Zero Divisors and Integral Domains 📂Abstract Algebra

Zero Divisors and Integral Domains

Definitions¹

For a ring $R$, nonzero elements $a,b \in R$ satisfying $ab = 0$ are called zero divisors.
If $D$, having unity $1 \ne 0$, has no zero divisors, it is called an integral domain.

Explanation

Zero Divisors

Examples of nonzero things multiplying to $0$ 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 $2 \cdot 3 \equiv 0 \pmod{6}$. As this shows, in rings we do not always deal only with things that compute conveniently, so caution is required. This means, for instance, that one cannot make claims like “if $xy = 0$ and $x \ne 0$, then $y = 0$”.

Integral Domains

An integral domain is often abbreviated by its acronym ID.

As a simple example of an integral domain, one can take the set of integers $\mathbb{Z}$. This is only natural, since the very word “Integral” of integral domain comes from “Integer” in the first place. The nice thing about an integral domain is that when dividing, there is nothing to worry about other than $0$. In an integral domain, one can be certain that if $x y = 0$ then $x = 0$ or $y = 0$, which makes it quite useful as an algebraic structure.

That $R$ is an integral domain guarantees that the cancellation law holds for multiplication in $R$, and being a ring without zero divisors, it is closely related to fields. Let us look at the following useful theorems.

Theorem

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

Proof

[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, at the same time, $$ \left[ \left( {{1} \over {a}} \right) a \right] b =1 b = b $$ holds. This means that if $ab= 0$, then one of the two must be $0$, so an element of a field cannot be a zero divisor, and $F$ is an integral domain.

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[2]

Let the elements of a finite integral domain $D$ other than $0$ be $1, a_{1} , \cdots , a_{n}$. Multiplying these by $a \ne 0$ and considering $$ a, aa_{1} , \cdots , aa_{n} $$ none of them is $0$, since $D$ is an integral domain.

Moreover, since the cancellation law holds in an integral domain, we see that $aa_{i} = aa_{j}$ implies $a_{i} = a_{j}$. In other words, $$ a_{i} \ne a_{j} \implies aa_{i} \ne aa_{j} $$ and we obtain $$ \left\{ 1, a_{1} , \cdots , a_{n} \right\} = \left\{ a, aa_{1} , \cdots , aa_{n} \right\} $$ Therefore, for $a \ne 0$ there always exists $b \in \left\{ 1, a_{1} , \cdots , a_{n} \right\}$ satisfying $ab=1$. Since $b$ is the multiplicative inverse of $a$, $D$ is a field.

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[3]

Trivially, $\mathbb{Z}_{p} = \left\{ 0 , 1, \cdots , p-1 \right\}$ is a finite set. But since $p$ is a prime, there exist no nonzero $a,b \in \mathbb{Z}_{p}$ 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 $F$, since $0^2 = 0$ and $1^2 = 1$, $0$ and $1$ are idempotents of $F$. Suppose there exists an idempotent $a \in F$ that is neither $0$ nor $1$; then since $a^2 = a$, we must have $a( a-1) = 0$. However, by Theorem [1], $F$ is an integral domain and therefore has no zero divisors, which contradicts the assumption.

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