Reflection and Refraction 📂Abstract Algebra

Reflection and Refraction

Definitions ¹

In a ring $R$, a $a,b \in R$ that is not $0$ and satisfies $ab = 0$ is called a Zero Divisor.
A $D$ with a unit $1 \ne 0$ and without zero divisors is called an Integral Domain.

Description

Zero Divisors

Examples of non-$0$ elements whose product is $0$ include $$ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} $$ and $2 \cdot 3 \equiv 0 \pmod{6}$. Hence, it is essential to be cautious as not everything in a ring behaves conveniently. This implies that if $xy = 0$ and $x \ne 0$, we cannot assert that $y = 0$.

Integral Domains

ID is a common abbreviation for Integral Domains.

An example of an integral domain is simple to illustrate with the set of integers $\mathbb{Z}$. Naturally, the term “Integral” originates from the word “Integer”. A significant advantage of an integral domain is that there is no need to worry about division by anything other than $0$. In an integral domain, if $x y = 0$, it assures that either $x = 0$ or $y = 0$, making it highly useful as an algebraic structure.

That $R$ is an integral domain guarantees that the Cancellation law applies to multiplication in $R$, indicating that it is a ring without zero divisors, closely related to a field. Let’s explore the following useful theorems.

Theorems

[1]: A field is an integral domain.
[2]: A finite integral domain is a field.
[3] If $p$ is prime, then $\mathbb{Z}_{p}$ is a field.
[4]: A field has only two idempotents, $0$ and $1$.

Proofs

[1]

For a field $F$, if $a \ne 0$ and $ab = 0$, then $$ \left( {{1} \over {a}} \right) (ab) = \left( {{1} \over {a}} \right) 0 = 0 $$ and concurrently, $$ \left[ \left( {{1} \over {a}} \right) a \right] b =1 b = b $$ holds. This means that if $ab= 0$, one of them must be $0$, so elements of a field cannot be zero divisors, and $F$ is an integral domain.

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[2]

Let’s name the elements of a finite integral domain $D$, excluding $0$, as $1, a_{1} , \cdots , a_{n}$. Considering them multiplied by $a \ne 0$, $$ a, aa_{1} , \cdots , aa_{n} $$ since $D$ is an integral domain, none of these are $0$.

Since the cancellation law holds in an integral domain, if $aa_{i} = aa_{j}$, then $a_{i} = a_{j}$. This implies $$ a_{i} \ne a_{j} \implies aa_{i} \ne aa_{j} $$, leading to $$ \left\{ 1, a_{1} , \cdots , a_{n} \right\} = \left\{ a, aa_{1} , \cdots , aa_{n} \right\} $$. Therefore, there always exists a $b \in \left\{ 1, a_{1} , \cdots , a_{n} \right\}$ satisfying $ab=1$ for $a \ne 0$. Since $b$ is the multiplicative inverse of $a$, $D$ is a field.

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[3]

Obviously, $\mathbb{Z}_{p} = \left\{ 0 , 1, \cdots , p-1 \right\}$ is a finite set. Given $p$ is prime, there exist no non-$0$ $a,b \in \mathbb{Z}_{p}$ satisfying $$ ab \equiv 0 \pmod{p} $$, so $\mathbb{Z}_{p}$ is an integral domain and, by theorem [2], a field.

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[4]

For a field $F$, if $0^2 = 0$ and $1^2 = 1$, then $0$ and $1$ are idempotents of $F$. Assuming the existence of an idempotent $a \in F$ that is neither $0$ nor $1$ leads to $a^2 = a$, hence $a( a-1) = 0$. However, given theorem [1], since $F$ is an integral domain and does not have zero divisors, this assumption is contradictory.

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