## 数学代写|抽象代数作业代写abstract algebra代考|Abstract Rings and Ring Homomorphisms

Definition 3.1. A ring $R$ is a set with two operations, generally called addition and multiplication and written
$$\underbrace{a+b}{\text {addition }} \text { and } \underbrace{a \cdot b \text { or } a b}{\text {multiplication }},$$
satisfying the following axioms:
(1) The set $R$ with its addition law + is an abelian group. The identity element of this group is denoted 0 or $0_R$.

(2) The set $R$ with its multiplication law – is almost a group, but its elements are not required to have inverses. ${ }^2$ Explicitly, the multiplication law of a ring has the following properties:

• There is an element $1_R \in R$ satisfying ${ }^{3,4}$
$$1_R \cdot a=a \cdot 1_R=a \quad \text { for all } a \in R$$
As with $0_R$ versus 0 , if there is no ambiguity we often write 1 instead of $1_R$.
• The associative law holds,
$$a \cdot(b \cdot c)=(a \cdot b) \cdot c \quad \text { for all } a, b, c \in R$$
(3) [Distributive Law] For all $a, b, c \in R$ we have
$$a \cdot(b+c)=a \cdot b+a \cdot c \quad \text { and } \quad(b+c) \cdot a=b \cdot a+c \cdot a .$$
(4) [Optional Feature: Commutative Law] If further $a \cdot b=b \cdot a$ for all $a, b \in R$, then the ring is said to be commutative.

Your long experience with the ring of integers $\mathbb{Z}$ might lead you to assume that various “obvious” formulas are true in every ring. For example, the formulas
$$0_R \cdot a=0_R \quad \text { and } \quad(-a) \cdot(-b)=a \cdot b$$
must be true, right? But why should they be true? The definition of $0_R$ is as the identity element for addition; i.e., $a+0_R=0_R+a=a$ for every $a \in R$, so why should that tell us anything about $0_R$ when we switch to multiplication? Similarly, the definition of $-a$ is as the element that gives $0_R$ when it is added to $a$, which seems to tell us very little about the product of $-a$ with other elements of $R$. The only hope of proving multiplication properties for $0_R$ and $-a$ lies in the distributive law, which intertwines addition and multiplication. Study closely the use of the distributive law in the following proof that $0_R \cdot a=0_R$.

## 数学代写|抽象代数作业代写abstract algebra代考|Interesting Examples of Rings

Four of the rings that we described in Section 3.1 fit one into another,
$$\mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C} \text {. }$$
We say that $\mathbb{Z}$ is a subring of $\mathbb{Q}$, and similarly for the others. The fifth ring that we mentioned is the ring of integers modulo $m$.

Example 3.4 (Integers Modulo $m: \mathbb{Z} / m \mathbb{Z}$ ). We recall the construction from Section 1.8 of the ring $\mathbb{Z} / m \mathbb{Z}$. We start with the integers, and we simply pretend that two integers $a$ and $b$ are “the same” if their difference $a-b$ is a multiple of $m$. In fancier language, we define an equivalence relation on $\mathbb{Z}$ by the rule
$a$ is equivalent to $b$ if $a-b$ is a multiple of $m$.

This equivalence relation is so important that we gave it its own name and its own notation: $a$ is congruent to $b$ modulo $m, \quad a \equiv b(\bmod m)$.

We then define the ring $\mathbb{Z} / m \mathbb{Z}$ to be the set of congruence classes. However, as noted in (1.15) in Section 1.8 and Exercise 1.34, it takes some work to check that addition and multiplication of congruence classes makes sense. The ring $\mathbb{Z} / m \mathbb{Z}$ is not a subring of $\mathbb{C}$, but there is a natural homomorphism
$$\phi: \mathbb{Z} \longrightarrow \mathbb{Z} / m \mathbb{Z}, \quad \phi(a)=a \bmod m$$
that assigns to the integer $a$ its equivalence class consisting of all integers that are congruent to $a$ modulo $m$. The homomorphism $\phi$ is called, naturally enough, the reduction $\bmod m$ homomorphism. The kernel of $\phi$ is the set of all multiples of $m$.

Example 3.5 (Gaussian Integers: $\mathbb{Z}[i]$ ). Here is another interesting subring of $\mathbb{C}$. It is called the ring of Gaussian integers and consists of the set
$$\mathbb{Z}[i]={a+b i: a, b \in \mathbb{Z}}$$
The quantity $i$ is, as usual, a symbol that represents a square root of -1 . Addition and multiplication of elements in $\mathbb{Z}[i]$ follow the usual rules for adding and multiplying complex numbers,
\begin{aligned} \left(a_1+b_1 i\right)+\left(a_2+b_2 i\right) & =\left(a_1+a_2\right)+\left(b_1+b_2\right) i \ \left(a_1+b_1 i\right) \cdot\left(a_2+b_2 i\right) & =\left(a_1 a_2-b_1 b_2\right)+\left(a_1 b_2+a_2 b_1\right) i . \end{aligned}
Note that if we allowed $a$ and $b$ to be real numbers, then we would get the entire ring of complex numbers; but we’re restricting $a$ and $b$ to be integers.

# 抽象代数代考

## 数学代写|抽象代数作业代写abstract algebra代考|Abstract Rings and Ring Homomorphisms

$\underbrace{a+b}$ addition and $\underbrace{a \cdot b \text { or } a b}$ multiplication ,

(1) 集合 $R$ 加上它的法则 + 是阿贝尔群。该组的身份元素表示为 0 或 $0_R$.
(2) 集合 $R$ 及其乘法定律 – 几乎是一个群，但它的元素不需要有逆元素。 ${ }^2$ 明确地，环的 乘法定律具有以下性质：

• 有一个元素 $1_R \in R$ 令人满意 3,4
$$1_R \cdot a=a \cdot 1_R=a \quad \text { for all } a \in R$$
与 $0_R$ 与 0 相比，如果没有歧义，我们通常写 1 而不是 $1_R$.
• 结合律认为,
$$a \cdot(b \cdot c)=(a \cdot b) \cdot c \quad \text { for all } a, b, c \in R$$
(3)【分配律】 对所有人 $a, b, c \in R$ 我们有
$$a \cdot(b+c)=a \cdot b+a \cdot c \quad \text { and } \quad(b+c) \cdot a=b \cdot a+c \cdot a$$
(4)【可选特征: 交换律】如果进一步 $a \cdot b=b \cdot a$ 对全部 $a, b \in R$, 则称环是可交换 的。
你对整数环的长期经验Z可能会让您假设各种“显而易见”的公式在每个环中都是正确 的。例如，公式
$$0_R \cdot a=0_R \quad \text { and } \quad(-a) \cdot(-b)=a \cdot b$$
一定是真的吧? 但为什么它们应该是真的呢? 的定义 $0_R$ 是作为加法的标识元素； $I E$ ， $a+0_R=0_R+a=a$ 每一个 $a \in R$ ，那为什么要告诉我们关于 $0_R$ 当我们切换到乘 法? 同样，定义 $-a$ 是作为给出的元素 $0_R$ 当它被添加到 $a$ ，这似乎告诉我们很少关于产 品的信息 $-a$ 与其他元素 $R$. 证明乘法性质的唯一希望 $0_R$ 和 $-a$ 在于分配律，它把加法和 乘法交织在一起。仔细研究分配律在以下证明中的使用 $0_R \cdot a=0_R$.

## 数学代写|抽象代数作业代写abstract algebra代考|Interesting Examples of Rings

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

$$\phi: \mathbb{Z} \longrightarrow \mathbb{Z} / m \mathbb{Z}, \quad \phi(a)=a \bmod m$$例 3.5 (高斯整数: $\mathbb{Z}[i]$. 这是另一个有趣的子环C. 它被称为高斯整数环，由集合组成
$$\mathbb{Z}[i]=a+b i: a, b \in \mathbb{Z}$$

$$\left(a_1+b_1 i\right)+\left(a_2+b_2 i\right)=\left(a_1+a_2\right)+\left(b_1+b_2\right) i\left(a_1+b_1 i\right) \cdot\left(a_2+b_2 i\right) \quad=\left(a_1 a_2\right.$$

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