$ divides $a$ and is $R = \left( 1\right)$ otherwise. Hence ${pM} = \left( p\right) /\left( a\right)$ if $p$ divides $a$ and is $R/\left( a\right) = M$ otherwise. If $p$ divides $a$ then $M/{pM} = \left( {R/\left( a\right) }\right) /\left( {\left( p\right) /\left( a\right) }\right) \cong R/\left( p\right)$ ,and if $p$ does not divide $a$ then $M/{pM} = M/M = 0$ ,which proves (2).

(2) 这遵循同构定理：首先注意 $p\left( {R/\left( a\right) }\right)$ 是理想(p)在商 $R/\left( a\right)$ 中的像，因此是 $\left( p\right) + \left( a\right) /\left( a\right)$ 。理想 $\left( p\right) + \left( a\right)$ 由 $p$ 和 $a$ 的最大公约数生成，因此如果 $p$ 整除 $a$ ，则是 (p)，否则是 $R = \left( 1\right)$ 。因此如果 $p$ 整除 $a$ ，则是 ${pM} = \left( p\right) /\left( a\right)$ ，否则是 $R/\left( a\right) = M$ 。如果 $p$ 整除 $a$ ，那么 $M/{pM} = \left( {R/\left( a\right) }\right) /\left( {\left( p\right) /\left( a\right) }\right) \cong R/\left( p\right)$ ，如果 $p$ 不整除 $a$ ，那么 $M/{pM} = M/M = 0$ ，这证明了 (2)。

(3) This follows from (2) as in the proof of part (1) of Theorem 5.

(3) 这遵循 (2)，如同定理 5 第 (1) 部分证明中一样。

Theorem 9. (Fundamental Theorem,Uniqueness) Let $R$ be a P.I.D.

定理 9。（基本定理，唯一性）设 $R$ 是一个 P.I.D.。

(1) Two finitely generated $R$ -modules ${M}_{1}$ and ${M}_{2}$ are isomorphic if and only if they have the same free rank and the same list of invariant factors.

(1) 两个有限生成的 $R$ -模 ${M}_{1}$ 和 ${M}_{2}$ 是同构的，当且仅当它们具有相同的自由秩和相同的不变因子列表。

(2) Two finitely generated $R$ -modules ${M}_{1}$ and ${M}_{2}$ are isomorphic if and only if they have the same free rank and the same list of elementary divisors.

(2) 两个有限生成的 $R$ -模 ${M}_{1}$ 和 ${M}_{2}$ 是同构的，当且仅当它们具有相同的自由秩和相同的初等因子列表。

Proof: If ${M}_{1}$ and ${M}_{2}$ have the same free rank and list of invariant factors or the same free rank and list of elementary divisors then they are clearly isomorphic.

证明：如果 ${M}_{1}$ 和 ${M}_{2}$ 具有相同的自由秩和不变因子列表，或者具有相同的自由秩和初等因子列表，那么它们显然是同构的。

Suppose that ${M}_{1}$ and ${M}_{2}$ are isomorphic. Any isomorphism between ${M}_{1}$ and ${M}_{2}$ maps the torsion in ${M}_{1}$ to the torsion in ${M}_{2}$ so we must have $\operatorname{Tor}\left( {M}_{1}\right) \cong \operatorname{Tor}\left( {M}_{2}\right)$ . Then ${R}^{{r}_{1}} \cong {M}_{1}/\operatorname{Tor}\left( {M}_{1}\right) \cong {M}_{2}/\operatorname{Tor}\left( {M}_{2}\right) \cong {R}^{{r}_{2}}$ where ${r}_{1}$ is the free rank of ${M}_{1}$ and ${r}_{2}$ is the free rank of ${M}_{2}$ . Let $p$ be any nonzero prime in $R$ . Then from ${R}^{{r}_{1}} \cong {R}^{{r}_{2}}$ we obtain ${R}^{{r}_{1}}/p{R}^{{r}_{1}} \cong {R}^{{r}_{2}}/p{R}^{{r}_{2}}$ . By (1) of the previous lemma,this implies ${F}^{{r}_{1}} \cong {F}^{{r}_{2}}$ where $F$ is the field $R/{pR}$ . Hence we have an isomorphism of an ${r}_{1}$ -dimensional vector space over $F$ with an ${r}_{2}$ -dimensional vector space over $F$ ,so that ${r}_{1} = {r}_{2}$ and ${M}_{1}$ and ${M}_{2}$ have the same free rank.

假设 ${M}_{1}$ 和 ${M}_{2}$ 是同构的。任何 ${M}_{1}$ 和 ${M}_{2}$ 之间的同构都将 ${M}_{1}$ 中的扭元映射到 ${M}_{2}$ 中的扭元，因此我们必须有 $\operatorname{Tor}\left( {M}_{1}\right) \cong \operatorname{Tor}\left( {M}_{2}\right)$ 。那么 ${R}^{{r}_{1}} \cong {M}_{1}/\operatorname{Tor}\left( {M}_{1}\right) \cong {M}_{2}/\operatorname{Tor}\left( {M}_{2}\right) \cong {R}^{{r}_{2}}$ ，其中 ${r}_{1}$ 是 ${M}_{1}$ 的自由秩，${r}_{2}$ 是 ${M}_{2}$ 的自由秩。设 $p$ 是 $R$ 中的任意非零素数。那么从 ${R}^{{r}_{1}} \cong {R}^{{r}_{2}}$ 我们得到 ${R}^{{r}_{1}}/p{R}^{{r}_{1}} \cong {R}^{{r}_{2}}/p{R}^{{r}_{2}}$ 。根据前一个引理的 (1) ，这暗示了 ${F}^{{r}_{1}} \cong {F}^{{r}_{2}}$ ，其中 $F$ 是域 $R/{pR}$ 。因此我们有一个 ${r}_{1}$ 维向量空间在 $F$ 上的同构与一个 ${r}_{2}$ 维向量空间在 $F$ 上的同构，从而 ${r}_{1} = {r}_{2}$ 以及 ${M}_{1}$ 和 ${M}_{2}$ 具有相同的自由秩。

We are reduced to showing that ${M}_{1}$ and ${M}_{2}$ have the same lists of invariant factors and elementary divisors. To do this we need only work with the isomorphic torsion modules $\operatorname{Tor}\left( {M}_{1}\right)$ and $\operatorname{Tor}\left( {M}_{2}\right)$ ,i.e.,we may as well assume that both ${M}_{1}$ and ${M}_{2}$ are torsion $R$ -modules.

我们需要证明 ${M}_{1}$ 和 ${M}_{2}$ 具有相同的不变因子和 elementary divisors 的列表。为此，我们只需要处理同构的扭元模块 $\operatorname{Tor}\left( {M}_{1}\right)$ 和 $\operatorname{Tor}\left( {M}_{2}\right)$ ，即，我们可以假设 ${M}_{1}$ 和 ${M}_{2}$ 都是扭元 $R$ -模块。

We first show they have the same elementary divisors. It suffices to show that for any fixed prime $p$ the elementary divisors which are a power of $p$ are the same for both ${M}_{1}$ and ${M}_{2}$ . If ${M}_{1} \cong {M}_{2}$ then the $p$ -primary submodule of ${M}_{1}$ ( = the direct sum of the cyclic factors whose elementary divisors are powers of $p$ ) is isomorphic to the $p$ -primary submodule of ${M}_{2}$ ,since these are the submodules of elements which are annihilated by some power of $p$ . We are therefore reduced to the case of proving that if two modules ${M}_{1}$ and ${M}_{2}$ which have annihilator a power of $p$ are isomorphic then they have the same elementary divisors.

我们首先证明它们有相同的初等因子。只需证明对于任意固定的素数 $p$，${M}_{1}$ 和 ${M}_{2}$ 的初等因子中是 $p$ 的幂次的因子是相同的。如果 ${M}_{1} \cong {M}_{2}$ ，那么 ${M}_{1}$ 的 $p$ -主子模（即初等因子为 $p$ 的幂次的循环因子的直接和）与 ${M}_{2}$ 的 $p$ -主子模同构，因为这些是能被 $p$ 的某个幂次消去的元素的子模。因此我们只需证明如果两个模 ${M}_{1}$ 和 ${M}_{2}$ 的消去子是 $p$ 的某个幂次，且它们同构，那么它们有相同的初等因子。

We proceed by induction on the power of $p$ in the annihilator of ${M}_{1}$ (which is the same as the annihilator of ${M}_{2}$ since ${M}_{1}$ and ${M}_{2}$ are isomorphic). If this power is 0, then both ${M}_{1}$ and ${M}_{2}$ are 0 and we are done. Otherwise ${M}_{1}$ (and ${M}_{2}$ ) have nontrivial elementary divisors. Suppose the elementary divisors of ${M}_{1}$ are given by

我们通过对 ${M}_{1}$ 的消去子中的 $p$ 的幂次进行归纳来继续。如果这个幂次是 0，那么 ${M}_{1}$ 和 ${M}_{2}$ 都是 0，我们就完成了证明。否则 ${M}_{1}$（和 ${M}_{2}$）有非平凡的初等因子。假设 ${M}_{1}$ 的初等因子由以下给出

where $2 \leq {\alpha }_{1} \leq {\alpha }_{2} \leq \cdots \leq {\alpha }_{s}$ ,i.e., ${M}_{1}$ is the direct sum of cyclic modules with generators ${x}_{1},{x}_{2},\ldots ,{x}_{m},{x}_{m + 1},\ldots ,{x}_{m + s}$ ,say,whose annihilators are $\left( p\right) ,\left( p\right) ,\ldots ,\left( p\right)$ , $\left( {p}^{{\alpha }_{1}}\right) ,\ldots ,\left( {p}^{{\alpha }_{s}}\right)$ ,respectively. Then the submodule $p{M}_{1}$ has elementary divisors

其中 $2 \leq {\alpha }_{1} \leq {\alpha }_{2} \leq \cdots \leq {\alpha }_{s}$ ，即 ${M}_{1}$ 是生成元为 ${x}_{1},{x}_{2},\ldots ,{x}_{m},{x}_{m + 1},\ldots ,{x}_{m + s}$ 的循环模的直接和，其消去子分别是 $\left( p\right) ,\left( p\right) ,\ldots ,\left( p\right)$ ，$\left( {p}^{{\alpha }_{1}}\right) ,\ldots ,\left( {p}^{{\alpha }_{s}}\right)$。那么子模 $p{M}_{1}$ 的初等因子是

since $p{M}_{1}$ is the direct sum of the cyclic modules with generators $p{x}_{1},p{x}_{2},\ldots ,p{x}_{m}$ , $p{x}_{m + 1},\ldots ,p{x}_{m + s}$ whose annihilators are $\left( 1\right) ,\left( 1\right) ,\ldots ,\left( 1\right) ,\left( {p}^{{\alpha }_{1} - 1}\right) ,\ldots ,\left( {p}^{{\alpha }_{s} - 1}\right)$ ,respectively. Similarly,if the elementary divisors of ${M}_{2}$ are given by

因为 $p{M}_{1}$ 是生成元分别为 $p{x}_{1},p{x}_{2},\ldots ,p{x}_{m}$ ，$p{x}_{m + 1},\ldots ,p{x}_{m + s}$ 的循环模的直接和，其消去子分别是 $\left( 1\right) ,\left( 1\right) ,\ldots ,\left( 1\right) ,\left( {p}^{{\alpha }_{1} - 1}\right) ,\ldots ,\left( {p}^{{\alpha }_{s} - 1}\right)$。类似地，如果 ${M}_{2}$ 的初等因子由以下给出

where $2 \leq {\beta }_{1} \leq {\beta }_{2} \leq \cdots \leq {\beta }_{t}$ ,then $p{M}_{2}$ has elementary divisors

其中 $2 \leq {\beta }_{1} \leq {\beta }_{2} \leq \cdots \leq {\beta }_{t}$ ，那么 $p{M}_{2}$ 具有初等因子

Since ${M}_{1} \cong {M}_{2}$ ,also $p{M}_{1} \cong p{M}_{2}$ and the power of $p$ in the annihilator of $p{M}_{1}$ is one less than the power of $p$ in the annihilator of ${M}_{1}$ . By induction,the elementary divisors for $p{M}_{1}$ are the same as the elementary divisors for $p{M}_{2}$ ,i.e., $s = t$ and ${\alpha }_{i} - 1 = {\beta }_{i} - 1$ for $i = 1,2,\ldots ,s$ ,hence ${\alpha }_{i} = {\beta }_{i}$ for $i = 1,2,\ldots ,s$ . Finally,since also ${M}_{1}/p{M}_{1} \cong {M}_{2}/p{M}_{2}$ we see from (3) of the lemma above that ${F}^{m + s} \cong {F}^{n + t}$ , which shows that $m + s = n + t$ hence $m = n$ since we have already seen $s = t$ . This proves that the set of elementary divisors for ${M}_{1}$ is the same as the set of elementary divisors for ${M}_{2}$ .

由于 ${M}_{1} \cong {M}_{2}$ ，同样 $p{M}_{1} \cong p{M}_{2}$ ，且 $p$ 在 $p{M}_{1}$ 的消灭子中的幂次比在 ${M}_{1}$ 的消灭子中的幂次少一。通过归纳，$p{M}_{1}$ 的初等因子与 $p{M}_{2}$ 的初等因子相同，即 $s = t$ 和 ${\alpha }_{i} - 1 = {\beta }_{i} - 1$ 对于 $i = 1,2,\ldots ,s$ ，因此 ${\alpha }_{i} = {\beta }_{i}$ 对于 $i = 1,2,\ldots ,s$ 。最后，由于同样 ${M}_{1}/p{M}_{1} \cong {M}_{2}/p{M}_{2}$ ，我们从上述引理的 (3) 中看到 ${F}^{m + s} \cong {F}^{n + t}$ ，这表明 $m + s = n + t$ ，因此 $m = n$ ，因为我们已经看到 $s = t$ 。这证明了 ${M}_{1}$ 的初等因子集合与 ${M}_{2}$ 的初等因子集合相同。

We now show that ${M}_{1}$ and ${M}_{2}$ must have the same invariant factors. Suppose ${a}_{1}\left| {a}_{2}\right| \cdots \mid {a}_{m}$ are invariant factors for ${M}_{1}$ . We obtain a set of elementary divisors for ${M}_{1}$ by taking the prime power factors of these elements. Note that then the divisibility relations on the invariant factors imply that ${a}_{m}$ is the product of the largest of the prime powers among these elementary divisors, ${a}_{m - 1}$ is the product of the largest prime powers among these elementary divisors once the factors for ${a}_{m}$ have been removed,and so on. If ${b}_{1}\left| {b}_{2}\right| \cdots \mid {b}_{n}$ are invariant factors for ${M}_{2}$ then we similarly obtain a set of elementary divisors for ${M}_{2}$ by taking the prime power factors of these elements. But we showed above that the elementary divisors for ${M}_{1}$ and ${M}_{2}$ are the same,and it follows that the same is true of the invariant factors.

我们现在证明 ${M}_{1}$ 和 ${M}_{2}$ 必须具有相同的不变因子。假设 ${a}_{1}\left| {a}_{2}\right| \cdots \mid {a}_{m}$ 是 ${M}_{1}$ 的不变因子。我们通过对这些元素的素数次幂取因子来得到 ${M}_{1}$ 的一组初等因子。注意，不变因子上的可整除关系意味着 ${a}_{m}$ 是这些初等因子中最大的素数次幂的乘积，${a}_{m - 1}$ 是在移除 ${a}_{m}$ 的因子后这些初等因子中最大的素数次幂的乘积，依此类推。如果 ${b}_{1}\left| {b}_{2}\right| \cdots \mid {b}_{n}$ 是 ${M}_{2}$ 的不变因子，那么我们同样通过对这些元素的素数次幂取因子来得到 ${M}_{2}$ 的一组初等因子。但我们已经证明 ${M}_{1}$ 和 ${M}_{2}$ 的初等因子是相同的，因此不变因子也是如此。

Corollary 10. Let $R$ be a P.I.D. and let $M$ be a finitely generated $R$ -module.

推论 10. 设 $R$ 是一个 P.I.D.，并且 $M$ 是一个有限生成的 $R$ -模。

(1) The elementary divisors of $M$ are the prime power factors of the invariant factors of $M$ .

(1) $M$ 的初等因子是 $M$ 不变因子的素数次幂因子。

(2) The largest invariant factor of $M$ is the product of the largest of the distinct prime powers among the elementary divisors of $M$ ,the next largest invariant factor is the product of the largest of the distinct prime powers among the remaining elementary divisors of $M$ ,and so on.

(2) $M$ 的最大不变因子是 $M$ 的初等因子中不同的素数次幂中最大的乘积，次大的不变因子是 $M$ 的剩余初等因子中不同的素数次幂中最大的乘积，依此类推。

Proof: The procedure in (1) gives $a$ set of elementary divisors and since the elementary divisors for $M$ are unique by the theorem,it follows that the procedure in (1) gives the set of elementary divisors. Similarly for (2).

证明：过程 (1) 给出了 $a$ 的一组初等因子，由于根据定理 $M$ 的初等因子是唯一的，因此过程 (1) 给出了初等因子的集合。对于 (2) 也是类似的情况。

Corollary 11. (The Fundamental Theorem of Finitely Generated Abelian Groups) See Theorem 5.3 and Theorem 5.5.

推论 11.（有限生成阿贝尔群的基本定理）参见定理 5.3 和定理 5.5。

Proof: Take $R = \mathbb{Z}$ in Theorems 5,6 and 9 (note however that the invariant factors are listed in reverse order in Chapter 5 for computational convenience).

证明：取定理 5、6 和 9 中的 $R = \mathbb{Z}$（然而注意，在第五章中为了计算方便，不变因子的列出顺序是相反的）。

The procedure for passing between elementary divisors and invariant factors in Corollary 10 is described in some detail in Chapter 5 in the case of finitely generated abelian groups.

推论 10 中从初等因子到不变因子的转换过程在第五章中对于有限生成阿贝尔群的情况有详细的描述。

Note also that if a finitely generated module $M$ is written as a direct sum of cyclic modules of the form $R/\left( a\right)$ then the ideals (a) which occur are not in general unique unless some additional conditions are imposed (such as the divisibility condition for the invariant factors or the condition that $a$ be the power of a prime in the case of the elementary divisors). To decide whether two modules are isomorphic it is necessary to first write them in such a standard (or canonical) form.

注意，如果一个有限生成模块 $M$ 可以写成形式为 $R/\left( a\right)$ 的循环模块的直接和，那么出现的理想 (a) 通常不是唯一的，除非附加某些条件（例如，对于不变因子的可除性条件，或者在素数幂的情况下 $a$ 的条件）。为了判断两个模块是否同构，首先需要将它们写成这样的标准（或典型）形式。

EXERCISES

练习

1. Let $M$ be a module over the integral domain $R$ .

(a) Suppose $x$ is a nonzero torsion element in $M$ . Show that $x$ and 0 are "linearly dependent." Conclude that the rank of $\operatorname{Tor}\left( M\right)$ is 0,so that in particular any torsion $R$ -module has rank 0 .

(b) Show that the rank of $M$ is the same as the rank of the (torsion free) quotient $M/\operatorname{Tor}M$ .

2. Let $M$ be a module over the integral domain $R$ .

(a) Suppose that $M$ has rank $n$ and that ${x}_{1},{x}_{2},\ldots ,{x}_{n}$ is any maximal set of linearly independent elements of $M$ . Let $N = R{x}_{1} + \ldots + R{x}_{n}$ be the submodule generated by ${x}_{1},{x}_{2},\ldots ,{x}_{n}$ . Prove that $N$ is isomorphic to ${R}^{n}$ and that the quotient $M/N$ is a torsion $R$ -module (equivalently,the elements ${x}_{1},\ldots ,{x}_{n}$ are linearly independent and for any $y \in M$ there is a nonzero element $r \in R$ such that ${ry}$ can be written as a linear combination ${r}_{1}{x}_{1} + \ldots + {r}_{n}{x}_{n}$ of the ${x}_{i}$ ).

(b) Prove conversely that if $M$ contains a submodule $N$ that is free of rank $n$ (i.e., $N \cong$ ${R}^{n}$ ) such that the quotient $M/N$ is a torsion $R$ -module then $M$ has rank $n$ . [Let ${y}_{1},{y}_{2},\ldots ,{y}_{n + 1}$ be any $n + 1$ elements of $M$ . Use the fact that $M/N$ is torsion to write ${r}_{i}{y}_{i}$ as a linear combination of a basis for $N$ for some nonzero elements ${r}_{1},\ldots ,{r}_{n + 1}$ of $R$ . Use an argument as in the proof of Proposition 3 to see that the ${r}_{i}{y}_{i}$ ,and hence also the ${y}_{i}$ ,are linearly dependent.]

3. Let $R$ be an integral domain and let $A$ and $B$ be $R$ -modules of ranks $m$ and $n$ ,respectively. Prove that the rank of $A \oplus B$ is $m + n$ . [Use the previous exercise.]

4. 设 $R$ 为一个整环，并且设 $A$ 和 $B$ 是 $R$ -秩分别为 $m$ 和 $n$ 的模块。证明 $A \oplus B$ 的秩是 $m + n$ 。[使用前一个练习。]

5. Let $R$ be an integral domain,let $M$ be an $R$ -module and let $N$ be a submodule of $M$ . Suppose $M$ has rank $n,N$ has rank $r$ and the quotient $M/N$ has rank $s$ . Prove that $n = r + s$ . [Let ${x}_{1},{x}_{2},\ldots ,{x}_{s}$ be elements of $M$ whose images in $M/N$ are a maximal set of independent elements and let ${x}_{s + 1},{x}_{s + 2},\ldots ,{x}_{s + r}$ be a maximal set of independent elements in $N$ . Prove that ${x}_{1},{x}_{2},\ldots ,{x}_{s + r}$ are linearly independent in $M$ and that for any element $y \in M$ there is a nonzero element $r \in R$ such that ${ry}$ is a linear combination of these elements. Then use Exercise 2.]

6. 设 $R$ 为一个整环，$M$ 为一个 $R$ -模块，并且 $N$ 是 $M$ 的一个子模块。假设 $M$ 的秩为 $n,N$，$N$ 的秩为 $r$，并且商 $M/N$ 的秩为 $s$。证明 $n = r + s$。[设 ${x}_{1},{x}_{2},\ldots ,{x}_{s}$ 是 $M$ 中的元素，它们在 $M/N$ 中的像是独立元素的最大集合，并且设 ${x}_{s + 1},{x}_{s + 2},\ldots ,{x}_{s + r}$ 是 $N$ 中独立元素的最大集合。证明 ${x}_{1},{x}_{2},\ldots ,{x}_{s + r}$ 在 $M$ 中是线性无关的，并且对于任何元素 $y \in M$，存在一个非零元素 $r \in R$，使得 ${ry}$ 是这些元素的线性组合。然后使用练习2。]

7. Let $R = \mathbb{Z}\left\lbrack x\right\rbrack$ and let $M = \left( {2,x}\right)$ be the ideal generated by 2 and $x$ ,considered as a submodule of $R$ . Show that $\{ 2,x\}$ is not a basis of $M$ . [Find a nontrivial $R$ -linear dependence between these two elements.] Show that the rank of $M$ is 1 but that $M$ is not free of rank 1 (cf. Exercise 2).

8. 设 $R = \mathbb{Z}\left\lbrack x\right\rbrack$ 并且设 $M = \left( {2,x}\right)$ 是由 2 和 $x$ 生成的理想，将其视为 $R$ 的子模。证明 $\{ 2,x\}$ 不是 $M$ 的基。[在两个元素之间找到一个非平凡的 $R$ -线性相关关系。] 证明 $M$ 的秩为 1 但 $M$ 不是秩为 1 的自由模（参见练习 2）。

9. Show that if $R$ is an integral domain and $M$ is any nonprincipal ideal of $R$ then $M$ is torsion free of rank 1 but is not a free $R$ -module.

10. 证明如果 $R$ 是一个整环且 $M$ 是 $R$ 的任意非主理想，那么 $M$ 是秩为 1 的无扭模但不是自由的 $R$ -模。

11. Let $R$ be any ring,let ${A}_{1},{A}_{2},\ldots ,{A}_{m}$ be $R$ -modules and let ${B}_{i}$ be a submodule of ${A}_{i}$ , $1 \leq i \leq m$ . Prove that

12. 设 $R$ 是任意环，${A}_{1},{A}_{2},\ldots ,{A}_{m}$ 是 $R$ -模，${B}_{i}$ 是 ${A}_{i}$ 的子模，$1 \leq i \leq m$ 。证明

8. Let $R$ be a P.I.D.,let $B$ be a torsion $R$ -module and let $p$ be a prime in $R$ . Prove that if ${pb} = 0$ for some nonzero $b \in B$ ,then $\operatorname{Ann}\left( B\right) \subseteq \left( p\right)$ .

9. 设 $R$ 是一个 P.I.D.，$B$ 是一个扭 $R$ -模，$p$ 是 $R$ 中的一个素数。证明如果 ${pb} = 0$ 对于某个非零 $b \in B$ 成立，那么 $\operatorname{Ann}\left( B\right) \subseteq \left( p\right)$ 。

10. Give an example of an integral domain $R$ and a nonzero torsion $R$ -module $M$ such that $\operatorname{Ann}\left( M\right) = 0$ . Prove that if $N$ is a finitely generated torsion $R$ -module then $\operatorname{Ann}\left( N\right) \neq 0$ .

11. 举一个整环 $R$ 和一个非零扭 $R$ -模 $M$ 的例子，使得 $\operatorname{Ann}\left( M\right) = 0$ 。证明如果 $N$ 是一个有限生成的扭 $R$ -模，那么 $\operatorname{Ann}\left( N\right) \neq 0$ 。

12. For $p$ a prime in the P.I.D. $R$ and $N$ an $R$ -module prove that the $p$ -primary component of $N$ is a submodule of $N$ and prove that $N$ is the direct sum of its $p$ -primary components (there need not be finitely many of them).

13. 对于 $p$ 是 P.I.D. $R$ 中的素数以及 $N$ 是一个 $R$ -模，证明 $p$ -主成分是 $N$ 的子模，并证明 $N$ 是其 $p$ -主成分的直和（这些成分不必是有限的）。

14. Let $R$ be a P.I.D.,let $a$ be a nonzero element of $R$ and let $M = R/\left( a\right)$ . For any prime $p$ of $R$ prove that

15. 设 $R$ 是一个 P.I.D.，$a$ 是 $R$ 中的非零元素，且 $M = R/\left( a\right)$ 。对于 $R$ 的任意素数 $p$ 证明

where $n$ is the power of $p$ dividing $a$ in $R$ .

其中 $n$ 是 $p$ 在 $R$ 中除 $a$ 的幂。

12. Let $R$ be a P.I.D. and let $p$ be a prime in $R$ .

13. 设 $R$ 是一个 P.I.D.，$p$ 是 $R$ 中的一个素数。

(a) Let $M$ be a finitely generated torsion $R$ -module. Use the previous exercise to prove that ${p}^{k - 1}M/{p}^{k}M \cong {F}^{{n}_{k}}$ where $F$ is the field $R/\left( p\right)$ and ${n}_{k}$ is the number of elementary divisors of $M$ which are powers ${p}^{\alpha }$ with $\alpha \geq k$ .

(a) 设 $M$ 是一个有限生成的扭 $R$ -模。使用前一个练习证明 ${p}^{k - 1}M/{p}^{k}M \cong {F}^{{n}_{k}}$ ，其中 $F$ 是域 $R/\left( p\right)$ ，${n}_{k}$ 是 $M$ 中是 ${p}^{\alpha }$ 的幂的初等除子的个数 $\alpha \geq k$ 。

(b) Suppose ${M}_{1}$ and ${M}_{2}$ are isomorphic finitely generated torsion $R$ -modules. Use (a) to prove that,for every $k \geq 0,{M}_{1}$ and ${M}_{2}$ have the same number of elementary divisors ${p}^{\alpha }$ with $\alpha \geq k$ . Prove that this implies ${M}_{1}$ and ${M}_{2}$ have the same set of elementary divisors.

(b) 假设 ${M}_{1}$ 和 ${M}_{2}$ 是同构的有限生成扭 $R$ -模。使用 (a) 证明，对于每一个 $k \geq 0,{M}_{1}$ ，${M}_{2}$ 和 ${M}_{1}$ 有相同数量的 ${p}^{\alpha }$ 的初等除子 $\alpha \geq k$ 。证明这意味着 ${M}_{1}$ 和 ${M}_{2}$ 有相同的初等除子集合。

13. If $M$ is a finitely generated module over the P.I.D. $R$ ,describe the structure of $M/\operatorname{Tor}\left( M\right)$ .

14. 如果 $M$ 是 P.I.D. $R$ 上的有限生成模，描述 $M/\operatorname{Tor}\left( M\right)$ 的结构。

15. Let $R$ be a P.I.D. and let $M$ be a torsion $R$ -module. Prove that $M$ is irreducible (cf. Exercises 9 to 11 of Section 10.3) if and only if $M = {Rm}$ for any nonzero element $m \in M$ where the annihilator of $m$ is a nonzero prime ideal(p).

16. 设 $R$ 是一个 P.I.D.，并且设 $M$ 是一个扭转 $R$ -模。证明 $M$ 是不可约的（参见第10.3节的练习9至11）当且仅当 $M = {Rm}$ 对于任何非零元素 $m \in M$，其中 $m \in M$ 的消去子是 $R$ 的一个非零素理想(p)。

17. Prove that if $R$ is a Noetherian ring then ${R}^{n}$ is a Noetherian $R$ -module. [Fix a basis of ${R}^{n}$ . If $M$ is a submodule of ${R}^{n}$ show that the collection of first coordinates of elements of $M$ is a submodule of $R$ hence is finitely generated. Let ${m}_{1},{m}_{2},\ldots ,{m}_{k}$ be elements of $M$ whose first coordinates generate this submodule of $R$ . Show that any element of $M$ can be written as an $R$ -linear combination of ${m}_{1},{m}_{2},\ldots ,{m}_{k}$ plus an element of $M$ whose first coordinate is 0 . Prove that $M \cap {R}^{n - 1}$ is a submodule of ${R}^{n - 1}$ where ${R}^{n - 1}$ is the set of elements of ${R}^{n}$ with first coordinate 0 and then use induction on $n$ .

18. 证明如果 $R$ 是一个诺特环，那么 ${R}^{n}$ 是一个诺特 $R$ -模。[固定 ${R}^{n}$ 的一个基。如果 $M$ 是 ${R}^{n}$ 的一个子模，证明 $M$ 中元素的第一个坐标的集合是 $R$ 的一个子模，因此是有限生成的。设 ${m}_{1},{m}_{2},\ldots ,{m}_{k}$ 是 $M$ 的元素，其第一个坐标生成 $R$ 的这个子模。证明 $M$ 的任何元素都可以写成 ${m}_{1},{m}_{2},\ldots ,{m}_{k}$ 的 $R$ -线性组合加上一个第一个坐标为0的 $M$ 中的元素。证明 $M \cap {R}^{n - 1}$ 是 ${R}^{n - 1}$ 的子模，其中 ${R}^{n - 1}$ 是 ${R}^{n}$ 中第一个坐标为0的元素的集合，然后对 $n$ 进行归纳。]

The following set of exercises outlines a proof of Theorem 5 in the special case where $R$ is a Euclidean Domain using a matrix argument involving row and column operations. This applies in particular to the cases $R = \mathbb{Z}$ and $R = F\left\lbrack x\right\rbrack$ of interest in the applications and is computationally useful.

以下练习集概述了在 $R$ 是欧几里得域的特殊情况下，使用涉及行和列操作的矩阵论证来证明定理5的证明过程。这特别适用于应用中感兴趣的 $R = \mathbb{Z}$ 和 $R = F\left\lbrack x\right\rbrack$ 的情况，并且在计算上是有效的。

Let $R$ be a Euclidean Domain and let $M$ be an $R$ -module.

设 $R$ 是一个欧几里得域，并且设 $M$ 是一个 $R$ -模。

16. Prove that $M$ is finitely generated if and only if there is a surjective $R$ -homomorphism $\varphi : {R}^{n} \rightarrow M$ for some integer $n$ (this is true for any ring $R$ ).

17. 证明 $M$ 是有限生成的当且仅当存在一个满射 $R$ -同态 $\varphi : {R}^{n} \rightarrow M$ 对于某个整数 $n$（这对于任何环 $R$ 都成立）。

Suppose $\varphi : {R}^{n} \rightarrow M$ is a surjective $R$ -module homomorphism. By Exercise 15, $\ker \varphi$ is finitely generated. If ${x}_{1},{x}_{2},\ldots ,{x}_{n}$ is a basis for ${R}^{n}$ and ${y}_{1},\ldots ,{y}_{m}$ are generators for $\ker \varphi$ we have

假设 $\varphi : {R}^{n} \rightarrow M$ 是一个满射 $R$ -模同态。由练习15，$\ker \varphi$ 是有限生成的。如果 ${x}_{1},{x}_{2},\ldots ,{x}_{n}$ 是 ${R}^{n}$ 的一个基，${y}_{1},\ldots ,{y}_{m}$ 是 $\ker \varphi$ 的生成元，那么我们有

with coefficients ${a}_{ij} \in R$ . It follows that the homomorphism $\varphi$ (hence the module structure of $M)$ is determined by the choice of generators for ${R}^{n}$ and the matrix $A = \left( {a}_{ij}\right)$ . Such a matrix $A$ will be called a relations matrix.

系数为 ${a}_{ij} \in R$ 。因此，同态 $\varphi$ （因此 $M)$ 的模结构由 ${R}^{n}$ 的生成元的选择和矩阵 $A = \left( {a}_{ij}\right)$ 决定）。这样的矩阵 $A$ 将被称为关系矩阵。

17. (a) Show that interchanging ${x}_{i}$ and ${x}_{j}$ in the basis for ${R}^{n}$ interchanges the ${i}^{\text{th }}$ column with the ${j}^{\text{th }}$ column in the corresponding relations matrix.

18. (a) 证明在 ${R}^{n}$ 的基中交换 ${x}_{i}$ 和 ${x}_{j}$ 会交换对应关系矩阵中的 ${i}^{\text{th }}$ 列与 ${j}^{\text{th }}$ 列。

(b) Show that,for any $a \in R$ ,replacing the element ${x}_{j}$ by ${x}_{j} - a{x}_{i}$ in the basis for ${R}^{n}$ gives another basis for ${R}^{n}$ and that the corresponding relations matrix for this basis is the same as the original relations matrix except that $a$ times the ${j}^{\text{th }}$ column has been added to the ${i}^{\text{th }}$ column. [Note that $\cdots + {a}_{i}{x}_{i} + \cdots + {a}_{j}{x}_{j} + \cdots = \cdots + \left( {{a}_{i} + }\right.$ $\left. {a{a}_{j}){x}_{i} + \cdots + {a}_{j}\left( {{x}_{j} - a{x}_{i}}\right) + \ldots .}\right\rbrack$

(b) 证明，对于任何 $a \in R$ ，在 ${R}^{n}$ 的基中用 ${x}_{j} - a{x}_{i}$ 替换元素 ${x}_{j}$ 给出另一个基，并且这个基的对应关系矩阵与原关系矩阵相同，除了 $a$ 乘以 ${j}^{\text{th }}$ 列被加到了 ${i}^{\text{th }}$ 列上。 [注意 $\cdots + {a}_{i}{x}_{i} + \cdots + {a}_{j}{x}_{j} + \cdots = \cdots + \left( {{a}_{i} + }\right.$ $\left. {a{a}_{j}){x}_{i} + \cdots + {a}_{j}\left( {{x}_{j} - a{x}_{i}}\right) + \ldots .}\right\rbrack$]

18. (a) Show that interchanging the generators ${y}_{i}$ and ${y}_{j}$ interchanges the ${i}^{\text{th }}$ row with the ${j}^{\text{th }}$ row in the relations matrix.

19. (a) 证明交换生成元 ${y}_{i}$ 和 ${y}_{j}$ 会交换关系矩阵中的 ${i}^{\text{th }}$ 行与 ${j}^{\text{th }}$ 行。

(b) Show that,for any $a \in R$ ,replacing the element ${y}_{j}$ by ${y}_{j} - a{y}_{i}$ gives another set of generators for $\ker \varphi$ and that the corresponding relations matrix for this choice of generators is the same as the original relations matrix except that $- a$ times the ${i}^{\text{th }}$ row has been added to the ${j}^{\text{th }}$ row.

(b) 证明，对于任意的 $a \in R$ ，将元素 ${y}_{j}$ 替换为 ${y}_{j} - a{y}_{i}$ 会得到另一组 $\ker \varphi$ 的生成元，并且对于这个生成元选择的相应关系矩阵与原始关系矩阵相同，除了 $- a$ 乘以 ${i}^{\text{th }}$ 行被加到了 ${j}^{\text{th }}$ 行。

19. By the previous two exercises we may perform elementary row and column operations on a given relations matrix by choosing different generators for ${R}^{n}$ and $\ker \varphi$ . If all relation matrices are the zero matrix then $\ker \varphi = 0$ and $M \cong {R}^{n}$ . Otherwise let ${a}_{1}$ be the (nonzero) g.c.d. (recall $R$ is a Euclidean Domain) of all the entries in a fixed initial relations matrix for $M$ .

20. 根据前两个练习，我们可以通过为 ${R}^{n}$ 和 $\ker \varphi$ 选择不同的生成元来对给定关系矩阵执行行和列的基本操作。如果所有关系矩阵都是零矩阵，那么 $\ker \varphi = 0$ 和 $M \cong {R}^{n}$ 。否则，设 ${a}_{1}$ 为所有条目（非零）的最大公约数（回顾 $R$ 是一个欧几里得域）在一个固定的初始关系矩阵的 $M$ 中。

(a) Prove that by elementary row and column operations we may assume ${a}_{1}$ occurs in a relations matrix of the form

(a) 证明通过行和列的基本操作，我们可以假设 ${a}_{1}$ 出现在形式为的关系矩阵中

where ${a}_{1}$ divides ${a}_{ij},i = 1,2,\ldots ,m,j = 1,2,\ldots ,n$ .

其中 ${a}_{1}$ 整除 ${a}_{ij},i = 1,2,\ldots ,m,j = 1,2,\ldots ,n$ 。

(b) Prove that there is a relations matrix of the form

(b) 证明存在形式为的关系矩阵

where ${a}_{1}$ divides all the entries.

其中 ${a}_{1}$ 整除所有条目。

(c) Let ${a}_{2}$ be a g.c.d. of all the entries except the element ${a}_{1}$ in the relations matrix in (b). Prove that there is a relations matrix of the form

(c) 设 ${a}_{2}$ 为除了元素 ${a}_{1}$ 之外所有条目的最大公约数，在 (b) 中的关系矩阵中。证明存在形式为的关系矩阵

where ${a}_{1}$ divides ${a}_{2}$ and ${a}_{2}$ divides all the other entries of the matrix.

其中 ${a}_{1}$ 整除 ${a}_{2}$ 且 ${a}_{2}$ 整除矩阵中的其他所有条目。

(d) Prove that there is a relations matrix of the form $\left( \begin{matrix} D & 0 \\ 0 & 0 \end{matrix}\right)$ where $D$ is a diagonal matrix with nonzero entries ${a}_{1},{a}_{2},\ldots ,{a}_{k},k \leq n$ ,satisfying

(d) 证明存在形式为 $\left( \begin{matrix} D & 0 \\ 0 & 0 \end{matrix}\right)$ 的关系矩阵，其中 $D$ 是一个对角矩阵，其非零条目为 ${a}_{1},{a}_{2},\ldots ,{a}_{k},k \leq n$ ，满足

Conclude that

结论是

If $n$ is not the minimal number of generators required for $M$ then some of the initial elements ${a}_{1},{a}_{2},\ldots$ above will be units,so the corresponding direct summands above will be 0 . If we remove these irrelevant factors we have produced the invariant factors of the module $M$ . Further,the image of the new generators for ${R}^{n}$ corresponding to the direct summands above will then be a set of $R$ -generators for the cyclic submodules of $M$ in its invariant factor decomposition (note that the image in $M$ of the generators corresponding to factors with ${a}_{i}$ a unit will be 0 ). The column operations performed in the relations matrix reduction correspond to changing the basis used for ${R}^{n}$ as described in Exercise 17:

如果 $n$ 不是 $M$ 所需的最小生成元数目，那么上述某些初始元素 ${a}_{1},{a}_{2},\ldots$ 将为单位元，因此上述相应的直和项将为 0。如果我们移除这些无关因子，就得到了模 $M$ 的不变因子。进一步地，对应于上述直和项的新生成元在 ${R}^{n}$ 下的像将是一组 $R$ -生成元，用于 $M$ 在其不变因子分解中的循环子模（注意，$M$ 中对应于单位元因子的生成元的像将为 0）。在关系矩阵化简中执行的列操作对应于改变用于 ${R}^{n}$ 的基，如练习 17 中所述：

(a) Interchanging the ${i}^{\text{th }}$ column with the ${j}^{\text{th }}$ column corresponds to interchanging the ${i}^{\text{th }}$ and ${j}^{\text{th }}$ elements in the basis for ${R}^{n}$ .

(a) 交换 ${i}^{\text{th }}$ 列与 ${j}^{\text{th }}$ 列对应于在 ${R}^{n}$ 的基中交换 ${i}^{\text{th }}$ 和 ${j}^{\text{th }}$ 元素。

(b) For any $a \in R$ ,adding $a$ times the ${j}^{\text{th }}$ column to the ${i}^{\text{th }}$ column corresponds to subtracting $a$ times the ${i}^{\text{th }}$ basis element from the ${j}^{\text{th }}$ basis element.

(b) 对于任何 $a \in R$ ，将 $a$ 倍的 ${j}^{\text{th }}$ 列加到 ${i}^{\text{th }}$ 列上对应于从 ${j}^{\text{th }}$ 基元素中减去 $a$ 倍的 ${i}^{\text{th }}$ 基元素。

Keeping track of the column operations performed and changing the initial choice of generators for $M$ in the same way therefore gives a set of $R$ -generators for the cyclic submodules of $M$ in its invariant factor decomposition.

记录执行的列操作并按照相同方式改变 $M$ 的初始生成元选择，因此可以得到 $M$ 在其不变因子分解中的循环子模的一组 $R$ -生成元。

This process is quite fast computationally once an initial set of generators for $M$ and initial relations matrix are determined. The element ${a}_{1}$ is determined using the Euclidean Algorithm as the g.c.d. of the elements in the initial relations matrix. Using the row and column operations we can obtain the appropriate linear combination of the entries to produce this g.c.d. in the (1,1)-position of a new relations matrix. One then subtracts the appropriate multiple of the first column and first row to obtain a matrix as in Exercise 19(b), then iterates this process. Some examples of this procedure in a special case are given at the end of the following section.

这个过程在确定了$M$的一组初始生成元和初始关系矩阵后，在计算上相当快速。元素${a}_{1}$通过欧几里得算法确定为初始关系矩阵中元素的最大公约数（g.c.d.）。使用行列操作，我们可以得到适当的线性组合，使得这个最大公约数出现在新关系矩阵的（1,1）位置。然后减去第一列和第一行的适当倍数，得到如练习19(b)中的矩阵，然后重复这个过程。在下一节的末尾给出了这个程序在特殊情况下的例子。

20. Let $R$ be an integral domain with quotient field $F$ and let $M$ be any $R$ -module. Prove that the rank of $M$ equals the dimension of the vector space $F{ \otimes }_{R}M$ over $F$ .

21. 设$R$是一个整环，其分式域为$F$，设$M$为任意$R$-模。证明$M$的秩等于向量空间$F{ \otimes }_{R}M$在$F$上的维数。

22. Prove that a finitely generated module over a P.I.D. is projective if and only if it is free.

23. 证明在主理想整环上的有限生成模是投射的当且仅当它是自由的。

24. Let $R$ be a P.I.D. that is not a field. Prove that no finitely generated $R$ -module is injective. [Use Exercise 4, Section 10.5 to consider torsion and free modules separately.]

25. 设$R$是一个不是域的主理想整环。证明没有有限生成的$R$-模是可内射的。[使用10.5节的练习4，分别考虑扭元和自由模。]