# §18 向量空间
## §18.1 张成、线性无关与基
**定义 18.1** 固定 $x_{1},\ldots,x_{n}\in M$。

(1) 如果映射 $X: R^{n}\to M$ 是满射，那么称该集合**张成(span)** $M$。

(2) 如果映射 $X: R^{n}\to M$ 是单射，那么称该集合在 $M$ 中是**线性无关(linearly independent)** 的。

(3) 如果 $X$ 同时是单射和满射，那么称该集合是 $M$ 的**基(basis)**。



你会在线性代数中认出这些术语。用方程来表达，这些定义正是你所预想的：

**命题 18.1** 设 $M$ 是一个左 $R$-模，$x_{1},\ldots,x_{n}\in M$ 是一个有序集合。

(1) 如果对每个 $y\in M$，存在一个集合 $a_{1},\ldots,a_{n}\in R$ 使得
$$
y=a_{1}x_{1}+\cdots+a_{n}x_{n}，
$$
那么该集合张成 $M$。

(2) 如果方程
$$
0=a_{1} x_{1}+\cdots+a_{n} x_{n}
$$
仅有唯一解 $\left(a_{1}, \ldots, a_{n}\right)=(0, \ldots, 0)$，那么该集合是线性无关的。

(3) 如果对任何 $y \in M$，方程
$$
y=a_1 x_1+\ldots+a_n x_n
$$
仅有唯一的解 $\left(a_{1}, \ldots, a_{n}\right)$，那么该集合是基。

证明：第一条是满射的定义。后一条结论成立，因为同态映射是单射当且仅当核是平凡的，而 $(0, \ldots, 0) \in R^{n}$ 是 $R^n$ 的加法单位元。最后一条是双射的定义。
$$
~\tag*{$\square$}
$$

**定义 18.2** 如果存在某个整数 $n \in \mathbb{Z}_{\geqslant 0}$ 和一个满 $R$-模同态 $R^{n} \rightarrow M$，称一个模是**有限生成的(finitely generated)**。

这与群的情况类似。一个群 $G$ 是有限生成的当且仅当存在有限个元素 $g_{i}$ 使得所有其他元素都可以表示为 $g_{i}$ 和它们的逆元的乘积。同样地，如果存在有限个 $x_{i}$，使得 $M$ 的每个元素都可以通过 $x_{i}$ 的线性组合得到，那么 $M$ 是有限生成的。

**例 18.1** 并非所有的 $R$-模都承认基。这与向量空间不同。例如，若 $R=\mathbb{Z}$ 且 $M=\mathbb{Z}/n\mathbb{Z}$，则对于任何 $x\in M$，方程
$$
a x=0
$$
有许多解——$a$ 可以是 $n, 2n, \ldots$。

**要点**: 并非每个有限生成的 $R$-模都承认基。

## §18.2 向量空间与子空间
**定义 18.3** 如果 $R-\{0\}$ 在乘法下构成群，则称一个交换环是**域(field)**。

**定义 18.4** 设 $F$ 是一个域。$F$ 上的模称为 $F$ 上的**向量空间(vector space)**。

**定义 18.5** 设 $V$ 是一个向量空间。那么 $V$ 的一个子模称为 $V$ 的**线性子空间(linear subspace)**。

## §18.3 张成集大于无关集
以下是域与环的最重要区别之一：

**定理 18.2** 设 $F$ 是一个域，$M$ 是一个 $F$ 上的向量空间。如果 $v_{1}, \ldots, v_{n}$ 张成 $M$，而 $w_{1}, \ldots, w_{m}$ 是线性无关的，那么 $n \geqslant m$。

证明：设 $y_{1}, \ldots, y_{m}$ 是线性无关的，$v_{1}, \ldots, v_{n}$ 是张成的。若有必要，可以对 $v_{i}$ 重新排序，以使
$$
y_{1}=a_{1} v_{1}+\ldots+a_{n} v_{n}
$$
且 $a_{1} \neq 0$。于是 $y_{1}, v_{2}, \ldots, v_{n}$ 也张成 $M$，因为我们可以通过 $y_{1}$ 和 $v_{i}$ 的线性组合得到 $v_{1}$——只需将上述方程除以 $a_{1} \neq 0$ 并重新排列即可。

让 $M_{1} \subset M$ 是由 $y_{1}$ 生成的子模——即 $R \rightarrow M$ 的像，由 $1 \mapsto y_{1}$ 定义——并考虑商模
$$
M/M_{1}
$$
（可以证明这是一个 $R$-模——即，一个向量空间。）那么 $\overline{y_{2}}, \ldots, \overline{y_{m}}$ 仍然是线性无关的，因为它们的线性组合等于零当且仅当
$$
a_{1} y_{1}=a_{2} y_{2}+\ldots+a_{m} y_{m}
$$
对于某个 $a_{1} \in F$ 成立，而此类方程仅当所有 $a_{i}=0$ 时成立，因为假定 $y_{i}$ 是线性无关的。注意 $\overline{y_{1}}=0, \overline{v_{2}}, \ldots, \overline{v_{n}}$ 仍是张成的，因此 $\overline{v_{2}}, \ldots, \overline{v_{n}}$ 是张成的。所以我们有 $n-1$ 个向量张成 $M/M_{1}$，且其中有 $m-1$ 个线性无关的向量。

通过重复上述技巧，如果我们在一个向量空间中有 $m$ 个线性无关的元素且有 $n$ 个元素张成它，我们可以在商向量空间中得到 $m-k$ 个线性无关的元素和 $n-k$ 个张成它的元素。那么这些数哪个会先到 0 呢？如果 $n-k=0$ 先到零，那么我们位于一个由 $0$ 个元素张成的商向量空间——即零向量空间——因此我们必须得出 $m-k=0$，因为在零向量空间中没有线性无关的向量。在这种情况下，$m=n$。如果 $m-k$ 在 $n-k$ 之前到零，那么 $m \leqslant n$。
$$
~\tag*{$\square$}
$$

## §18.4 维数
**推论 18.3** 如果 $M$ 是一个有限生成的向量空间，那么 $M$ 的任意两个基的元素数相同。

证明：如果 $\left\{v_{i}\right\}$ 和 $\left\{w_{i}\right\}$ 均是张成且线性无关的，我们有 $n \geqslant m$ 且 $m \geqslant n$。因此 $m=n$。
$$
~\tag*{$\square$}
$$

**定义 18.6** 设 $V$ 是一个有限生成的 $F$-模——即有限生成的向量空间。称此类 $V$ 为**有限维**向量空间，并定义 $V$ 的**维数(dimension)**
$$
\dim_{F} V
$$
为 $V$ 的任意基中的元素数。

**注** 这是线性代数中最重要的事实：我们有维数的概念。人类花了几千年才理解什么是 $n$ 维空间，因此不要轻视这一概念！

**例 18.2** $0$ 维向量空间是由平凡Abel群 $M=\{0\}$ 给出的模。

**推论 18.4** 如果 $M$ 是一个有限生成的向量空间，那么任意线性无关集合 $w_{1}, \ldots, w_{m}$ 可以扩展为一个基——也就是说，我们可以找到 $w_{m+1}, \ldots, w_{n}$ 使得所得集合 $w_{1}, \ldots, w_{n}$ 同时线性无关且张成。

证明：由于 $M$ 是有限生成的，存在某个 $N$ 使得我们有一个满射 $F^{N} \rightarrow M$。因此，任意线性无关向量集的大小必须 $\leqslant N$。如果 $X_{m}: F^{m} \rightarrow M$ 是由 $w_{1}, \ldots, w_{m}$ 决定的映射，且 $X_{m}$ 不是满射，则选择一个不在 $\mathrm{im}\left(X_{m}\right)$ 中的元素 $w_{m+1}$。注意所得集合 $w_{1}, \ldots, w_{m+1}$ 仍然是线性无关的，因为如果
$$
a_{1} w_{1}+\ldots+a_{m+1} w_{m+1}=0
$$
则
$$
a_{1} w_{1}+\ldots+a_{m} w_{m}=-a_{m+1} w_{m+1}
$$
如果 $a_{m+1}=0$，根据 $w_{i}$ 的线性无关性，我们知道所有 $a_{i}=0$。另一方面，如果 $a_{m+1} \neq 0$，我们通过除法得出矛盾：
$$
\frac{a_{1}}{-a_{m+1}} w_{1}+\ldots+\frac{a_{m}}{-a_{m+1}} w_{m}=w_{m+1}
$$
左边在 $X_{m}$ 的像中，但 $w_{m+1}$ 被选择为不在其中。

因此我们有一个单射同态 $X_{m+1}: F^{m+1} \rightarrow M$。如果 $X_{m+1}$ 不是满射，我们重复上述论证。根据定理，$X_{m+k}$ 在某个 $m+k \leqslant N$ 处必须成为满射。让 $k$ 是 $X_{m+k}$ 首次成为满射的整数。根据上述论证，它仍然是单射，因此我们有一个由生成元 $w_{1}, \ldots, w_{m+k}$ 决定的基。
$$
~\tag*{$\square$}
$$

## §18.5 若干推论
我们将要做什么？你以前已经研究过元素为实数的矩阵。你对它们进行了多种操作——相乘、相加，以及判断它们何时是可逆的。我声称几乎所有关于实数矩阵的操作都可以在任何域的矩阵上完成。

**推论 18.5** 任意有限生成的域 $F$ 上的模与 $F^{n}$ 对某个 $n$ 同构。

证明：从线性无关集合 0 开始扩展为基。一个基定义了从 $F^{n}$ 到你的模的同构。
$$
~\tag*{$\square$}
$$

**注** 如果 $R$ 不是域，这一结论对于 $R$-模不成立——毕竟，任何有限Abel群都是一个 $\mathbb{Z}$-模，但任何自由 $\mathbb{Z}$-模要么是零模，要么是无限模。

**推论 18.6** 如果 $V^{\prime}\subset V$ 是子空间，
$$
\dim V^{\prime}=\

dim V\quad\Leftrightarrow\quad V=V^{\prime}.
$$

证明：一方向显然成立。对于另一方向，令 $y_{1},\ldots,y_{n}$ 是 $V^{\prime}$ 的基。由于这些向量是线性无关的，它们可以通过上述推论之一扩展为 $V$ 的基。但根据维数的定义，该基必须恰好有 $n$ 个元素——换句话说，$y_{i}$ 已经是一个基。
$$
~\tag*{$\square$}
$$

**推论 18.7** 设 $V^{\prime} \subset V$ 是一个子空间。则有 $\dim V^{\prime}+\dim V/V^{\prime}=\dim V$。

证明：设 $v_{1}, \ldots, v_{\dim V^{\prime}}$ 是 $V^{\prime}$ 的基。设 $\overline{u_{1}}, \ldots, \overline{u_{\dim V/V^{\prime}}}$ 是 $V/V^{\prime}$ 的基。选择代表 $u_{i}$ 作为 $\overline{u_{i}}$，则集合
$$
v_{1}, \ldots, v_{\dim V^{\prime}}, u_{1}, \ldots, u_{\dim V/V^{\prime}}
$$
是 $V$ 的基。

它显然是张成的，因为对于每个 $a \in V$，$\overline{a}$ 是 $\overline{u_{i}}$ 的线性组合，因此 $a$ 位于某个线性组合的 $V^{\prime}$-轨道内。它也是线性无关的，因为如果
$$
0 = a_{1} v_{1} + \cdots + a_{\dim V^{\prime}} v_{\dim V^{\prime}} + b_{1} u_{1} + \cdots + b_{\dim V/V^{\prime}} u_{\dim V/V^{\prime}}
$$
那么
$$
\overline{0} = a_{1} \overline{v_{1}} + \cdots + a_{\dim V^{\prime}} \overline{v_{\dim V^{\prime}}} + b_{1} \overline{u_{1}} + \cdots + b_{\dim V/V^{\prime}} \overline{u_{\dim V/V^{\prime}}}
$$
因为 $\overline{v_{i}} = 0$，$a_{i}$ 项消失，因此我们得到一个等式，表示 $\overline{u_{i}}$ 的线性组合等于零。由于 $\overline{u_{i}}$ 是线性无关的，得出每个 $b_{i}$ 必须为零。原等式表明 $0 = \sum a_{i} v_{i}$，因此根据 $v_{i}$ 的线性无关性，所有 $a_{i}$ 也必须为零。
$$
~\tag*{$\square$}
$$

**推论 18.8 (秩-零度定理)**
设 $f: V \rightarrow W$ 是一个 $F$-模之间的映射，且假设 $V$ 是有限生成的。那么有 $\dim \ker f + \dim \mathrm{im} f = \dim V$。

证明：根据同态基本定理，我们知道存在一个群同构 $V/\ker f \cong \mathrm{im} f$。但这个同态映射也是一个 $F$-模映射，可以手工验证。因此 $\mathrm{im} f \cong V/\ker f$。
$$
~\tag*{$\square$}
$$

**推论 18.9 (同构的判别准则)**
设 $f: V \rightarrow W$ 是有限维向量空间之间的线性映射。那么 $f$ 是同构当且仅当 $f$ 是单射且 $\dim V = \dim W$。

证明：根据秩-零度定理，$f$ 的像的维数等于 $V$ 的维数，因为 $f$ 是单射。
$$
~\tag*{$\square$}
$$

## §18.6 总结
从上述所有内容中得到的总结是维数概念的强大。无论你的域是 $\mathbb{R}$ 这样熟悉的，还是类似 $\mathbb{Z}/p\mathbb{Z}$ 这样陌生的；无论线性映射是熟悉的矩阵，还是像多项式函数的值这样的线性映射，我们都有一种强大的方法来研究线性映射。

## §18.7 行列式
我们从线性代数中得到的另一个强大工具是行列式的概念。行列式只需要一个关于乘以 $-1$（取加法逆元）、矩阵元素的乘法和加法的概念。因此我们应该能够定义任何系数在环 $R$ 上的矩阵的行列式。

事实证明，如果环 $R$ 不是交换的，有些公式可能不成立——因为乘法的顺序很重要——因此我们将限制自己使用交换环。

**定义 18.7** 设 $R$ 是一个交换环。一个 $k \times k$ 的**矩阵(matrix)** 是元素的集合
$$
A_{ij} \in R
$$
其中 $i \in 1, \ldots, k$，$j \in 1, \ldots, k$。我们用符号表示矩阵
$$
A = \left(A_{ij}\right).
$$

**例 18.3** 一个 $3 \times 3$ 的 $R$ 上的矩阵可以用通常的方式表示：
$$
\begin{pmatrix}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{pmatrix}
$$

**定义 18.8** $R$ 中 $k \times k$ 矩阵的**环(ring)**，记作 $M_{k \times k}(R)$，其加法定义为
$$
\left(A_{i j}\right)+\left(B_{i j}\right)=\left(A_{i j}+B_{i j}\right)
$$
且乘法定义为
$$
\left(A_{i j}\right)\left(B_{i j}\right)=\left(\sum_{l=1}^k A_{i l} B_{l j}\right)
$$
即加法是逐项相加，而乘法中第 $i,j$ 项是 $B$ 的第 $j$ 列与 $A$ 的第 $i$ 行的配对。

**定义 18.9 (余子式矩阵)**
设 $A$ 是一个 $k \times k$ 矩阵。$A$ 的第 $(i,j)$ 项的**余子式矩阵(cofactor matrix)** 是删除 $A$ 的第 $i$ 行和第 $j$ 列所得的矩阵。当 $A$ 是已知时，我们记作
$$
C_{i, j}
$$
表示由 $A$ 的第 $(i, j)$ 项余子式得到的 $(k-1) \times (k-1)$ 矩阵。

**定义 18.10** $R$ 中 $1 \times 1$ 矩阵的**行列式(determinant)** 是矩阵的唯一元素 $A_{11}$。

递归定义：设 $A$ 是一个 $k \times k$ 矩阵。则 $A$ 的行列式定义为以下求和：
$$
\det A = A_{11} \det C_{1,1} - A_{21} \det C_{2,1} + \cdots + (-1)^{1+k} A_{k 1} \det C_{k, 1}。
$$
使用求和符号表示为：
$$
\det A := \sum_{i=1}^k (-1)^{i+1} A_{i 1} \det C_{i, 1}。
$$
这定义了一个函数
$$
\det: M_{k \times k}(R) \rightarrow R。
$$

**例 18.4** 如果 $A$ 是一个 $2 \times 2$ 矩阵，
$$
\det(A) = A_{11} A_{22} - A_{12} A_{21}。
$$

我们不证明以下定理，但你对实数进行的证明同样适用于一般情形：

**定理 18.10** 设 $A$ 和 $B$ 是 $k \times k$ 矩阵。那么
$$
\det(A) \det(B) = \det(AB)
$$
且
$$
\det\left(A^\mathrm{T}\right) = \det(A).
$$

**定理 18.11** 设 $\mathrm{adj}(A)$ 是一个 $k \times k$ 矩阵，其第 $(i, j)$ 项由
$$
(-1)^{i+j} \det C_{j, i}
$$
给出。那么
$$
A \cdot (\mathrm{adj} A) = (\mathrm{adj} A) \cdot A = \det A \cdot I
$$
其中 $\det A \cdot I$ 是对角矩阵，对角元素由元素 $\det A \in R$ 给出。

**注** 如果你之前没有见过这个定理的最后一条说明，我会简要说明一下证明思路。第一乘法的 $(i, j)$ 项由以下给出：
$$
\sum_{l=1}^k A_{i l} (\mathrm{adj} A)_{l j} = \sum_{l=1}^k A_{i l} (-1)^{j+l} \det C_{j, l}.
$$
例如，$(1,1)$ 项就是 $A$ 的行列式的定义。通过使用关于交换行只改变行列式符号的性质，可以证明每个对角线项都是 $A$ 的行列式。

对于非对角线项，注意到上述求和式变为具有两行相等的矩阵的行列式，因此为零。

**推论 18.12** 设 $A \in M_{k \times k}(R)$。则 $A$ 是可逆矩阵当且仅当 $\det A \in R$ 有乘法逆元。

证明：令 $B = \det A^{-1} \mathrm{adj} A$。则有
$$
B A = \det A^{-1} \mathrm{adj} A \cdot A = \det A^{-1} \det A \cdot I = I.
$$
同样可以证明 $BA = I$。
$$
~\tag*{$\square$}
$$

**例 18.5** 如果 $A$ 是一个仅包含整数元素的矩阵，那么当且仅当 $\det A = \pm 1$ 时存在一个仅包含整数元素的逆矩阵。

**例 18.6** 设 $A$ 是 $\mathbb{Z}/n\mathbb{Z}$ 上的一个矩阵。则当且仅当其行列式与 $n$ 互素时，$A$ 是可逆的。

# §18 Vector Spaces

## §18.1 Spanning Sets, Linear Independence, and Bases

**Definition 18.1** Fix

$$
x_{1},\ldots,x_{n}\in M.
$$

(1) If the map

$$
X:R^{n}\to M
$$

is surjective, then we say that the set **spans** $M$.

(2) If the map

$$
X:R^{n}\to M
$$

is injective, then we say that the set is **linearly independent** in $M$.

(3) If $X$ is both injective and surjective, then we say that the set is a **basis** of $M$.



You will recognize these terms from linear algebra. Expressed in terms of equations, these definitions are exactly what you would expect.

**Proposition 18.1** Let $M$ be a left $R$-module, and let

$$
x_{1},\ldots,x_{n}\in M
$$

be an ordered set.

(1) If for every

$$
y\in M
$$

there exist

$$
a_{1},\ldots,a_{n}\in R
$$

such that

$$
y=a_{1}x_{1}+\cdots+a_{n}x_{n},
$$

then the set spans $M$.

(2) If the equation

$$
0=a_{1}x_{1}+\cdots+a_{n}x_{n}
$$

has the unique solution

$$
(a_{1},\ldots,a_{n})=(0,\ldots,0),
$$

then the set is linearly independent.

(3) If for every

$$
y\in M,
$$

the equation

$$
y=a_{1}x_{1}+\ldots+a_{n}x_{n}
$$

has a unique solution

$$
(a_{1},\ldots,a_{n}),
$$

then the set is a basis.

Proof: The first statement is simply the definition of surjectivity.

The second follows because a homomorphism is injective if and only if its kernel is trivial, and

$$
(0,\ldots,0)\in R^{n}
$$

is the additive identity of
$R^n$.

The final statement is the definition of a bijection.

$$
~\tag*{$\square$}
$$

**Definition 18.2** A module is called **finitely generated** if there exists some integer

$$
n\in\mathbb{Z}_{\geqslant0}
$$

and a surjective $R$-module homomorphism

$$
R^{n}\to M.
$$

This is analogous to the case of groups.

A group $G$ is finitely generated if and only if there exist finitely many elements
$g_i$
such that every other element can be expressed as a product of the
$g_i$
and their inverses.

Similarly, a module $M$ is finitely generated if there exist finitely many
$x_i$
such that every element of $M$ can be obtained as a linear combination of the
$x_i$.

**Example 18.1** Not every $R$-module admits a basis.

This is different from the situation for vector spaces.

For example, let

$$
R=\mathbb{Z}
$$

and

$$
M=\mathbb{Z}/n\mathbb{Z}.
$$

Then for any

$$
x\in M,
$$

the equation

$$
ax=0
$$

has many solutions: $a$ may be

$$
n,2n,\ldots.
$$

**Key Point:** Not every finitely generated $R$-module admits a basis.

## §18.2 Vector Spaces and Subspaces

**Definition 18.3** A commutative ring is called a **field** if

$$
R-\{0\}
$$

forms a group under multiplication.

**Definition 18.4** Let $F$ be a field. A module over $F$ is called a **vector space over $F$**.

**Definition 18.5** Let $V$ be a vector space. A submodule of $V$ is called a **linear subspace** of $V$.

## §18.3 A Spanning Set Is at Least as Large as an Independent Set

The following is one of the most important differences between fields and general rings.

**Theorem 18.2** Let $F$ be a field and let $M$ be a vector space over $F$.

If

$$
v_{1},\ldots,v_{n}
$$

span $M$, and

$$
w_{1},\ldots,w_{m}
$$

are linearly independent, then

$$
n\geqslant m.
$$

Proof: Let

$$
y_{1},\ldots,y_{m}
$$

be linearly independent, and let

$$
v_{1},\ldots,v_{n}
$$

span.

If necessary, reorder the $v_i$ so that

$$
y_{1}
=
a_{1}v_{1}
+\ldots+
a_{n}v_{n}
$$

with

$$
a_{1}\neq0.
$$

Then

$$
y_{1},v_{2},\ldots,v_{n}
$$

also span $M$, because we can express $v_1$ as a linear combination of
$y_1$
and the other $v_i$ simply by dividing the equation above by
$a_1\neq0$
and rearranging.

Let

$$
M_{1}\subset M
$$

be the submodule generated by
$y_1$,
i.e. the image of

$$
R\to M
$$

defined by

$$
1\mapsto y_1,
$$

and consider the quotient module

$$
M/M_1.
$$

One can prove that this is an $R$-module, hence a vector space.

Then

$$
\overline{y_2},\ldots,\overline{y_m}
$$

remain linearly independent, because a linear combination of them is zero if and only if

$$
a_1y_1
=
a_2y_2+\ldots+a_my_m
$$

for some

$$
a_1\in F,
$$

and such an equation can hold only when all the
$a_i$
are zero, since the
$y_i$
are assumed to be linearly independent.

Notice that

$$
\overline{y_1}=0,
\qquad
\overline{v_2},\ldots,\overline{v_n}
$$

still span.

Therefore,

$$
\overline{v_2},\ldots,\overline{v_n}
$$

span

$$
M/M_1.
$$

Thus we have
$n-1$
vectors spanning
$M/M_1$
and
$m-1$
linearly independent vectors in it.

Repeating this argument, if a vector space has $m$ linearly independent elements and is spanned by $n$ elements, then in a quotient vector space we obtain

$$
m-k
$$

linearly independent elements and

$$
n-k
$$

spanning elements.

Which of these numbers reaches $0$ first?

If

$$
n-k=0
$$

first, then we are in a quotient vector space spanned by $0$ elements, i.e. the zero vector space.

Therefore,

$$
m-k=0,
$$

because there are no linearly independent vectors in the zero vector space.

In this case,

$$
m=n.
$$

If

$$
m-k
$$

reaches zero before
$n-k$,
then

$$
m\leqslant n.
$$

$$
~\tag*{$\square$}
$$

## §18.4 Dimension

**Corollary 18.3** If $M$ is a finitely generated vector space, then any two bases of $M$ contain the same number of elements.

Proof: If

$$
\{v_i\}
$$

and

$$
\{w_i\}
$$

are both spanning and linearly independent, then

$$
n\geqslant m
$$

and

$$
m\geqslant n.
$$

Therefore,

$$
m=n.
$$

$$
~\tag*{$\square$}
$$

**Definition 18.6** Let $V$ be a finitely generated $F$-module, i.e. a finitely generated vector space.

Such a $V$ is called a **finite-dimensional** vector space, and the **dimension** of $V$ is defined by

$$
\dim_F V
$$

to be the number of elements in any basis of $V$.

**Remark** This is one of the most important facts in linear algebra: we have a notion of dimension.

It took humanity thousands of years to understand what an $n$-dimensional space is, so do not take this concept lightly!

**Example 18.2** The $0$-dimensional vector space is the module given by the trivial Abelian group

$$
M=\{0\}.
$$

**Corollary 18.4** If $M$ is a finitely generated vector space, then any linearly independent set

$$
w_{1},\ldots,w_{m}
$$

can be extended to a basis.

That is, we can find

$$
w_{m+1},\ldots,w_{n}
$$

such that

$$
w_{1},\ldots,w_{n}
$$

is both linearly independent and spanning.

Proof: Since $M$ is finitely generated, there exists some $N$ such that we have a surjection

$$
F^N\to M.
$$

Therefore, the size of any linearly independent set of vectors must be

$$
\leqslant N.
$$

If

$$
X_m:F^m\to M
$$

is the map determined by

$$
w_1,\ldots,w_m
$$

and $X_m$ is not surjective, choose an element

$$
w_{m+1}
$$

not lying in

$$
\mathrm{im}(X_m).
$$

Notice that the resulting set

$$
w_1,\ldots,w_{m+1}
$$

is still linearly independent.

Indeed, if

$$
a_1w_1+\ldots+a_{m+1}w_{m+1}
=
0,
$$

then

$$
a_1w_1+\ldots+a_mw_m
=
-a_{m+1}w_{m+1}.
$$

If

$$
a_{m+1}=0,
$$

then the linear independence of the $w_i$ implies that all
$a_i=0$.

On the other hand, if

$$
a_{m+1}\neq0,
$$

then dividing gives the contradiction

$$
\frac{a_1}{-a_{m+1}}w_1
+\ldots+
\frac{a_m}{-a_{m+1}}w_m
=
w_{m+1}.
$$

The left-hand side lies in the image of
$X_m$,
while $w_{m+1}$ was chosen not to lie in it.

Therefore, we obtain an injective homomorphism

$$
X_{m+1}:F^{m+1}\to M.
$$

If $X_{m+1}$ is not surjective, repeat the argument.

By the theorem, $X_{m+k}$ must become surjective for some

$$
m+k\leqslant N.
$$

Let $k$ be the first integer for which
$X_{m+k}$
is surjective.

By the argument above, it is still injective.

Therefore, the generators

$$
w_1,\ldots,w_{m+k}
$$

determine a basis.

$$
~\tag*{$\square$}
$$

## §18.5 Some Corollaries

What are we going to do with this?

You have already studied matrices with real entries.

You performed many operations on them: multiplication, addition, and determining when they are invertible.

I claim that almost everything you do with real matrices can also be done with matrices over any field.

**Corollary 18.5** Every finitely generated module over a field $F$ is isomorphic to

$$
F^n
$$

for some $n$.

Proof: Starting with the linearly independent set containing $0$ elements, extend it to a basis.

A basis defines an isomorphism from

$$
F^n
$$

to your module.

$$
~\tag*{$\square$}
$$

**Remark** If $R$ is not a field, this statement fails for $R$-modules.

After all, every finite Abelian group is a
$\mathbb{Z}$-module,
but every free
$\mathbb{Z}$-module
is either the zero module or infinite.

**Corollary 18.6** If

$$
V^{\prime}\subset V
$$

is a subspace, then

$$
\dim V^{\prime}
=
\dim V
\quad\Leftrightarrow\quad
V=V^{\prime}.
$$

Proof: One direction is obvious.

For the other direction, let

$$
y_1,\ldots,y_n
$$

be a basis of
$V^{\prime}$.

Since these vectors are linearly independent, one of the previous corollaries allows us to extend them to a basis of $V$.

But by the definition of dimension, this basis must have exactly
$n$
elements.

In other words, the
$y_i$
already form a basis of $V$.

$$
~\tag*{$\square$}
$$

**Corollary 18.7** Let

$$
V^{\prime}\subset V
$$

be a subspace.

Then

$$
\dim V^{\prime}
+
\dim V/V^{\prime}
=
\dim V.
$$

Proof: Let

$$
v_1,\ldots,v_{\dim V^{\prime}}
$$

be a basis of
$V^{\prime}$.

Let

$$
\overline{u_1},
\ldots,
\overline{u_{\dim V/V^{\prime}}}
$$

be a basis of

$$
V/V^{\prime}.
$$

Choose representatives
$u_i$
for the
$\overline{u_i}$.

Then the set

$$
v_1,\ldots,v_{\dim V^{\prime}},
u_1,\ldots,u_{\dim V/V^{\prime}}
$$

is a basis of $V$.

It clearly spans, because for every

$$
a\in V,
$$

the class

$$
\overline{a}
$$

is a linear combination of the
$\overline{u_i}$.

Therefore, $a$ lies in the
$V^{\prime}$-orbit of some such linear combination.

It is also linearly independent.

Indeed, if

$$
0
=
a_1v_1
+\cdots+
a_{\dim V^{\prime}}v_{\dim V^{\prime}}
+
b_1u_1
+\cdots+
b_{\dim V/V^{\prime}}u_{\dim V/V^{\prime}},
$$

then

$$
\overline{0}
=
a_1\overline{v_1}
+\cdots+
a_{\dim V^{\prime}}\overline{v_{\dim V^{\prime}}}
+
b_1\overline{u_1}
+\cdots+
b_{\dim V/V^{\prime}}\overline{u_{\dim V/V^{\prime}}}.
$$

Since

$$
\overline{v_i}=0,
$$

the terms involving the
$a_i$
vanish.

Thus we obtain an equation saying that a linear combination of the
$\overline{u_i}$
is zero.

Since the
$\overline{u_i}$
are linearly independent, every
$b_i$
must be zero.

The original equation then becomes

$$
0
=
\sum a_iv_i.
$$

Since the
$v_i$
are linearly independent, all
$a_i$
must also be zero.

$$
~\tag*{$\square$}
$$

**Corollary 18.8 (Rank-Nullity Theorem)**

Let

$$
f:V\to W
$$

be a map of $F$-modules, and assume that $V$ is finitely generated.

Then

$$
\dim\ker f
+
\dim\mathrm{im}f
=
\dim V.
$$

Proof: By the fundamental theorem of homomorphisms, we know that there is a group isomorphism

$$
V/\ker f
\cong
\mathrm{im}f.
$$

But this homomorphism is also an $F$-module map, as can be verified directly.

Therefore,

$$
\mathrm{im}f
\cong
V/\ker f.
$$

$$
~\tag*{$\square$}
$$

**Corollary 18.9 (Criterion for Isomorphisms)**

Let

$$
f:V\to W
$$

be a linear map between finite-dimensional vector spaces.

Then $f$ is an isomorphism if and only if $f$ is injective and

$$
\dim V
=
\dim W.
$$

Proof: By the Rank-Nullity Theorem, the dimension of the image of $f$ equals the dimension of $V$ because $f$ is injective.

$$
~\tag*{$\square$}
$$

## §18.6 Summary

The main lesson from everything above is the power of the concept of dimension.

Whether your field is something familiar like
$\mathbb{R}$
or something more unfamiliar like
$\mathbb{Z}/p\mathbb{Z}$;
whether your linear map is a familiar matrix or something like evaluation of polynomial functions, we have a powerful method for studying linear maps.

## §18.7 Determinants

Another powerful tool from linear algebra is the concept of the determinant.

The determinant requires only notions of multiplication by
$-1$,
i.e. taking additive inverses, multiplication of matrix entries, and addition.

Therefore, we should be able to define the determinant of a matrix whose entries lie in any ring $R$.

It turns out that if the ring $R$ is not commutative, some formulas may fail because the order of multiplication matters.

Therefore, we restrict ourselves to commutative rings.

**Definition 18.7** Let $R$ be a commutative ring.

A
$k\times k$
**matrix** is a collection of elements

$$
A_{ij}\in R
$$

where

$$
i\in1,\ldots,k,
\qquad
j\in1,\ldots,k.
$$

We denote the matrix by

$$
A=(A_{ij}).
$$

**Example 18.3** A
$3\times3$
matrix over $R$ may be written in the usual way:

$$
\begin{pmatrix}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{pmatrix}.
$$

**Definition 18.8** The **ring** of
$k\times k$
matrices over $R$, denoted

$$
M_{k\times k}(R),
$$

has addition defined by

$$
(A_{ij})+(B_{ij})
=
(A_{ij}+B_{ij}),
$$

and multiplication defined by

$$
(A_{ij})(B_{ij})
=
\left(
\sum_{l=1}^{k}
A_{il}B_{lj}
\right).
$$

Thus addition is performed entrywise, while the
$(i,j)$
entry of the product is the pairing of the $i$th row of $A$ with the $j$th column of $B$.

**Definition 18.9 (Cofactor Matrix)**

Let $A$ be a
$k\times k$
matrix.

The **cofactor matrix** associated with the
$(i,j)$
entry of $A$ is the matrix obtained by deleting the $i$th row and the $j$th column of $A$.

When $A$ is understood, we write

$$
C_{i,j}
$$

for the
$(k-1)\times(k-1)$
matrix obtained from the cofactor associated with the
$(i,j)$
entry of $A$.

**Definition 18.10** The **determinant** of a
$1\times1$
matrix over $R$ is its unique entry

$$
A_{11}.
$$

Recursively, let $A$ be a
$k\times k$
matrix.

The determinant of $A$ is defined by the sum

$$
\det A
=
A_{11}\det C_{1,1}
-
A_{21}\det C_{2,1}
+
\cdots
+
(-1)^{1+k}
A_{k1}\det C_{k,1}.
$$

Using summation notation,

$$
\det A
:=
\sum_{i=1}^{k}
(-1)^{i+1}
A_{i1}
\det C_{i,1}.
$$

This defines a function

$$
\det:
M_{k\times k}(R)
\to
R.
$$

**Example 18.4** If $A$ is a
$2\times2$
matrix, then

$$
\det(A)
=
A_{11}A_{22}
-
A_{12}A_{21}.
$$

We will not prove the following theorem, but the proofs you know over the real numbers work equally well in this general setting.

**Theorem 18.10** Let $A$ and $B$ be
$k\times k$
matrices.

Then

$$
\det(A)\det(B)
=
\det(AB),
$$

and

$$
\det(A^{\mathrm{T}})
=
\det(A).
$$

**Theorem 18.11** Let

$$
\mathrm{adj}(A)
$$

be the
$k\times k$
matrix whose
$(i,j)$
entry is

$$
(-1)^{i+j}
\det C_{j,i}.
$$

Then

$$
A\cdot(\mathrm{adj}A)
=
(\mathrm{adj}A)\cdot A
=
\det A\cdot I,
$$

where

$$
\det A\cdot I
$$

is the diagonal matrix whose diagonal entries are the element

$$
\det A\in R.
$$

**Remark** If you have not seen the final statement of this theorem before, here is a brief sketch of the proof.

The
$(i,j)$
entry of the first product is

$$
\sum_{l=1}^{k}
A_{il}(\mathrm{adj}A)_{lj}
=
\sum_{l=1}^{k}
A_{il}
(-1)^{j+l}
\det C_{j,l}.
$$

For example, the
$(1,1)$
entry is exactly the definition of the determinant of $A$.

Using the fact that exchanging two rows only changes the sign of the determinant, one can prove that every diagonal entry is equal to
$\det A$.

For the off-diagonal entries, observe that the sum above becomes the determinant of a matrix having two identical rows, and is therefore zero.

**Corollary 18.12** Let

$$
A\in M_{k\times k}(R).
$$

Then $A$ is invertible if and only if

$$
\det A\in R
$$

has a multiplicative inverse.

Proof: Let

$$
B
=
\det A^{-1}
\mathrm{adj}A.
$$

Then

$$
BA
=
\det A^{-1}
\mathrm{adj}A\cdot A
=
\det A^{-1}
\det A\cdot I
=
I.
$$

Similarly, one can prove

$$
BA=I.
$$

$$
~\tag*{$\square$}
$$

**Example 18.5** If $A$ is a matrix with integer entries, then it has an inverse with integer entries if and only if

$$
\det A=\pm1.
$$

**Example 18.6** Let $A$ be a matrix over

$$
\mathbb{Z}/n\mathbb{Z}.
$$

Then $A$ is invertible if and only if its determinant is relatively prime to $n$.

§ DS-18 向量空间

§ DS-18 Vector Space

# §17 模

## §17.1 模
正如群作用在集合上一样，环作用在Abel群上。当一个环作用于一个Abel群时，该Abel群被称为该环上的**模**。

当群作用于集合时，必须通过双射作用，因此必须尊重集合的**势(cardinality)** 等性质。但环作用于Abel群时，它尊重的是Abel群内部的加法结构，这便是下面定义中的条件(1)。

在此情境下，我们将像对待群一样，建立模代数的所有定义。

**定义 17.1** 设$R$为一个环，$M$为一个Abel群。$R$在$M$上的**左作用(left action)** 是一个函数
$$
\begin{aligned}
R\times M&\to M,\\
(r,m)&\mapsto rm
\end{aligned}
$$
使得对于所有$r, s\in R$和$m, m^{\prime}\in M$，满足：

(1) $r(m+m^{\prime})=rm+rm^{\prime}$。

(2) $(r+s)m=rm+sm$。

(3) $s(rm)=(sr)m$。

(4) $1m=m$。

当$R$在$M$上的左作用确定后，我们称$M$为**左$R$-模(left $R$-module)**。



**注** 这里，“乘以$r$”可以理解为由某个环元素缩放。因此，一个模就是一个带有加法的集合，并附带一个由$r$缩放的概念。

有一种更简洁的表达方式。上面所有内容等价于一个环同态$R\to\mathrm{End}(M)$。

一个右$R$-模是一个Abel群$M$，带有一个满足(1)-(4)类似条件的函数$M\times R\to M$。

现在可能感觉数据量很大，因为我们同时涉及$R$和$M$。实际上，我们通常固定一个环$R$，只研究不同$M$之间的关系。

**练习** 设$M$为左$R$-模，则
$$
0m=m\quad\text{and}\quad (-r)m=-(rm).
$$

证明：为清晰起见，我们用$0_{R}$表示$R$的零元。根据条件(2)，我们有
$$
0_{R}m=(0_{R}+0_{R})m=0_{R}m+0_{R}m
$$
因此根据消去律，我们得到
$$
0=0_{R}m
$$
左边的$0$是$M$的加法单位元。同样地，
$$
rm+(-r)m=(r-r)m=0_{R}m=0.
$$
$$
~\tag*{$\square$}
$$

**例 17.1** (a) 设$R=\mathbb{R}$，$M=\mathbb{R}^{n}$，即$n$维的实向量空间。那么我们定义一个函数
$$
\begin{aligned}
\mathbb{R}\times M&\to M,\\
(t,v)&\mapsto t\vec{v}
\end{aligned}
$$
即通过$t$进行缩放。如果$\vec{v}=(v_{1},\ldots,v_{n})$，
$$
t\vec{v}=(tv_{1},\ldots,tv_{n}).
$$
这满足上述所有性质。

(b) 任意环$R$自身是其上的一个左模。

(c) 这是一个远离“缩放”概念的例子，更贴近“作用”这一概念。设$\mathbb{R}[t]$为变量$t$上的实系数多项式环。选取一个$m\times m$矩阵$T$，我们将其视为一个线性映射$T: \mathbb{R}^{m}\to\mathbb{R}^{m}$。那么，$\mathbb{R}^{m}$是一个左$\mathbb{R}[t]$-模，其作用定义为
$$
(a_{0}+a_{1}t+\cdots+a_{k}t^{k})v:=a_{0}v+a_{1}T(v)+\cdots+a_{k}(T\circ\cdots\circ T)(v)
$$
其中$T\circ \cdots\circ T$表示$T$复合$k$次。

## §17.2 子模
**定义 17.2** 设$R$为一个环，$M$为一个左$R$-模。若$M$的一个Abel子群$M^{\prime}\subset M$满足对于所有$r\in R$，$x\in M^{\prime}$时有$rx\in M^{\prime}$，则称$M^{\prime}$为$M$的**子模(submodule)**。

**例 17.2** (a) 若$R=\mathbb{R}$且$M=\mathbb{R}^{n}$，则子模是闭合于加法、取逆及缩放的子集。这与$\mathbb{R}^{n}$的线性子空间相同。

(b) 设$R$为交换环，则$I\subset R$是理想当且仅当$I$是$R$的子模。

(c) 设$M=\mathbb{R}^{m}$为线性变换$T: \mathbb{R}^{k}\to\mathbb{R}^{k}$定义的$\mathbb{R}[t]$-模。那么，子模是一个线性子空间$V$，满足$T(V)\subset V$。换言之，它是$T$不变的子空间。

## §17.3 模同态
**定义 17.3** 设$M$和$N$为左$R$-模。则一个$R$-模同态，或$R$-模映射，是一个函数
$$
f: M\to N.
$$
使得$f$是一个群同态，并且
$$
f(rx)=rf(x)
$$
对于所有$r\in R$, $x\in M$。

**定义 17.4** 一个$R$-模**同构(isomorphism)**是一个双射的同态。

**定义 17.5** 一个$R$-模同态$f: M\to N$的**核(kernel)** 和**像(image)** 分别是$f$作为群同态的核和像。

这意味着$\ker(f)=\{x~|~f(x)=0\}$，$\mathrm{im}(f)=\{y\in N~|~y=f(x), x\in M\}$。

**例 17.3** (a) 若$M\cong\mathbb{R}^{n}$和$N\cong\mathbb{R}^{m}$，具有$\mathbb{R}$上的缩放模结构，则线性映射$M\to N$是一个$\mathbb{R}$-模同态。

(b) (没有从上面(b)来的明显的例子)

(c) 设$M=\mathbb{R}^{m}$为左模，定义由线性变换$T$。令$N=\mathbb{R}^{n}$为线性变换$S: \mathbb{R}^{n}\to\mathbb{R}^{n}$定义的左模。则一个$R$-模同态是一个线性映射
$$
f: \mathbb{R}^{m}\to\mathbb{R}^{n}.
$$
具有性质
$$
f(T(v))=S(f(v)).
$$

**定义 17.6** 设$M$和$N$为左$R$-模，则所有$R$-模同态的集合记作
$$
\mathrm{Hom}_{R}(M, N).
$$

**命题 17.1** $\mathrm{Hom}_{R}(M, N)$是一个左$R$-模。

证明：**标量乘法的定义**：

对于$r \in R$和$f \in \mathrm{Hom}_R(M, N)$，定义标量乘法$r \cdot f$为：
$$
(r \cdot f)(m) = r f(m), \quad \forall m \in M.
$$

**验证$r \cdot f$是一个$R$-模同态**：

1. 加法性：
$$
 \begin{aligned}
 (r \cdot f)(m_1 + m_2) &= r f(m_1 + m_2) \\
 &= r [f(m_1) + f(m_2)] \\
 &= r f(m_1) + r f(m_2) \\
 &= (r \cdot f)(m_1) + (r \cdot f)(m_2).
 \end{aligned}
$$
因此，$r \cdot f$是加法性的。

2. 标量乘法的相容性：对于所有$s \in R$和$m \in M$，
$$
\begin{align*}
 (r \cdot f)(s m) &= r f(s m) \\
 &= r [s f(m)] \\
 &= s [r f(m)] \tag{因为$R$是交换的} \\
 &= s (r \cdot f)(m).
\end{align*}
$$
因此，$r \cdot f$是$R$-线性的。

**验证模公理**：

1. 关于环加法的分配性：
$$
 \begin{aligned}
 ((r + s) \cdot f)(m) &= (r + s) f(m) \\
 &= r f(m) + s f(m) \\
 &= (r \cdot f + s \cdot f)(m).
 \end{aligned}
$$

2. 关于模加法的分配性：
$$
 \begin{aligned}
 (r \cdot (f + g))(m) &= r (f + g)(m) \\
 &= r [f(m) + g(m)] \\
 &= r f(m) + r g(m) \\
 &= (r \cdot f)(m) + (r \cdot g)(m).
 \end{aligned}
$$

3. 标量乘法的结合性：
$$
 \begin{aligned}
 ((r s) \cdot f)(m) &= (r s) f(m) \\
 &= r (s f(m)) \\
 &= r \cdot (s \cdot f)(m).
 \end{aligned}
$$

4. 单位元：
$$
 (1_R \cdot f)(m) = 1_R f(m) = f(m).
$$
 
**结论**：

由于满足所有模的公理，$\mathrm{Hom}_R(M, N)$在上述定义的标量乘法下是一个左$R$-模。
$$
~\tag*{$\square$}
$$

## §17.4 直和与自由模
**定义 17.7** 设$M$和$N$为左$R$-模，则定义**直和(direct sum)**
$$
M\oplus N
$$
等同于作为群的$M\times N$，并具有$R$-模结构
$$
r(m,n):=(rm,rn).
$$

**命题 17.2** $M\oplus N$是一个$R$-模。

证明：我们已知$M\oplus N$是一个Abel群。另一方面，
$$
1(m,n):=(1m,1n)=(m,n)
$$
因为$M$和$N$都是模。而且我们有
$$
\begin{align*}
r((m,n)+(m^{\prime},n^{\prime}))&=r(m+m^{\prime},n+n^{\prime})\\
&=(r(m+m^{\prime}), r(n+n^{\prime}))\\
&=(rm+rm^{\prime},rn+rn^{\prime})\tag{3}\\
&=(rm,rn)+(rm^{\prime},rn^{\prime})\\
&=r(m,n)+r(m^{\prime},n^{\prime}).
\end{align*}
$$
其中(3)我们运用了$M$和$N$都是左$R$-模的性质。
$$
~\tag*{$\square$}
$$

**例 17.4** 若$R=\mathbb{R}$，且$M$和$N$也是$\mathbb{R}$，作为其自身的模，那么
$$
\mathbb{R}\oplus\mathbb{R}\cong\mathbb{R}\times\mathbb{R}\cong\mathbb{R}^{2}
$$
作为群，我们有通常的缩放操作
$$
r(x_{1},x_{2})=(rx_{1},rx_{2})。
$$

**注** 存在一个显然的同构
$$
(M\oplus N)\oplus O\cong M\oplus(N\oplus O),\quad (m,n,o)\mapsto (m,n,o).
$$

**定义 17.8** 设$R$为一个环，则直和模
$$
R^{n}:=R\oplus \cdots\oplus R
$$
称为**秩为$n$的自由$R$-模(free $R$-module of rank $n$)**。

## §17.5 自由模的泛性质
**问题**：为什么称之为自由$R$-模？

**命题 17.3** 设$M$是一个$R$-模。则任意的元素$n$元组$x_{1},\ldots,x_{n}\in M$唯一地确定一个$R$-模同态
$$
X: R^{n}\to M
$$
定义为
$$
(0,\ldots,0,1,0,\ldots,0)\mapsto x_{i}
$$
其中第$i$个坐标为$1$。

**注** 这与具有$n$个生成元的自由群的性质相同：任意一个群$G$的元素$n$元组唯一地确定从$F_{n}$到$G$的映射。

证明：给定$(x_{1},\ldots,x_{n})$，定义$X: R^{n}\to M$为
$$
X(a_{1},\ldots,a_{n}):=a_{1}x_{1}+\cdots+a_{n}x_{n}\in M。
$$
这是一个群同态，因为
$$
\begin{aligned}
X((a_{1},\ldots,a_{n})+(b_{1},\ldots,b_{n}))&=(a_{1}+b_{1})
x_{1}+\cdots+(a_{n}+b_{n})x_{n}\\
&=(a_{1}x_{1}+\cdots+a_{n}x_{x})+(b_{1}x_{1}+\cdots+b_{n}x_{n})\\
&=X(a_{1},\ldots,a_{n})+X(b_{1},\ldots,b_{n}).
\end{aligned}
$$
其中中间的等号使用了$M$为$R$-模的性质。这也是一个$R$-模同态，因为
$$
\begin{aligned}
X(r(a_{1},\ldots,a_{n}))&=X((ra_{1},\ldots,ra_{n}))\\
&=(ra_{1})x_{1}+\cdots+(ra_{n})x_{n}\\
&=r(a_{1}x_{1}+\cdots+a_{n}x_{n})\\
&=rX((a_{1},\ldots,a_{n})).
\end{aligned}
$$
同样，倒数第二个等号使用了$M$是$R$-模的性质。
$$
~\tag*{$\square$}
$$

## §17.6 0个生成元的自由模
**练习** 设$M$为左$R$-模，证明
$$
r0_{M}=0_{M},\quad\text{and}\quad r(-x)=-rx.
$$

证明：$R$在$M$上的作用等价于一个环同态$R\to\mathrm{End}(M)$。特别地，$R$中的每个元素$r$决定了一个Abel群同态。因此，以$r$为倍数的缩放保留了$M$的加法单位元和加法逆元。

如果你喜欢更计算化的证明，可以观察到
$$
r0_{M}+r0_{M}=r(0_{M}+0_{M})=r0_{M}.
$$
因此根据Abel群的消去律，我们可以从两边减去$r0_{M}$得到
$$
r0_{M}=0_{M}.
$$
因此
$$
r(-x)+rx=r(-x+x)=r0_{M}=0_{M},
$$
这表明$r(-x)$是$rx$的加法逆元。
$$
~\tag*{$\square$}
$$

**注** 我们知道$R^{\oplus n}$对于$n\geqslant 1$的情况。但对于$n=0$呢？

我们应该寻找一个$R$-模$R^{\oplus 0}$，使得存在一个双射
$$
\mathrm{Hom}_{R}(R^{\oplus 0},M)\cong\mathrm{Map}_{\text{Sets}}(\varnothing,M).
$$
但从空集到任意集合的函数唯一。因此我们必须寻找一个$R^{\oplus 0}$的模，使得只有一个模同态到任意$M$。唯一满足此性质的模是零模，即具有$r0=0$的平凡Abel群。

# §17 Modules

## §17.1 Modules

Just as groups act on sets, rings act on Abelian groups. When a ring acts on an Abelian group, that Abelian group is called a **module** over the ring.

When a group acts on a set, it must act by bijections, and therefore it must preserve properties such as the **cardinality** of the set. But when a ring acts on an Abelian group, it respects the additive structure inside the Abelian group. This is condition (1) in the definition below.

In this setting, we will build all the definitions of module algebra in much the same way as we did for groups.

**Definition 17.1** Let $R$ be a ring and let $M$ be an Abelian group. A **left action** of $R$ on $M$ is a function

$$
\begin{aligned}
R\times M&\to M,\\
(r,m)&\mapsto rm
\end{aligned}
$$

such that for all $r,s\in R$ and $m,m^{\prime}\in M$:

(1) $r(m+m^{\prime})=rm+rm^{\prime}$.

(2) $(r+s)m=rm+sm$.

(3) $s(rm)=(sr)m$.

(4) $1m=m$.

Once a left action of $R$ on $M$ is specified, we call $M$ a **left $R$-module**.



**Remark** Here, “multiplication by $r$” may be interpreted as scaling by some element of the ring. Thus, a module is a set equipped with addition together with a notion of scaling by $r$.

There is a more concise way to express this. Everything above is equivalent to giving a ring homomorphism

$$
R\to\mathrm{End}(M).
$$

A right $R$-module is an Abelian group $M$ equipped with a function

$$
M\times R\to M
$$

satisfying analogues of conditions (1)–(4).

At first, this may seem like a large amount of data, since we are simultaneously dealing with both $R$ and $M$.

In practice, however, we usually fix a ring $R$ and study only the relationships between different modules $M$.

**Exercise** Let $M$ be a left $R$-module. Then

$$
0m=m\quad\text{and}\quad(-r)m=-(rm).
$$

Proof: For clarity, write

$$
0_R
$$

for the zero element of $R$.

By condition (2),

$$
0_Rm
=
(0_R+0_R)m
=
0_Rm+0_Rm.
$$

Therefore, by the cancellation law,

$$
0=0_Rm.
$$

The $0$ on the left is the additive identity of $M$.

Similarly,

$$
rm+(-r)m
=
(r-r)m
=
0_Rm
=
0.
$$

$$
~\tag*{$\square$}
$$

**Example 17.1**

(a) Let

$$
R=\mathbb{R},
\qquad
M=\mathbb{R}^n,
$$

the $n$-dimensional real vector space.

Define a function

$$
\begin{aligned}
\mathbb{R}\times M&\to M,\\
(t,v)&\mapsto t\vec{v}
\end{aligned}
$$

by scaling by $t$.

If

$$
\vec{v}
=
(v_1,\ldots,v_n),
$$

then

$$
t\vec{v}
=
(tv_1,\ldots,tv_n).
$$

This satisfies all of the properties above.

(b) Every ring $R$ is naturally a left module over itself.

(c) Here is an example that is further removed from the idea of “scaling” and closer to the idea of an “action”.

Let

$$
\mathbb{R}[t]
$$

be the ring of real-coefficient polynomials in the variable $t$.

Choose an
$m\times m$
matrix $T$, viewed as a linear map

$$
T:\mathbb{R}^m\to\mathbb{R}^m.
$$

Then
$\mathbb{R}^m$
is a left
$\mathbb{R}[t]$-module
with action defined by

$$
(a_0+a_1t+\cdots+a_kt^k)v
:=
a_0v+a_1T(v)+\cdots+a_k(T\circ\cdots\circ T)(v),
$$

where

$$
T\circ\cdots\circ T
$$

denotes $T$ composed with itself $k$ times.

## §17.2 Submodules

**Definition 17.2** Let $R$ be a ring and let $M$ be a left $R$-module.

An Abelian subgroup

$$
M^{\prime}\subset M
$$

is called a **submodule** of $M$ if

$$
x\in M^{\prime}
\quad\Rightarrow\quad
rx\in M^{\prime}
$$

for all
$r\in R$.

**Example 17.2**

(a) If

$$
R=\mathbb{R}
$$

and

$$
M=\mathbb{R}^n,
$$

then a submodule is a subset closed under addition, additive inverses, and scalar multiplication.

This is exactly the same thing as a linear subspace of
$\mathbb{R}^n$.

(b) Let $R$ be a commutative ring. Then

$$
I\subset R
$$

is an ideal if and only if $I$ is a submodule of $R$.

(c) Let

$$
M=\mathbb{R}^m
$$

be the
$\mathbb{R}[t]$-module
defined by a linear transformation

$$
T:\mathbb{R}^k\to\mathbb{R}^k.
$$

Then a submodule is a linear subspace $V$ satisfying

$$
T(V)\subset V.
$$

In other words, it is a $T$-invariant subspace.

## §17.3 Module Homomorphisms

**Definition 17.3** Let $M$ and $N$ be left $R$-modules.

An **$R$-module homomorphism**, or **$R$-module map**, is a function

$$
f:M\to N
$$

such that $f$ is a group homomorphism and

$$
f(rx)=rf(x)
$$

for all
$r\in R$
and
$x\in M$.

**Definition 17.4** An $R$-module **isomorphism** is a bijective homomorphism.

**Definition 17.5** The **kernel** and **image** of an $R$-module homomorphism

$$
f:M\to N
$$

are respectively the kernel and image of $f$ viewed as a group homomorphism.

Thus

$$
\ker(f)
=
\{x\mid f(x)=0\},
$$

and

$$
\mathrm{im}(f)
=
\{y\in N\mid y=f(x),~x\in M\}.
$$

**Example 17.3**

(a) If

$$
M\cong\mathbb{R}^n
$$

and

$$
N\cong\mathbb{R}^m
$$

with their usual
$\mathbb{R}$-module structures, then a linear map

$$
M\to N
$$

is an
$\mathbb{R}$-module homomorphism.

(b) (There is no obvious example corresponding to (b) above.)

(c) Let

$$
M=\mathbb{R}^m
$$

be a left module defined by a linear transformation $T$.

Let

$$
N=\mathbb{R}^n
$$

be a left module defined by a linear transformation

$$
S:\mathbb{R}^n\to\mathbb{R}^n.
$$

Then an $R$-module homomorphism is a linear map

$$
f:\mathbb{R}^m\to\mathbb{R}^n
$$

with the property

$$
f(T(v))
=
S(f(v)).
$$

**Definition 17.6** Let $M$ and $N$ be left $R$-modules. The set of all $R$-module homomorphisms is denoted

$$
\mathrm{Hom}_R(M,N).
$$

**Proposition 17.1**

$$
\mathrm{Hom}_R(M,N)
$$

is a left $R$-module.

Proof: **Definition of scalar multiplication:**

For
$r\in R$
and
$f\in\mathrm{Hom}_R(M,N)$,
define the scalar multiple
$r\cdot f$
by

$$
(r\cdot f)(m)
=
rf(m),
\qquad
\forall m\in M.
$$

**Verify that $r\cdot f$ is an $R$-module homomorphism:**

1. Additivity:

$$
\begin{aligned}
(r\cdot f)(m_1+m_2)
&=
rf(m_1+m_2)\\
&=
r[f(m_1)+f(m_2)]\\
&=
rf(m_1)+rf(m_2)\\
&=
(r\cdot f)(m_1)
+
(r\cdot f)(m_2).
\end{aligned}
$$

Thus
$r\cdot f$
is additive.

2. Compatibility with scalar multiplication:

For all
$s\in R$
and
$m\in M$,

$$
\begin{align*}
(r\cdot f)(sm)
&=
rf(sm)\\
&=
r[sf(m)]\\
&=
s[rf(m)]
\tag{since $R$ is commutative}\\
&=
s(r\cdot f)(m).
\end{align*}
$$

Thus
$r\cdot f$
is $R$-linear.

**Verify the module axioms:**

1. Distributivity with respect to addition in the ring:

$$
\begin{aligned}
((r+s)\cdot f)(m)
&=
(r+s)f(m)\\
&=
rf(m)+sf(m)\\
&=
(r\cdot f+s\cdot f)(m).
\end{aligned}
$$

2. Distributivity with respect to addition in the module:

$$
\begin{aligned}
(r\cdot(f+g))(m)
&=
r(f+g)(m)\\
&=
r[f(m)+g(m)]\\
&=
rf(m)+rg(m)\\
&=
(r\cdot f)(m)
+
(r\cdot g)(m).
\end{aligned}
$$

3. Associativity of scalar multiplication:

$$
\begin{aligned}
((rs)\cdot f)(m)
&=
(rs)f(m)\\
&=
r(sf(m))\\
&=
r\cdot(s\cdot f)(m).
\end{aligned}
$$

4. Identity:

$$
(1_R\cdot f)(m)
=
1_Rf(m)
=
f(m).
$$

**Conclusion:**

Since all the module axioms are satisfied,

$$
\mathrm{Hom}_R(M,N)
$$

is a left $R$-module under the scalar multiplication defined above.

$$
~\tag*{$\square$}
$$

## §17.4 Direct Sums and Free Modules

**Definition 17.7** Let $M$ and $N$ be left $R$-modules.

The **direct sum**

$$
M\oplus N
$$

is defined to be the group

$$
M\times N
$$

equipped with the $R$-module structure

$$
r(m,n)
:=
(rm,rn).
$$

**Proposition 17.2**

$$
M\oplus N
$$

is an $R$-module.

Proof: We already know that

$$
M\oplus N
$$

is an Abelian group.

On the other hand,

$$
1(m,n)
:=
(1m,1n)
=
(m,n)
$$

because both $M$ and $N$ are modules.

Moreover,

$$
\begin{align*}
r((m,n)+(m^{\prime},n^{\prime}))
&=
r(m+m^{\prime},n+n^{\prime})\\
&=
(r(m+m^{\prime}),
r(n+n^{\prime}))\\
&=
(rm+rm^{\prime},
rn+rn^{\prime})
\tag{3}\\
&=
(rm,rn)
+
(rm^{\prime},rn^{\prime})\\
&=
r(m,n)
+
r(m^{\prime},n^{\prime}).
\end{align*}
$$

In (3), we used the fact that both $M$ and $N$ are left $R$-modules.

$$
~\tag*{$\square$}
$$

**Example 17.4** If

$$
R=\mathbb{R},
$$

and both $M$ and $N$ are also
$\mathbb{R}$
viewed as modules over themselves, then

$$
\mathbb{R}\oplus\mathbb{R}
\cong
\mathbb{R}\times\mathbb{R}
\cong
\mathbb{R}^2
$$

as groups, with the usual scalar multiplication

$$
r(x_1,x_2)
=
(rx_1,rx_2).
$$

**Remark** There is an obvious isomorphism

$$
(M\oplus N)\oplus O
\cong
M\oplus(N\oplus O),
\quad
(m,n,o)\mapsto(m,n,o).
$$

**Definition 17.8** Let $R$ be a ring. The direct-sum module

$$
R^n
:=
R\oplus\cdots\oplus R
$$

is called the **free $R$-module of rank $n$**.

## §17.5 Universal Property of Free Modules

**Question:** Why is it called a free $R$-module?

**Proposition 17.3** Let $M$ be an $R$-module.

Any $n$-tuple of elements

$$
x_1,\ldots,x_n\in M
$$

uniquely determines an $R$-module homomorphism

$$
X:R^n\to M
$$

defined by

$$
(0,\ldots,0,1,0,\ldots,0)
\mapsto
x_i,
$$

where the $i$th coordinate is $1$.

**Remark** This is the same property as that of a free group on $n$ generators: any $n$-tuple of elements of a group $G$ uniquely determines a map

$$
F_n\to G.
$$

Proof: Given

$$
(x_1,\ldots,x_n),
$$

define

$$
X:R^n\to M
$$

by

$$
X(a_1,\ldots,a_n)
:=
a_1x_1+\cdots+a_nx_n
\in M.
$$

This is a group homomorphism because

$$
\begin{aligned}
X((a_1,\ldots,a_n)+(b_1,\ldots,b_n))
&=
(a_1+b_1)x_1
+\cdots
+
(a_n+b_n)x_n\\
&=
(a_1x_1+\cdots+a_nx_x)
+
(b_1x_1+\cdots+b_nx_n)\\
&=
X(a_1,\ldots,a_n)
+
X(b_1,\ldots,b_n).
\end{aligned}
$$

The middle equality uses the fact that $M$ is an $R$-module.

It is also an $R$-module homomorphism because

$$
\begin{aligned}
X(r(a_1,\ldots,a_n))
&=
X((ra_1,\ldots,ra_n))\\
&=
(ra_1)x_1+\cdots+(ra_n)x_n\\
&=
r(a_1x_1+\cdots+a_nx_n)\\
&=
rX((a_1,\ldots,a_n)).
\end{aligned}
$$

Again, the penultimate equality uses the fact that $M$ is an $R$-module.

$$
~\tag*{$\square$}
$$

## §17.6 The Free Module on 0 Generators

**Exercise** Let $M$ be a left $R$-module. Prove that

$$
r0_M=0_M,
\quad\text{and}\quad
r(-x)=-rx.
$$

Proof: An action of $R$ on $M$ is equivalent to a ring homomorphism

$$
R\to\mathrm{End}(M).
$$

In particular, every element
$r\in R$
determines an Abelian-group homomorphism.

Therefore, scaling by $r$ preserves the additive identity and additive inverses of $M$.

If you prefer a more computational proof, observe that

$$
r0_M+r0_M
=
r(0_M+0_M)
=
r0_M.
$$

Therefore, by the cancellation law in the Abelian group, we may subtract
$r0_M$
from both sides to obtain

$$
r0_M
=
0_M.
$$

Hence

$$
r(-x)+rx
=
r(-x+x)
=
r0_M
=
0_M.
$$

This shows that
$r(-x)$
is the additive inverse of
$rx$.

$$
~\tag*{$\square$}
$$

**Remark** We understand

$$
R^{\oplus n}
$$

for
$n\geqslant1$.

But what about

$$
n=0?
$$

We should look for an $R$-module

$$
R^{\oplus0}
$$

such that there is a bijection

$$
\mathrm{Hom}_R(R^{\oplus0},M)
\cong
\mathrm{Map}_{\text{Sets}}(\varnothing,M).
$$

But there is a unique function from the empty set to any set.

Therefore, we must find an $R$-module

$$
R^{\oplus0}
$$

such that there is exactly one module homomorphism from it to every module $M$.

The unique module with this property is the zero module, i.e. the trivial Abelian group satisfying

$$
r0=0.
$$

§ DS-17 模

§ DS-17 Module

# §16 域
## §16.1 域的基本概念
**定义 16.1** 一个交换环称为**域(field)**，如果 $R-\{0\}$ 在乘法下是一个群。



**例 16.1** $\mathbb{R}, \mathbb{Q}, \mathbb{C}$ 是域，因为每个非零元素都有一个乘法逆元。

**例 16.2** $\mathbb{Z}$ 不是一个域，因为任何不是 $\pm 1$ 的整数都没有乘法逆元。
## §16.2 子域
## §16.3 素域
## §16.4 域的特征
## §16.5 域扩张

# §16 Fields

## §16.1 Basic Concepts of Fields

**Definition 16.1** A commutative ring is called a **field** if

$$
R-\{0\}
$$

is a group under multiplication.



**Example 16.1** $\mathbb{R}$, $\mathbb{Q}$, and $\mathbb{C}$ are fields because every nonzero element has a multiplicative inverse.

**Example 16.2** $\mathbb{Z}$ is not a field because any integer other than $\pm1$ has no multiplicative inverse.

## §16.2 Subfields

## §16.3 Prime Fields

## §16.4 Characteristic of a Field

## §16.5 Field Extensions

§ DS-16 域

§ DS-16 Field

# §15理想与商环
## §15.1 理想
你可能会这样想：如果 $S \subset R$ 是一个“正常”的子环，那么 $R/S$ 将是某个环。这是对群的盲目类比。然而，这个类比是错误的。

**定义 15.1** 设 $R$ 是一个交换环。一个子集 $I \subset R$ 被称为一个**理想(ideal)**，如果

(1) $I$ 在加法下是一个子群，并且

(2) $x \in I$ 蕴含 $rx \in I$ 对于所有 $r \in R$。



**注** 注意 (2) 表明如果 $x, y \in I$，那么 $xy \in I$。所以看起来像是一个关于成为子对象的闭合条件。但是 $I$ 不一定包含 $R$ 的乘法单位元，因此 $I$ 绝对不是一个子环。从启发性的角度看，(2) 实际上是说 $I$ 通过乘法将 $R$ 的每个元素都吸引到 $I$ 中。

**命题 15.1** 对于每个非零整数 $n$，令 $n\mathbb{Z} \subset \mathbb{Z}$ 为那些是 $n$ 的倍数的整数。$n\mathbb{Z}$ 是环 $\mathbb{Z}$ 中的一个理想。

证明：(1) $n\mathbb{Z}$ 包含 $0$，并且如果两个数可被 $n$ 整除，那么它们的和也可以。类似地，如果 $a$ 可被 $n$ 整除，那么 $-a$ 也可以。因此 $n\mathbb{Z}$ 在加法下是一个子群。

(2) 最后，如果 $r$ 是任意整数且 $x$ 可被 $n$ 整除，那么 $rx$ 也可被 $n$ 整除。
$$
~\tag*{$\square$}
$$

**注** 由于 $R$ 是Abel群，注意任何子群 $I$ 都是正规子群。因此存在一个Abel群 $R/I$。

**命题 15.2** 设 $R$ 是一个交换环，$I \subset R$ 是一个理想。那么运算
$$
\times: R/I \times R/I \to R/I,\qquad \overline{r} \cdot \overline{s} = \overline{rs}
$$
连同 $R/I$ 上的通常加法，使得 $R/I$ 成为一个交换环。

证明：我们需要证明此运算不依赖于代表元 $r \in \overline{r}$ 和 $s \in \overline{s}$ 的选择。

令 $r^{\prime} = r + x$ 和 $s^{\prime} = s + y$，其中 $x, y \in I$。（这意味着 $\overline{r^{\prime}} = \overline{r} \in R/I$，且 $\overline{s^{\prime}} = \overline{s} \in R/I$。）

则
$$
r^{\prime}s^{\prime} = (r + x)(s + y) = rs + xs + ry + xy.
$$
注意最后三项在 $I$ 中，因为 $I$ 是一个理想，因此它们的和也在 $I$ 中，因为 $I$ 是一个子群。所以 $\overline{r^{\prime}s^{\prime}} = \overline{rs}$。也就是说，该运算是良定义的。

我们已经知道 $(R/I, +)$ 是一个Abel群。因此我们需要证明 $(R/I, \times)$ 是一个Abel幺半群，并且乘法对加法分配。

首先，乘法是结合的，因为
$$
(\overline{a}\overline{b})\overline{c} = \overline{ab}\overline{c} = \overline{(ab)c} = \overline{a(bc)} = \overline{a}(\overline{b}\overline{c}).
$$
注意这里关键的一步是使用了 $(R, \times)$ 是结合的这个事实。

乘法是交换的，因为
$$
\overline{a}\overline{b} = \overline{ab} = \overline{ba} = \overline{b}\overline{a}
$$
再次使用了 $(R, \times)$ 是交换的。

乘法单位元是 $\overline{1}$：
$$
\overline{1}\overline{a} = \overline{1a} = \overline{a},\quad \overline{a}\overline{1} = \overline{a1} = \overline{a}.
$$
最后，乘法对加法分配，因为
$$
\overline{a}(\overline{b} + \overline{c}) = \overline{a(b + c)} = \overline{ab + ac} = \overline{a}\overline{b} + \overline{a}\overline{c}.
$$
$$
~\tag*{$\square$}
$$
因此，为了得到新的有趣的环，我们可以寻找理想然后取商环。

**例 15.1** 环 $\mathbb{Z}/n\mathbb{Z}$ 是 $\mathbb{Z}$ 对理想 $I = n\mathbb{Z}$ 的商环。

**例 15.2** $\mathbb{Z} \subset \mathbb{Q}$ 是一个子群，实际上也是一个子环，但它绝不是一个理想。这是因为如果 $x$ 是一个整数且 $r$ 是一个有理数，$rx$ 不一定是一个整数。实际上，子环通常不是理想。

## §15.2 理想与商环的例子
**定义 15.2** 设 $x \in R$ 是一个交换环的元素。由 $x$ 生成的理想是所有形如 $rx$ 的元素的集合，其中 $r \in R$。我们用 $(x)$ 表示这个理想。

**命题 15.3** 这是一个理想。

证明：设 $I = (x)$。$I$ 在加法下是封闭的，因为 $rx + sx = (r+s)x \in I$。它包含加法单位元，因为 $0x = 0$。它包含逆元，因为 $-(rx) = (-r)x$。因此 $I$ 在加法下是一个子群。最后，如果 $s \in R$ 且 $rx \in I$，我们有 $s(rx) = (sr)x \in I$。
$$
~\tag*{$\square$}
$$

**例 15.3** 设 $R = \mathbb{R}[t]$ 为一元多项式环。考虑由多项式 $t^{2}+1$ 生成的理想 $I$。因此
$$
I = \{f(t)~\text{使得}~f(t) = g(t)(t^{2}+1)~\text{对于某个多项式}~g(t)\in \mathbb{R}[t]\}.
$$

那么环 $R/I$ 是什么？

**命题 15.4** 环 $\mathbb{R}[t]/(t^{2}+1)$ 同构于 $\mathbb{C}$。

是不是很酷？通常，当你有一个环 $R$ 并且你通过某个等式对它的多项式环取商时，你实际上是向 $R$ 中“添加”了一个满足这个多项式方程的元素。这就是Galois理论的开端。

## §15.3 理想的几何解释
**问题**：你应该如何理解理想？

代数上：一个在倍数下闭合的子群。$I \subset R$，满足
- $rx \in I$，对所有 $x \in I$，$r \in R$。
- $(I, +) \subset (R, +)$ 子群。

你可能觉得这不太有启发性。因此，

几何上：设 $R = \{\text{从某空间}~X~\text{到}~\mathbb{R}~\text{的连续函数}\}$。这个 $R$ 是一个环，因为
- 连续函数之和是连续的，
- 连续函数之积是连续的，
- 零函数是加法单位元
 $$
 (0 + f)(x) = 0(x) + f(x) = f(x)
 $$
 因此 $0 + f = f$。（类似地，$f = f + 0$。）
- 常数函数 $1: x \mapsto 1_{\mathbb{R}}$ 是乘法单位元
 $$
 (1 \cdot f)(x) = 1(x) \cdot f(x) = 1_{\mathbb{R}} \cdot f(x) = f(x)
 $$
 因此 $1 \cdot f = f$。（类似地，$f \cdot 1 = f$。）
- $-f$ 发送 $x \mapsto -f(x)$，是加法逆元。
- $\begin{aligned}
 f \cdot (g + h): x \mapsto f(x)((g + h)(x)) &= f(x)(g(x) + h(x))\\
 &= f(x)g(x) + f(x)h(x)
 \end{aligned}$，因此 $f(g + h) = fg + fh$。
- 结合律也可以很容易地验证。

好的，因此 $R = \{\text{连续函数}\}$。

令 $Y \subset X$ 是一个子集并定义
$$
I_{Y} = \{\text{函数}~f \in R~\text{使得}~f(y) = 0, \forall y \in Y\}.
$$
即，$I_{Y} = \{\text{在}~Y~\text{上消失的函数}\}$。

**命题 15.5** $I_{Y} \subset R$ 是一个理想。

证明：令 $f_{1}, f_{2} \in I_{Y}$。那么 $\forall y \in Y$，
$$
(f_{1} + f_{2})(y) = f_{1}(y) + f_{2}(y) = 0 + 0 = 0
$$
因此 $f_{1} + f_{2} \in I_{Y}$。类似地，
$$
(-f_{1})(y) = -f_{1}(y) = -0 = 0.
$$
因此 $-f_{1} \in I_{Y}$。（注意这也意味着 $0 \in I_{Y}$。但更直接地，$0(y) = 0$，$\forall y \in Y$，所以 $0 \in I_{Y}$。）所以 $I_{Y} \subset R$ 是一个子群。我们只需要验证它在 $R$ 的缩放下是封闭的。

给定 $g \in R$，$f \in I_{Y}$，
$$
(\underbrace{g \cdot f}_{\in R})(y) = \underbrace{g(y) \cdot f(y)}_{\in R} = g(y) \cdot 0 = 0
$$
因此 $gf \in I_{Y}$。
$$
~\tag*{$\square$}
$$
**结论**：每个子集 $Y \subset X$ 产生一个理想 $I_{Y} \subset R$。

**注** 这个例子应该成为范式的事实并不显然。例如，你如何将 $\mathbb{Z}$ 理解为“某空间 $X$ 上的函数”？理想 $p\mathbb{Z}$ 如何由 $X$ 的某个“子集”给出？

无论如何，这种思维方式对微分几何、数论（想象一下能够讨论素数的几何！）等领域有着巨大的影响。

此外，如果我们有 $Y \subset X$，我们应该能够讨论 $Y$ 上的函数——另一个环！

**哲学**：令 $Y$ 产生理想 $I_{Y}$。那么
$$
\{\text{在}~Y~\text{上的函数}\} \cong R/I_{Y}。
$$

**例 15.4 （我们偏离所有连续函数，只考察多项式函数。）**
设 $R = \mathbb{R}[x,y] = \{\text{在}~\mathbb{R}^{2}~\text{上的多项式函数}\}$。令 $Y = \{(x,y)~\text{使得}~y^{2} - x = 0\}$。那么
$$
I_{Y} = \{\text{多项式}~f~\text{使得}~f(y^{2}, y) = 0\} \xlongequal{\text{不显然}} (y^{2} - x)
$$
其中 $(y^{2} - x)$ 是由元素 $y^{2} - x \in R$ 生成的理想。大致来说，如果 $f(x, y)$ 在 $Y$ 上消失，它必须由 $y^{2} - x$ 因式分解。然后
$$
\{\text{在}~Y~\text{上的代数/多项式函数}\} \cong R/(y^{2} - x).
$$
那么为什么？如果 $f_{1}$，$f_{2}$ 是 $X$ 上的函数，它们限制在 $Y$ 上。但
$$
f_{1}(y)= f_{2}(y),\quad \forall y \in Y
$$
$\Leftrightarrow$ 
$$
f_{1}(y) - f_{2}(y) = 0,\quad \forall y \in Y
$$
$\Leftrightarrow$ 
$$
f_{1} - f_{2} \in I_{Y}
$$
即，$f_{1}$ 和 $f_{2}$ 定义了相同的 $Y$ 上的函数当且仅当 $[f_{1}] = [f_{2}] \in R/I_{Y}$。

# §15 Ideals and Quotient Rings

## §15.1 Ideals

You might think as follows: if $S\subset R$ is a “normal” subring, then $R/S$ should be some kind of ring. This is a blind analogy with groups. However, this analogy is wrong.

**Definition 15.1** Let $R$ be a commutative ring. A subset $I\subset R$ is called an **ideal** if

(1) $I$ is a subgroup under addition, and

(2) $x\in I$ implies $rx\in I$ for every $r\in R$.



**Remark** Notice that (2) implies that if $x,y\in I$, then

$$
xy\in I.
$$

So this looks like a closure condition for being a subobject.

However, $I$ need not contain the multiplicative identity of $R$, so $I$ is certainly not necessarily a subring.

Heuristically, condition (2) really says that $I$ absorbs every element of $R$ under multiplication.

**Proposition 15.1** For every nonzero integer $n$, let

$$
n\mathbb{Z}\subset\mathbb{Z}
$$

be the set of integers divisible by $n$.

Then

$$
n\mathbb{Z}
$$

is an ideal in the ring
$\mathbb{Z}$.

Proof:

(1)
$n\mathbb{Z}$
contains $0$, and if two numbers are divisible by $n$, then their sum is also divisible by $n$.

Similarly, if $a$ is divisible by $n$, then $-a$ is also divisible by $n$.

Therefore,
$n\mathbb{Z}$
is a subgroup under addition.

(2) Finally, if $r$ is any integer and $x$ is divisible by $n$, then

$$
rx
$$

is also divisible by $n$.

$$
~\tag*{$\square$}
$$

**Remark** Since $R$ is an Abelian group, every subgroup $I$ is normal.

Therefore there exists an Abelian group

$$
R/I.
$$

**Proposition 15.2** Let $R$ be a commutative ring and let

$$
I\subset R
$$

be an ideal.

Then the operation

$$
\times:
R/I\times R/I
\to
R/I,
\qquad
\overline{r}\cdot\overline{s}
=
\overline{rs}
$$

together with the usual addition on $R/I$, makes $R/I$ into a commutative ring.

Proof: We need to prove that this operation does not depend on the choice of representatives

$$
r\in\overline{r}
$$

and

$$
s\in\overline{s}.
$$

Let

$$
r^{\prime}=r+x
$$

and

$$
s^{\prime}=s+y,
$$

where

$$
x,y\in I.
$$

This means

$$
\overline{r^{\prime}}
=
\overline{r}
\in R/I,
$$

and

$$
\overline{s^{\prime}}
=
\overline{s}
\in R/I.
$$

Then

$$
r^{\prime}s^{\prime}
=
(r+x)(s+y)
=
rs+xs+ry+xy.
$$

Notice that the last three terms lie in $I$ because $I$ is an ideal.

Therefore their sum also lies in $I$, since $I$ is a subgroup.

Hence

$$
\overline{r^{\prime}s^{\prime}}
=
\overline{rs}.
$$

Thus the operation is well defined.

We already know that

$$
(R/I,+)
$$

is an Abelian group.

Therefore, we need to prove that

$$
(R/I,\times)
$$

is an Abelian monoid and that multiplication distributes over addition.

First, multiplication is associative because

$$
(\overline{a}\overline{b})\overline{c}
=
\overline{ab}\,\overline{c}
=
\overline{(ab)c}
=
\overline{a(bc)}
=
\overline{a}(\overline{b}\overline{c}).
$$

The key step here uses the fact that

$$
(R,\times)
$$

is associative.

Multiplication is commutative because

$$
\overline{a}\overline{b}
=
\overline{ab}
=
\overline{ba}
=
\overline{b}\overline{a},
$$

again using the commutativity of
$(R,\times)$.

The multiplicative identity is

$$
\overline{1}:
$$

$$
\overline{1}\,\overline{a}
=
\overline{1a}
=
\overline{a},
\qquad
\overline{a}\,\overline{1}
=
\overline{a1}
=
\overline{a}.
$$

Finally, multiplication distributes over addition because

$$
\overline{a}
(\overline{b}+\overline{c})
=
\overline{a(b+c)}
=
\overline{ab+ac}
=
\overline{a}\,\overline{b}
+
\overline{a}\,\overline{c}.
$$

$$
~\tag*{$\square$}
$$

Thus, in order to obtain new and interesting rings, we can look for ideals and then take quotient rings.

**Example 15.1** The ring

$$
\mathbb{Z}/n\mathbb{Z}
$$

is the quotient of
$\mathbb{Z}$
by the ideal

$$
I=n\mathbb{Z}.
$$

**Example 15.2**

$$
\mathbb{Z}\subset\mathbb{Q}
$$

is a subgroup and is in fact also a subring, but it is certainly not an ideal.

Indeed, if $x$ is an integer and $r$ is a rational number, then

$$
rx
$$

need not be an integer.

In fact, subrings are usually not ideals.

## §15.2 Examples of Ideals and Quotient Rings

**Definition 15.2** Let

$$
x\in R
$$

be an element of a commutative ring.

The ideal generated by $x$ is the set of all elements of the form

$$
rx,
$$

where
$r\in R$.

We denote this ideal by

$$
(x).
$$

**Proposition 15.3** This is an ideal.

Proof: Let

$$
I=(x).
$$

$I$ is closed under addition because

$$
rx+sx
=
(r+s)x
\in I.
$$

It contains the additive identity because

$$
0x=0.
$$

It contains additive inverses because

$$
-(rx)
=
(-r)x.
$$

Thus $I$ is a subgroup under addition.

Finally, if

$$
s\in R
$$

and

$$
rx\in I,
$$

then

$$
s(rx)
=
(sr)x
\in I.
$$

$$
~\tag*{$\square$}
$$

**Example 15.3** Let

$$
R=\mathbb{R}[t]
$$

be the polynomial ring in one variable.

Consider the ideal $I$ generated by the polynomial

$$
t^2+1.
$$

Thus

$$
I
=
\{
f(t)
~\text{such that}~
f(t)=g(t)(t^2+1)
~\text{for some polynomial}~
g(t)\in\mathbb{R}[t]
\}.
$$

What is the ring

$$
R/I?
$$

**Proposition 15.4** The ring

$$
\mathbb{R}[t]/(t^2+1)
$$

is isomorphic to

$$
\mathbb{C}.
$$

Isn't that cool?

In general, when you have a ring $R$ and quotient its polynomial ring by some polynomial equation, you are effectively “adjoining” to $R$ an element satisfying that polynomial equation.

This is the beginning of Galois theory.

## §15.3 Geometric Interpretation of Ideals

**Question:** How should you think about ideals?

Algebraically: an ideal is a subgroup closed under multiplication by arbitrary elements of the ring.

That is,

$$
I\subset R,
$$

with

* $rx\in I$ for every

$$
x\in I,
\qquad
r\in R;
$$

* $(I,+)\subset(R,+)$ is a subgroup.

You may find this rather uninspiring.

So let us look at the geometric interpretation.

Let

$$
R
=
\{
\text{continuous functions from some space }X\text{ to }\mathbb{R}
\}.
$$

This $R$ is a ring because

* the sum of continuous functions is continuous;

* the product of continuous functions is continuous;

* the zero function is the additive identity:

$$
(0+f)(x)
=
0(x)+f(x)
=
f(x),
$$

so

$$
0+f=f.
$$

Similarly,

$$
f=f+0.
$$

* The constant function

$$
1:x\mapsto1_{\mathbb{R}}
$$

is the multiplicative identity:

$$
(1\cdot f)(x)
=
1(x)\cdot f(x)
=
1_{\mathbb{R}}\cdot f(x)
=
f(x),
$$

so

$$
1\cdot f=f.
$$

Similarly,

$$
f\cdot1=f.
$$

* The function $-f$ sends

$$
x\mapsto-f(x)
$$

and is the additive inverse of $f$.

* We have

$$
\begin{aligned}
f\cdot(g+h):
x\mapsto
f(x)((g+h)(x))
&=
f(x)(g(x)+h(x))\\
&=
f(x)g(x)+f(x)h(x),
\end{aligned}
$$

so

$$
f(g+h)
=
fg+fh.
$$

* Associativity is also easy to verify.

Thus

$$
R
=
\{\text{continuous functions}\}.
$$

Let

$$
Y\subset X
$$

be a subset and define

$$
I_Y
=
\{
\text{functions }f\in R
~\text{such that}~
f(y)=0,
\forall y\in Y
\}.
$$

That is,

$$
I_Y
=
\{
\text{functions vanishing on }Y
\}.
$$

**Proposition 15.5**

$$
I_Y\subset R
$$

is an ideal.

Proof: Let

$$
f_1,f_2\in I_Y.
$$

Then for every

$$
y\in Y,
$$

$$
(f_1+f_2)(y)
=
f_1(y)+f_2(y)
=
0+0
=
0.
$$

Therefore,

$$
f_1+f_2
\in
I_Y.
$$

Similarly,

$$
(-f_1)(y)
=
-f_1(y)
=
-0
=
0.
$$

Hence

$$
-f_1\in I_Y.
$$

Notice that this also implies

$$
0\in I_Y.
$$

More directly,

$$
0(y)=0,
\qquad
\forall y\in Y,
$$

so

$$
0\in I_Y.
$$

Therefore,

$$
I_Y\subset R
$$

is a subgroup.

We only need to verify that it is closed under multiplication by elements of $R$.

Given

$$
g\in R,
\qquad
f\in I_Y,
$$

we have

$$
(\underbrace{g\cdot f}_{\in R})(y)
=
\underbrace{g(y)\cdot f(y)}_{\in R}
=
g(y)\cdot0
=
0.
$$

Therefore,

$$
gf\in I_Y.
$$

$$
~\tag*{$\square$}
$$

**Conclusion:** Every subset

$$
Y\subset X
$$

gives rise to an ideal

$$
I_Y\subset R.
$$

**Remark** It is far from obvious that this example should be viewed as a paradigm.

For example, how should one interpret

$$
\mathbb{Z}
$$

as “functions on some space $X$”?

How is the ideal

$$
p\mathbb{Z}
$$

given by some “subset” of $X$?

In any case, this way of thinking has had an enormous influence on areas such as differential geometry and number theory.

Imagine being able to discuss the geometry of prime numbers!

Moreover, if

$$
Y\subset X,
$$

then we should be able to talk about functions on $Y$—another ring.

**Philosophy:** Let $Y$ give rise to the ideal

$$
I_Y.
$$

Then

$$
\{
\text{functions on }Y
\}
\cong
R/I_Y.
$$

**Example 15.4 (We now move away from all continuous functions and consider only polynomial functions.)**

Let

$$
R
=
\mathbb{R}[x,y]
=
\{
\text{polynomial functions on }\mathbb{R}^2
\}.
$$

Let

$$
Y
=
\{
(x,y)
~\text{such that}~
y^2-x=0
\}.
$$

Then

$$
I_Y
=
\{
\text{polynomials }f
~\text{such that}~
f(y^2,y)=0
\}
\xlongequal{\text{not obvious}}
(y^2-x),
$$

where

$$
(y^2-x)
$$

is the ideal generated by the element

$$
y^2-x\in R.
$$

Roughly speaking, if

$$
f(x,y)
$$

vanishes on $Y$, then it must factor through

$$
y^2-x.
$$

Then

$$
\{
\text{algebraic/polynomial functions on }Y
\}
\cong
R/(y^2-x).
$$

Why is this?

If
$f_1$
and
$f_2$
are functions on $X$, then they restrict to functions on $Y$.

But

$$
f_1(y)
=
f_2(y),
\qquad
\forall y\in Y
$$

$\Leftrightarrow$

$$
f_1(y)-f_2(y)
=
0,
\qquad
\forall y\in Y
$$

$\Leftrightarrow$

$$
f_1-f_2
\in
I_Y.
$$

That is,
$f_1$
and
$f_2$
define the same function on $Y$ if and only if

$$
[f_1]
=
[f_2]
\in
R/I_Y.
$$

§ DS-15 理想和商环

§ DS-15 Ideal and Quotient

# §14 环
## §14.1 环的定义
**定义 14.1** 一个**幺半群(monoid)** 是一个没有逆元的群。也就是说，幺半群是一个集合$M$，与之配合一个函数
$$
\cdot: M\times M\to M
$$
该函数具有单位元并且是结合的。我们称一个幺半群是**交换的(commutative)**，如果对所有$a, b\in M$，有$ab=ba$。



**定义 14.2** 一个**结合环(associative ring)** 是一个三元组$(R,+, \cdot)$，其中$R$是一个集合，且

(1) $+: R \times R \rightarrow R$是一个函数，使$(R,+)$成为一个Abel群。我们称此运算为加法，其单位元为$0$。

(2) $\cdot: R \times R \rightarrow R$是一个函数，使$R$成为一个幺半群。我们称$\cdot$运算为乘法，其单位元为$1$。

(3) 最后，我们要求乘法对加法是分配的：这意味着对所有$a, b, c \in R$，有
$$
a(b+c)=a b+a c \quad \text{且} \quad(b+c) a=b a+c a
$$
其中我们将$a \cdot b$写作$a b$。

我们通常只写$R$表示环，而隐去运算$+$和$\cdot$。

**定义 14.3** 如果$(R, \cdot)$是一个abel幺半群，我们称$R$为**交换环(commutative ring)**。

在讨论群时，我很快证明了消去律，因为它是有用的知识。这里是另一个有用的知识：

**命题 14.1** 设$R$是一个结合环，$0$是$R$的加法单位元。那么
$$
0\cdot a=a\cdot 0=0
$$
对所有$a\in R$成立。

证明：
$$
\begin{align*}
0\cdot a&=(0+0)\cdot a\tag{因为$0+0=0$}\\
&=0\cdot a+0\cdot a\tag{分配律}
\end{align*}
$$
利用Abel群的消去律，我们可以从等式的两边减去$0\cdot a$。剩下$0=0\cdot a$。类似地，$a\cdot 0=0$。
$$
~\tag*{$\square$}
$$
**注** 我们不会深入探讨原因，但环的行为在它们是否交换方面有着极大的不同。

## §14.2 交换环的例子
**例 14.1** 考虑三元组$(\mathbb{Z},+,\cdot)$。这使得$(\mathbb{Z},+)$成为阿贝尔群，而$(\mathbb{Z},\cdot)$显然是一个幺半群——乘法有一个称为$1$的单位元并且是结合的。分配律是你熟悉的分配律。

**例 14.2** 使用通常的加法和乘法，三元组$(\mathbb{Q},+,\cdot)$是一个环。对于$\mathbb{R}$和$\mathbb{C}$及其通常的乘法也是如此。这些是特殊类型的环，因为$R-\{0\}$中的任一元素在乘法下都有逆元。

**例 14.3 (多项式环)**
令$\mathbb{Z}[x]$表示系数为整数的$x$的多项式的集合。因此，一个元素是一个表达式
$$
p(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}=\sum\limits_{i=0}^{\infty}a_{i}x^{i}
$$
其中对于大于某个有限$n$的所有$i$，$a_{i}=0$。例如，以下是$\mathbb{Z}[x]$中的元素：
$$
0, 5, 3+x-5x^{4}, x。
$$
在第三个例子中，$a_{0}=3$，$a_{1}=1$，$a_{2}=0$，$a_{3}=0$，$a_{4}=-5$。

设$q(x)$是一个系数为$b_{i}$的多项式。多项式的加法定义如下
$$
p(x)+q(x):=\sum\limits_{i=0}^{\infty}(a_{i}+b_{i})x^{i}。
$$
注意，因为对于大于$n$的$i$，$a_{i}=0$，且对于大于某个$m$的所有$i$，$b_{i}=0$，因此和确实是一个多项式，因为$(a_{i}+b_{i})=0$对于所有大于$\max(m,n)$的$i$成立。

两个多项式的积定义为通常的方式
$$
\begin{aligned}
p(x)\cdot q(x)&=(a_{0}+a_{1}x+\cdots+a_{n}x^{n})(b_{0}+b_{1}x+\cdots+b_{m}x^{m})\\
&=a_{0}b_{0}+(a_{1}b_{0}+a_{0}b_{1})x+\cdots+a_{n}b_{m}x^{n+m}\\
&=\sum\limits_{k\geqslant 0}\left(\sum\limits_{i+j=k}a_{i}b_{j}\right)x^{k}.
\end{aligned}
$$

**命题 14.2** $\mathbb{Z}[x]$是一个交换环。更一般地说，如果$R$是一个交换环，那么系数在$R$中的多项式集合$R[x]$是一个交换环。

证明：
1. $R[x]$是一个环，因为

(1) 加法是封闭的：对于$R[x]$中的两个多项式$p(x)$和$q(x)$，它们的和$p(x)+q(x)$是一个多项式，其系数为$p(x)$和$q(x)$的相应系数之和。由于$R$是交换环并在加法下封闭，因此该运算在$R[x]$中是良定义的。

(2) 乘法是封闭的：对于两个多项式$p(x)$和$q(x)$，它们的积$p(x)\cdot q(x)$表示为
$$
p(x)\cdot q(x)=\left(\sum\limits_{i=0}^{n}a_{i}x^{i}\right)\left(\sum\limits_{j=0}^{m}b_{j}x^{j}\right)=\sum\limits_{k=0}^{n+m}c_{k}x^{k}
$$
其中系数$c_{k}$计算为
$$
c_{k}=\sum\limits_{i+j=k}a_{i}b_{j}。
$$
因为$R$在乘法下封闭，所以系数$c_{k}$是$R$的元素，因此积是$R[x]$中的多项式。

(3) 加法是结合且交换的：$R[x]$中的多项式加法是结合且交换的，因为$R$中的加法是结合且交换的，并且对各个系数的操作遵循$R$的性质。

(4) 乘法是结合的：$R[x]$中的多项式乘法是结合的，因为分配律成立，且$R$中的乘法是结合的。因此对于$R[x]$中的多项式$p(x), q(x)$和$r(x)$，有
$$
(p(x)\cdot q(x))\cdot r(x)=p(x)\cdot(q(x)\cdot r(x)).
$$

(5) 分配律：乘法对加法是分配的，因为对于$R[x]$中的多项式$p(x), q(x)$和$r(x)$，有
$$
p(x)\cdot(q(x)+r(x))=p(x)\cdot q(x)+p(x)\cdot r(x).
$$
这是因为分配律适用于$R$中的个别系数。

2. 乘法的交换性：对于任何多项式$p(x), q(x)\in R[x]$，有
$$
p(x)\cdot q(x)=q(x)\cdot p(x)。
$$
这是因为$R$中的乘法是交换的。特别地，如果$p(x)=\sum\limits_{i=0}^{n}a_{i}x^{i}$且$q(x)=\sum\limits_{j=0}^{m}b_{j}x^{j}$，则积为
$$
p(x)\cdot q(x)=\sum\limits_{k=0}^{n+m}\sum\limits_{i+j=k}a_{i}b_{j}x^{k}.
$$
由于$R$中的乘法是交换的，有$a_{i}b_{j}=b_{j}a_{i}$，因此
$$
p(x)\cdot q(x)=q(x)\cdot p(x).
$$
$$
~\tag*{$\square$}
$$

**例 14.4 (光滑函数)**
这是一个更难写出集合中所有元素的例子。令$C^{\infty}(\mathbb{R}^{n})$表示从$\mathbb{R}^{n}$到$\mathbb{R}$的所有无穷可微函数的集合。给定两个函数$f$和$g$，定义它们的和$f+g$为一个函数，该函数将$x\in\mathbb{R}^{n}$映射到$f(x)+g(x)$，其中加法发生在$\mathbb{R}$中。它们的乘积$fg$是一个将$x\in\mathbb{R}^{n}$映射到$f(x)\cdot g(x)\in\mathbb{R}$的函数。这也是一个环。

## §14.3 环$\mathbb{Z}/n\mathbb{Z}$
我们已经看到了三个我们熟悉的环的例子。它们都是无限的。现在来看一些有限的例子。

**引理 14.3** 令$\mathbb{Z}/n\mathbb{Z}$为整数模$n$的集合。函数
$$
\cdot: \mathbb{Z}/n\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z},\qquad \overline{a}\cdot\overline{b}:=\overline{a\cdot b}
$$
是良定义的。

**注** 我们用$\overline{a}$表示与$a\in\mathbb{Z}$相关联的等价类。这里，$a\cdot b$是整数的通常乘法，而$\overline{a\cdot b}$表示模$n$的等价类。

证明：我们需要证明，如果$\overline{a}=\overline{a}^{\prime}$且$\overline{b}=\overline{b}^{\prime}$，那么
$$
\overline{a\cdot b}=\overline{a^{\prime}\cdot b^{\prime}}。
$$
因为$a=a^{\prime}$模$n$当且仅当$a=a^{\prime}+An$（其中$A$是整数）。同样，$b=b^{\prime}+Bn$对于某个$B$。因此
$$
ab=(a^{\prime}+An)(b^{\prime}+Bn)=a^{\prime}b^{\prime}+(a^{\prime}B+Ab^{\prime}+AB)n
$$
因此$ab$等于$a^{\prime}b^{\prime}$模$n$。
$$
~\tag*{$\square$}
$$

**推论 14.4** 令$+$为$\mathbb{Z}/n\mathbb{Z}$上的通常加法，$\cdot$为上述运算。那么

(1) $(\mathbb{Z}/n\mathbb{Z},+)$是一个单位元为$\overline{0}$的Abel群。

(2) $(\mathbb{Z}/n\mathbb{Z},\cdot)$是一个单位元为$\overline{1}$的Abel幺半群。

(3) 运算$\cdot$对$+$是分配的。

证明：(1) 是我们早已证明的结论。

(2) 为了证明结合性，注意到
$$
\begin{aligned}
(\overline{a}\cdot\overline{b})\cdot\overline{c}&=\overline{a\cdot b}\cdot\overline{c}\\
&=\overline{(a\cdot b)\cdot c}\\
&=\overline{a\cdot(b\cdot c)}\\
&=\overline{a}\cdot\overline{b\cdot c}\\
&=\overline{a}\cdot(\overline{b}\cdot\overline{c}).\tag{14.1}
\end{aligned}
$$
除(14.1)外，每一行都是基于$\cdot$的定义，而(14.1)使用了整数乘法的结合性。交换性成立是因为
$$
\overline{a}\cdot\overline{b}=\overline{a\cdot b}=\overline{b\cdot a}=\overline{b}\cdot\overline{a}。
$$
注意中间等号只是整数乘法的交换性。单位元是$1$，因为$\overline{a}\cdot\overline{1}=\overline{a\cdot 1}=\overline{a}$。

(3) 分配律成立是因为
$$
\begin{aligned}
\overline{a}\cdot(\overline{b}+\overline{c})&=\overline{a}\cdot\overline{b+c}\\
&=\overline{a\cdot(b+c)}\\
&=\overline{ab+ac}\\
&=\overline{ab}+\overline{ac}\\
&=\overline{a}\cdot\overline{b}+\overline{a}\cdot\overline{c}.\tag{14.2}
\end{aligned}
$$
除了(14.2)，其他各行均基于定义，而(14.2)使用了通常整数的分配律。
$$
~\tag*{$\square$}
$$
## §14.4 交换环的动机
现代数学中有一种哲学认为，可以通过研究空间上的函数集的性质来推断空间本身的性质。例如，通过研究空间$X$上的多项式函数集，可以推断出空间$X$的某些性质。实际上，所有函数集始终是一个交换环。这些环的性质决定了空间$X$的某些特性。这并不是显而易见的，其中一些最有意义的相关发展直到19世纪80年代才出现——也就是在Descartes首先注意到代数方程可以描述具体几何近两百年后。因此，如果你认为代数（从伊斯兰黄金时期的800年代开始）和几何（源自希腊人）在Descartes将其结合起来之前经过了近千年，而我们又花了两百年才系统地理解环是研究几何的有力工具，你可能会意识到自己正在接触一些非常深刻的思想。我们无法深入探讨利用环研究几何的理论——但是如果你感兴趣，可以查阅一些关于交换代数和代数几何的资料。
## §14.5 非交换环的例子
**例 14.5 (矩阵环)**
确定一个整数$n\geqslant 0$，考虑所有元素为$\mathbb{R}$的$n\times n$矩阵的集合$M_{n\times n}(\mathbb{R})$。你可以对矩阵进行加法和乘法运算，并且矩阵乘法对加法是分配的。因此$M_{n\times n}(\mathbb{R})$是一个环。为了明确分配律，考虑三个矩阵$A, B, C$，其元素为$a_{ij}, b_{ij}, c_{ij}$。则$A(B+C)$的$ij$项为
$$
\sum\limits_{k=1}^{n}a_{ik}(b_{kj}+c_{kj})=\sum\limits_{k=1}^{n}a_{ik}b_{kj}+\sum\limits_{k=1}^{n}a_{ik}c_{kj}
$$
但这个右侧是$AB+AC$的$ij$项。你也可以类似地证明$(B+C)A=BA+CA$。

**例 14.6 (群环)**
设$G$是有限群，$R$是一个交换环。那么$R[G]$作为一个集合，是从$G$到$R$的所有函数的集合。（因此每个元素$g\in G$都对应一个$R$中的元素$r_{g}$。）我们用求和表示这样的一个函数：
$$
\sum\limits_{g\in G}r_{g}g。
$$
例如，下面是$\mathbb{Z}[S_{3}]$的一个元素：
$$
5(~)+3(12)-8(123)。
$$
加法是显然的——我们仅对每一项相加，因此
$$
\left(\sum r_{g}g\right)+\sum\left(s_{g}g\right)=\sum\limits_{g\in G}(r_{g}+s_{g})g。
$$
即，这是函数的加法。乘法不仅仅是函数的乘法。$\sum r_{g} g$和$\sum s_{g} g$的乘积在$g$的系数由以下公式给出
$$
\sum\limits_{(g_{1},g_{2})~\text{s.t.}~g_{1}g_{2}=g}r_{g_{1}}s_{g_{2}}.
$$
换句话说，
$$
\left(\sum\limits_{g\in G}r_{g}g\right)\left(\sum\limits_{h\in G}s_{h}h\right)=\sum\limits_{g\in G}\sum\limits_{h\in G}r_{g}s_{h}(gh)=\sum\limits_{k\in G}\left(\sum\limits_{(g_{1},g_{2})~\text{s.t.}~g_{1}g_{2}=g} r_{g}s_{h}\right)k.
$$
注意，这种乘法不是交换的。

## §14.6 环同态
**定义 14.4** 设$R$和$S$是环，且$f: R\to S$是一个函数。我们称$f$为一个**环同态(ring homomorphism)**，如果

(1) $f$是加法的群同态，

(2) $f(1)=1$（即$f$将$R$的乘法单位元映射到$S$的单位元），并且

(3) 对所有$a,b\in R$有$f(ab)=f(a)f(b)$。

**定义 14.5** 如果$f$是一个双射，我们称$f$是一个**同构(isomorphism)**。

现在我想进一步说明为什么$\mathbb{Z}/n\mathbb{Z}$是一个环。我们如何证明它是一个群的？通过应用一个普遍原则：如果$H\triangleleft G$，那么$G/H$是一个群。

我想对环做同样的事情。但在这个上下文中，当我们说到环时，我们将意味着一个交换环。

# §14 Rings

## §14.1 Definition of a Ring

**Definition 14.1** A **monoid** is a group without inverses. That is, a monoid is a set $M$ together with a function

$$
\cdot:M\times M\to M
$$

which has an identity element and is associative. We call a monoid **commutative** if

$$
ab=ba
$$

for all
$a,b\in M$.



**Definition 14.2** An **associative ring** is a triple

$$
(R,+,\cdot),
$$

where $R$ is a set and

(1)

$$
+:R\times R\to R
$$

is a function making $(R,+)$ into an Abelian group. We call this operation addition, and its identity element is $0$.

(2)

$$
\cdot:R\times R\to R
$$

is a function making $R$ into a monoid. We call the operation $\cdot$ multiplication, and its identity element is $1$.

(3) Finally, we require multiplication to distribute over addition. This means that for all
$a,b,c\in R$,

$$
a(b+c)=ab+ac
\quad\text{and}\quad
(b+c)a=ba+ca,
$$

where we write
$a\cdot b$
as
$ab$.

We usually write simply $R$ for a ring, suppressing the operations $+$ and $\cdot$.

**Definition 14.3** If $(R,\cdot)$ is an Abelian monoid, we call $R$ a **commutative ring**.

When discussing groups, I quickly proved the cancellation law because it is useful to know. Here is another useful fact.

**Proposition 14.1** Let $R$ be an associative ring, and let $0$ be the additive identity of $R$. Then

$$
0\cdot a=a\cdot0=0
$$

for every
$a\in R$.

Proof:

$$
\begin{align*}
0\cdot a
&=(0+0)\cdot a
\tag{since $0+0=0$}\\
&=0\cdot a+0\cdot a
\tag{distributivity}.
\end{align*}
$$

Using the cancellation law in the Abelian group, we may subtract
$0\cdot a$
from both sides of the equation. This leaves

$$
0=0\cdot a.
$$

Similarly,

$$
a\cdot0=0.
$$

$$
~\tag*{$\square$}
$$

**Remark** We will not explore the reason in detail, but rings behave very differently depending on whether or not they are commutative.

## §14.2 Examples of Commutative Rings

**Example 14.1** Consider the triple

$$
(\mathbb{Z},+,\cdot).
$$

This makes
$(\mathbb{Z},+)$
an Abelian group, while
$(\mathbb{Z},\cdot)$
is clearly a monoid: multiplication has an identity element called $1$ and is associative.

The distributive law is the familiar distributive law.

**Example 14.2** With the usual addition and multiplication, the triple

$$
(\mathbb{Q},+,\cdot)
$$

is a ring.

The same is true of
$\mathbb{R}$
and
$\mathbb{C}$
with their usual multiplication.

These are special kinds of rings because every element of

$$
R-\{0\}
$$

has a multiplicative inverse.

**Example 14.3 (Polynomial Rings)**

Let

$$
\mathbb{Z}[x]
$$

denote the set of polynomials in $x$ with integer coefficients.

Thus an element is an expression

$$
p(x)
=
a_0+a_1x+a_2x^2+\cdots+a_nx^n
=
\sum_{i=0}^{\infty}a_ix^i,
$$

where

$$
a_i=0
$$

for all $i$ larger than some finite $n$.

For example, the following are elements of
$\mathbb{Z}[x]$:

$$
0,\quad5,\quad3+x-5x^4,\quad x.
$$

In the third example,

$$
a_0=3,\qquad
a_1=1,\qquad
a_2=0,\qquad
a_3=0,\qquad
a_4=-5.
$$

Let $q(x)$ be a polynomial with coefficients $b_i$.

Polynomial addition is defined by

$$
p(x)+q(x)
:=
\sum_{i=0}^{\infty}(a_i+b_i)x^i.
$$

Notice that since
$a_i=0$
for
$i>n$,
and
$b_i=0$
for every $i$ larger than some $m$, the sum really is a polynomial because

$$
a_i+b_i=0
$$

for every

$$
i>\max(m,n).
$$

The product of two polynomials is defined in the usual way:

$$
\begin{aligned}
p(x)\cdot q(x)
&=
(a_0+a_1x+\cdots+a_nx^n)
(b_0+b_1x+\cdots+b_mx^m)\\
&=
a_0b_0
+
(a_1b_0+a_0b_1)x
+
\cdots
+
a_nb_mx^{n+m}\\
&=
\sum_{k\geqslant0}
\left(
\sum_{i+j=k}a_ib_j
\right)x^k.
\end{aligned}
$$

**Proposition 14.2**
$\mathbb{Z}[x]$
is a commutative ring.

More generally, if $R$ is a commutative ring, then the set

$$
R[x]
$$

of polynomials with coefficients in $R$ is a commutative ring.

Proof:

1. $R[x]$ is a ring because:

(1) Addition is closed.

For two polynomials
$p(x)$
and
$q(x)$
in
$R[x]$,
their sum

$$
p(x)+q(x)
$$

is a polynomial whose coefficients are the sums of the corresponding coefficients of
$p(x)$
and
$q(x)$.

Since $R$ is a commutative ring and is closed under addition, this operation is well defined in
$R[x]$.

(2) Multiplication is closed.

For two polynomials
$p(x)$
and
$q(x)$,
their product can be written as

$$
p(x)\cdot q(x)
=
\left(
\sum_{i=0}^{n}a_ix^i
\right)
\left(
\sum_{j=0}^{m}b_jx^j
\right)
=
\sum_{k=0}^{n+m}c_kx^k,
$$

where

$$
c_k
=
\sum_{i+j=k}a_ib_j.
$$

Since $R$ is closed under multiplication, each coefficient
$c_k$
is an element of $R$. Thus the product is a polynomial in
$R[x]$.

(3) Addition is associative and commutative.

Polynomial addition in
$R[x]$
is associative and commutative because addition in $R$ is associative and commutative, and the operations on individual coefficients obey the corresponding properties in $R$.

(4) Multiplication is associative.

Polynomial multiplication in
$R[x]$
is associative because the distributive laws hold and multiplication in $R$ is associative.

Thus, for polynomials
$p(x),q(x),r(x)\in R[x]$,

$$
(p(x)\cdot q(x))\cdot r(x)
=
p(x)\cdot(q(x)\cdot r(x)).
$$

(5) Distributivity.

Multiplication distributes over addition because, for
$p(x),q(x),r(x)\in R[x]$,

$$
p(x)\cdot(q(x)+r(x))
=
p(x)\cdot q(x)
+
p(x)\cdot r(x).
$$

This follows from the distributive law for the individual coefficients in $R$.

2. Commutativity of multiplication.

For any polynomials

$$
p(x),q(x)\in R[x],
$$

we have

$$
p(x)\cdot q(x)
=
q(x)\cdot p(x).
$$

This is because multiplication in $R$ is commutative.

In particular, if

$$
p(x)
=
\sum_{i=0}^{n}a_ix^i
$$

and

$$
q(x)
=
\sum_{j=0}^{m}b_jx^j,
$$

then

$$
p(x)\cdot q(x)
=
\sum_{k=0}^{n+m}
\sum_{i+j=k}
a_ib_jx^k.
$$

Since multiplication in $R$ is commutative,

$$
a_ib_j=b_ja_i.
$$

Therefore,

$$
p(x)\cdot q(x)
=
q(x)\cdot p(x).
$$

$$
~\tag*{$\square$}
$$

**Example 14.4 (Smooth Functions)**

Here is an example in which it is more difficult to explicitly list all the elements of the set.

Let

$$
C^{\infty}(\mathbb{R}^{n})
$$

denote the set of all infinitely differentiable functions from
$\mathbb{R}^{n}$
to
$\mathbb{R}$.

Given two functions $f$ and $g$, define their sum
$f+g$
to be the function sending

$$
x\in\mathbb{R}^{n}
$$

to

$$
f(x)+g(x),
$$

where the addition takes place in
$\mathbb{R}$.

Their product
$fg$
is the function sending

$$
x\in\mathbb{R}^{n}
$$

to

$$
f(x)\cdot g(x)\in\mathbb{R}.
$$

This is also a ring.

## §14.3 The Ring $\mathbb{Z}/n\mathbb{Z}$

We have already seen three examples of familiar rings.

They are all infinite.

Now let us look at some finite examples.

**Lemma 14.3** Let

$$
\mathbb{Z}/n\mathbb{Z}
$$

be the set of integers modulo $n$.

The function

$$
\cdot:
\mathbb{Z}/n\mathbb{Z}
\times
\mathbb{Z}/n\mathbb{Z}
\to
\mathbb{Z}/n\mathbb{Z},
\qquad
\overline{a}\cdot\overline{b}
:=
\overline{a\cdot b}
$$

is well defined.

**Remark** We use
$\overline{a}$
to denote the equivalence class associated with
$a\in\mathbb{Z}$.

Here,
$a\cdot b$
is ordinary multiplication of integers, while
$\overline{a\cdot b}$
denotes the corresponding equivalence class modulo $n$.

Proof: We need to prove that if

$$
\overline{a}
=
\overline{a}^{\prime}
$$

and

$$
\overline{b}
=
\overline{b}^{\prime},
$$

then

$$
\overline{a\cdot b}
=
\overline{a^{\prime}\cdot b^{\prime}}.
$$

Since
$a=a^{\prime}$
modulo $n$ if and only if

$$
a=a^{\prime}+An
$$

for some integer $A$.

Similarly,

$$
b=b^{\prime}+Bn
$$

for some integer $B$.

Therefore,

$$
ab
=
(a^{\prime}+An)(b^{\prime}+Bn)
=
a^{\prime}b^{\prime}
+
(a^{\prime}B+Ab^{\prime}+AB)n.
$$

Hence

$$
ab
$$

is equal to

$$
a^{\prime}b^{\prime}
$$

modulo $n$.

$$
~\tag*{$\square$}
$$

**Corollary 14.4** Let $+$ be the usual addition on
$\mathbb{Z}/n\mathbb{Z}$,
and let $\cdot$ be the operation above.

Then:

(1)

$$
(\mathbb{Z}/n\mathbb{Z},+)
$$

is an Abelian group with identity
$\overline{0}$.

(2)

$$
(\mathbb{Z}/n\mathbb{Z},\cdot)
$$

is an Abelian monoid with identity
$\overline{1}$.

(3) The operation
$\cdot$
distributes over
$+$.

Proof:

(1) This is a result we have already proved.

(2) To prove associativity, note that

$$
\begin{aligned}
(\overline{a}\cdot\overline{b})\cdot\overline{c}
&=
\overline{a\cdot b}\cdot\overline{c}\\
&=
\overline{(a\cdot b)\cdot c}\\
&=
\overline{a\cdot(b\cdot c)}\\
&=
\overline{a}\cdot\overline{b\cdot c}\\
&=
\overline{a}\cdot
(\overline{b}\cdot\overline{c}).
\tag{14.1}
\end{aligned}
$$

Every line except (14.1) follows from the definition of $\cdot$, while (14.1) uses associativity of integer multiplication.

Commutativity holds because

$$
\overline{a}\cdot\overline{b}
=
\overline{a\cdot b}
=
\overline{b\cdot a}
=
\overline{b}\cdot\overline{a}.
$$

The middle equality is simply commutativity of integer multiplication.

The identity is $1$ because

$$
\overline{a}\cdot\overline{1}
=
\overline{a\cdot1}
=
\overline{a}.
$$

(3) Distributivity holds because

$$
\begin{aligned}
\overline{a}\cdot
(\overline{b}+\overline{c})
&=
\overline{a}\cdot\overline{b+c}\\
&=
\overline{a\cdot(b+c)}\\
&=
\overline{ab+ac}\\
&=
\overline{ab}+\overline{ac}\\
&=
\overline{a}\cdot\overline{b}
+
\overline{a}\cdot\overline{c}.
\tag{14.2}
\end{aligned}
$$

Except for (14.2), each line follows from the definitions, while (14.2) uses the ordinary distributive law for integers.

$$
~\tag*{$\square$}
$$

## §14.4 Motivation for Commutative Rings

There is a philosophy in modern mathematics that the properties of a space can be inferred by studying the properties of collections of functions on that space.

For example, by studying the collection of polynomial functions on a space $X$, one can infer certain properties of the space $X$ itself.

In fact, collections of functions naturally form commutative rings.

Properties of these rings determine certain features of the space $X$.

This is far from obvious.

Some of the most important developments related to this idea did not appear until the 1880s—almost two hundred years after Descartes first observed that algebraic equations could describe concrete geometry.

So if you consider that algebra, beginning in the Islamic Golden Age around the 800s, and geometry, originating with the Greeks, developed for almost a thousand years before Descartes brought them together, and that it then took another two hundred years before we systematically understood rings as powerful tools for studying geometry, you may begin to appreciate that these are very deep ideas.

We will not be able to explore the theory of using rings to study geometry in detail.

But if you are interested, you can look into commutative algebra and algebraic geometry.

## §14.5 Examples of Noncommutative Rings

**Example 14.5 (Matrix Rings)**

Fix an integer

$$
n\geqslant0,
$$

and consider the set

$$
M_{n\times n}(\mathbb{R})
$$

of all
$n\times n$
matrices with entries in
$\mathbb{R}$.

You can add and multiply matrices, and matrix multiplication distributes over addition.

Therefore,

$$
M_{n\times n}(\mathbb{R})
$$

is a ring.

To make distributivity explicit, consider three matrices
$A,B,C$
with entries

$$
a_{ij},
\qquad
b_{ij},
\qquad
c_{ij}.
$$

The $ij$ entry of

$$
A(B+C)
$$

is

$$
\sum_{k=1}^{n}
a_{ik}(b_{kj}+c_{kj})
=
\sum_{k=1}^{n}a_{ik}b_{kj}
+
\sum_{k=1}^{n}a_{ik}c_{kj}.
$$

But the expression on the right is exactly the $ij$ entry of

$$
AB+AC.
$$

Similarly, one can prove

$$
(B+C)A
=
BA+CA.
$$

**Example 14.6 (Group Rings)**

Let $G$ be a finite group and let $R$ be a commutative ring.

As a set,

$$
R[G]
$$

is the set of all functions from $G$ to $R$.

Thus each element

$$
g\in G
$$

corresponds to an element

$$
r_g\in R.
$$

We write such a function using the notation

$$
\sum_{g\in G}r_gg.
$$

For example, the following is an element of

$$
\mathbb{Z}[S_3]:
$$

$$
5(~)+3(12)-8(123).
$$

Addition is straightforward: we simply add the corresponding terms:

$$
\left(
\sum r_gg
\right)
+
\sum(s_gg)
=
\sum_{g\in G}(r_g+s_g)g.
$$

That is, this is simply addition of functions.

Multiplication is not just multiplication of functions.

The coefficient of $g$ in the product of

$$
\sum r_gg
$$

and

$$
\sum s_gg
$$

is given by

$$
\sum_{(g_1,g_2)~\text{s.t.}~g_1g_2=g}
r_{g_1}s_{g_2}.
$$

In other words,

$$
\left(
\sum_{g\in G}r_gg
\right)
\left(
\sum_{h\in G}s_hh
\right)
=
\sum_{g\in G}
\sum_{h\in G}
r_gs_h(gh)
=
\sum_{k\in G}
\left(
\sum_{(g_1,g_2)~\text{s.t.}~g_1g_2=g}
r_gs_h
\right)k.
$$

Notice that this multiplication is not commutative.

## §14.6 Ring Homomorphisms

**Definition 14.4** Let $R$ and $S$ be rings, and let

$$
f:R\to S
$$

be a function.

We call $f$ a **ring homomorphism** if

(1) $f$ is a group homomorphism with respect to addition;

(2)

$$
f(1)=1,
$$

i.e. $f$ sends the multiplicative identity of $R$ to the multiplicative identity of $S$;

(3) for all
$a,b\in R$,

$$
f(ab)=f(a)f(b).
$$

**Definition 14.5** If $f$ is a bijection, we call $f$ an **isomorphism**.

Now I would like to explain further why

$$
\mathbb{Z}/n\mathbb{Z}
$$

is a ring.

How did we prove that it is a group?

By applying a general principle:

if

$$
H\triangleleft G,
$$

then

$$
G/H
$$

is a group.

I would like to do the same thing for rings.

But in this context, whenever we say “ring”, we will mean a commutative ring.