Divisors of Zero in a Ring: Definition, Examples, How to Find

A zero divisor of a ring (R, +, ⋅) is an element such that its product with a non-zero element is equal to 0. In this article, we will learn about the divisors of zero in a ring with examples, and how to find zero divisors.

Definition of a Zero Divisor

Let (R, +, ⋅) be a ring. A non-zero element a ∈ R is called a zero divisor if there exists a non-zero element b in R such that a⋅b = 0, or there exists a non-zero element c in R such that c⋅a = 0.

When a⋅b = 0, a is called a left divisor of zero whereas c⋅a = 0 implies that a is a right divisor of zero.

Example: As the product of two non-zero integers is always non-zero, so the ring (ℤ, +, ⋅) does not have any divisor of zero.

Examples

1. The ring (ℝ, +, ⋅) contains no zero divisors.
2. As $\bar{2} \cdot \bar{3}=\bar{0}$ in the ring Z₆, the elements $\bar{2}$ and $\bar{3}$ are zero divisors in Z₆.
3. (Z_p, +, ⋅) does not have zero divisors when p is a prime number.

Properties

The zero divisors in a ring have the following properties.

1. If R is a commutative ring, then every left zero divisor is also a right zero divisor.
2. For a prime p, the ring (Z_p, +, ⋅) of integers modulo p does not contain any zero divisors.
3. A ring R has no zero divisors if and only if the cancellation law holds in R.
4. Let R be a non-trivial ring such that it has no zero divisors. Then R must be a ring with unity.
5. If R is a finite ring with no zero divisors, then every non-zero element of R is a unit.

How to Find Zero Divisors

In the ring (Z_n, +, ⋅), an element $\bar{m}$ is a zero divisor if and only if gcd(m, n) ≠ 1.Using this fact, the divisors of zero in (Z_n, +, ⋅) are given by {$\bar{m}$ ∈ Z_n : m is a divisor of n}.

Hence, the number of zero divisors in (Z_n, +, ⋅) is equal to φ(n) where φ denotes the Euler-Phi function.

Example: the set of zero divisors of (Z₁₀, +, ⋅) = {$\bar{1}, \bar{2}, \bar{5}, \bar{10}$}

Related Topics: Introduction to Ring Theory

Units of a Ring

Characteristic of a Ring

Solved Problems

Question 1: Prove that the ring of matrices R = $\Big\{ \begin{pmatrix} a & b \\ 2b & a \end{pmatrix} : a, b \in \mathbb{R}\Big\}$ contains zero divisors.

Solution:

Let A = $\begin{pmatrix} a & b \\ 2b & a \end{pmatrix}$ be a non-zero element in R. Thus either a or b is non-zero.

To find zero divisors, let AB = O, the zero matrix, for some B = $\begin{pmatrix} x & y \\ 2y & x \end{pmatrix}$

The AB = O implies that

ax + 2by = 0

bx + ay = 0.

This is a homogeneous system of equations. Thus, for non-zero solutions, we must have that $\begin{vmatrix} a & 2b \\ b & a \end{vmatrix}=0$.

⇒ a² – 2b² = 0.

Take a=√2, and b=1.

So there exists a non-zero B in R such that AB=O. Hence, the ring R contains zero divisors.

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