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数学代写|现代代数代写Modern Algebra代考|Integral Domains and Fields
In the preceding section we defined the terms ring with unity, commutative ring, and zero divisors. All three of these terms are used in defining an integral domain.
Integral Domain
Let $D$ be a ring. Then $D$ is an integral domain provided these conditions hold:
- $D$ is a commutative ring.
- $D$ has a unity $e$, and $e \neq 0$.
- $D$ has no zero divisors.
Note that the requirement $e \neq 0$ means that an integral domain must have at least two elements.
Example 1 The ring $\mathbf{Z}$ of all integers is an integral domain, but the ring $\mathbf{E}$ of all even integers is not an integral domain, because it does not contain a unity. As familiar examples of integral domains, we can list the set of all rational numbers, the set of all real numbers, and the set of all complex numbers-all of these with their usual operations.
Example 2 The ring $\mathbf{Z}{10}$ is a commutative ring with a unity, but the presence of zero divisors such as [2] and [5] prevents $\mathbf{Z}{10}$ from being an integral domain. Considered as a possible integral domain, the ring $M_2(\mathbf{R})$ of all $2 \times 2$ matrices with real numbers as elements fails on two counts: Multiplication is not commutative, and it has zero divisors.
In Example 4 of Section 5.1, we saw that $\mathbf{Z}_n$ is a ring for every value of $n>1$. Moreover, $\mathbf{Z}_n$ is a commutative ring since
$$
[a] \cdot[b]=[a b]=[b a]=[b] \cdot[a]
$$
for all $[a],[b]$ in $\mathbf{Z}_n$. Since $\mathbf{Z}_n$ has $[1]$ as the unity, $\mathbf{Z}_n$ is an integral domain if and only if has no zero divisors. The following theorem characterizes the $\mathbf{Z}_n$ with no zero divisors, providing us with a large class of finite integral domains (that is, integral domains that have a finite number of elements).
数学代写|现代代数代写Modern Algebra代考|The Integral Domain Zn When n Is a Prime
For $n>1, \mathbf{Z}_n$ is an integral domain if and only if $n$ is a prime.
Proof From the previous discussion, it is clear that we need to only prove that $\mathbf{Z}_n$ has no zero divisors if and only if $n$ is a prime.
Suppose first that $n$ is a prime. Let $[a] \neq[0]$ in $\mathbf{Z}_n$, and suppose $[a][b]=[0]$ for some $[b]$ in $\mathbf{Z}_n$. Now $[a][b]=[0]$ implies that $[a b]=[0]$, and therefore, $n \mid a b$. However, $[a] \neq[0]$ means that $n \nmid a$. Thus $n \mid a b$ and $n \nmid a$. Since $n$ is a prime, this implies that $n \mid b$, by Theorem 2.16 (Euclid’s Lemma); that is, $[b]=[0]$. We have shown that if $[a] \neq[0]$, the only way that $[a][b]$ can be $[0]$ is for $[b]$ to be $[0]$. Therefore, $\mathbf{Z}_n$ has no zero divisors and is an integral domain.
Suppose now that $n$ is not a prime. Then $n$ has divisors other than \pm 1 and $\pm n$, so there are integers $a$ and $b$ such that
$$
n=a b \text { where } 1<a<n \text { and } 1<b<n .
$$
This means that $[a] \neq[0],[b] \neq[0]$, but
$$
[a][b]=[a b]=[n]=[0] .
$$
Therefore, $[a]$ is a zero divisor in $\mathbf{Z}_n$, and $\mathbf{Z}_n$ is not an integral domain.
Combining the two cases, we see that $n$ is a prime if and only if $\mathbf{Z}_n$ is an integral domain.
现代代数代考
数学代写|现代代数代写Modern Algebra代考|Integral Domains and Fields
在前一节中，我们定义了具有单位环、可交换环和零因子的环。这三个术语都用于定义一个积分域。
积分域
让$D$成为一个戒指。那么$D$是一个整域，只要满足以下条件:
$D$ 是一个交换环。
$D$ 有一个统一的$e$，和$e \neq 0$。
$D$ 没有零因子。
注意，需求$e \neq 0$意味着一个积分域必须至少有两个元素。
例1所有整数的环$\mathbf{Z}$是一个整域，但所有偶数的环$\mathbf{E}$不是一个整域，因为它不包含一个单位。作为熟悉的积分域的例子，我们可以列出所有有理数的集合，所有实数的集合，以及所有复数的集合——所有这些都有它们通常的运算。
例2环$\mathbf{Z}{10}$是一个具有单位的交换环，但是像[2]和[5]这样的零因子的存在使得$\mathbf{Z}{10}$不是一个整域。作为一个可能的积分域，所有以实数为元素的$2 \times 2$矩阵的环$M_2(\mathbf{R})$在两个方面失败:乘法是不可交换的，并且它没有除数。
在第5.1节的例4中，我们看到$\mathbf{Z}_n$是一个对应于$n>1$的每个值的环。而且，$\mathbf{Z}_n$是一个交换环，因为
$$
[a] \cdot[b]=[a b]=[b a]=[b] \cdot[a]
$$
所有的$[a],[b]$都在$\mathbf{Z}_n$中。因为$\mathbf{Z}_n$以$[1]$为单位，所以当且仅当没有零因子时，$\mathbf{Z}_n$是一个整域。下面的定理描述了没有零因子的$\mathbf{Z}_n$，为我们提供了一大类有限积分域(即具有有限个数元素的积分域)。
数学代写|现代代数代写Modern Algebra代考|The Integral Domain Zn When n Is a Prime
因为$n>1, \mathbf{Z}_n$是一个积分域当且仅当$n$是素数。
从前面的讨论，很明显，我们只需要证明$\mathbf{Z}_n$没有零因子当且仅当$n$是素数。
首先假设$n$是质数。在$\mathbf{Z}_n$中设置$[a] \neq[0]$，在$\mathbf{Z}_n$中设置$[a][b]=[0]$表示某些$[b]$。现在$[a][b]=[0]$意味着$[a b]=[0]$，因此是$n \mid a b$。然而，$[a] \neq[0]$表示$n \nmid a$。因此$n \mid a b$和$n \nmid a$。因为$n$是质数，这意味着$n \mid b$，根据定理2.16(欧几里得引理);也就是$[b]=[0]$。我们已经证明，如果$[a] \neq[0]$，那么$[a][b]$可以是$[0]$的唯一方法就是$[b]$可以是$[0]$。因此，$\mathbf{Z}_n$没有零因子，是一个积分域。
假设$n$不是质数。那么$n$有除\pm 1和$\pm n$之外的除数，所以有整数$a$和$b$，使得
$$
n=a b \text { where } 1<a<n \text { and } 1<b<n .
$$
这意味着$[a] \neq[0],[b] \neq[0]$，但是
$$
[a][b]=[a b]=[n]=[0] .
$$
因此，$[a]$在$\mathbf{Z}_n$中是一个零因子，而$\mathbf{Z}_n$不是一个积分域。
结合这两种情况，我们看到$n$是一个素数当且仅当$\mathbf{Z}_n$是一个积分域。
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