Zero Divisors in Rings

Definition: Let $(R, +, *)$ be a ring where $0 \in R$ is the identity of $+$. The element $a \in R \setminus \{ 0 \}$ is said to be a Zero-Divisor of $R$ if there exists a $b \in R \setminus \{ 0 \}$ such that $a * b = 0$ or $b * a = 0$.

For example, consider the ring $(M_{22}, +, *)$ of $2 \times 2$ matrices with real coefficients and with the operations of standard matrix addition $+$, and standard matrix multiplication $*$. Recall that the identity of $+$ is the $2 \times 2$ zero matrix $0 = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}$.

Further consider the matrices $A, B \in M_{22}$ given by:

(1)

\begin{align} \quad A = \begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix} \quad , \quad B = \begin{bmatrix} 0 & 1\\ 0 & -1 \end{bmatrix} \end{align}

When we multiply the matrices $A$ and $B$ together we have that:

(2)

\begin{align} \quad A * B = \begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1\\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} = 0 \end{align}

Notice that $A$ is not the identity for $+$ and $B$ is not the identity for $+$. Therefore, the matrices $A$ and $B$ are zero divisors of $M_{22}$.

We should be clear that a ring $(R, +, *)$ need not have any zero divisors. For example, consider the ring $(\mathbb{C}, +, *)$ of complex numbers where $+$ is standard addition and $*$ is standard multiplication. We note that the identity of $+$ is $0 = 0 + 0i \in \mathbb{C}$. Since $(\mathbb{C}, +, *)$ is a commutative ring, then for $x, y \in \mathbb{C} \setminus \{ 0 \}$ and where $x = a+bi$ and $y = c + di$ for $a, b, c, d \in \mathbb{R}$ we only need to consider the following equation:

(3)

\begin{align} \quad x * y = (a + bi)(c + di) = (ac - bd) + (ad + bc)i = 0 + 0i \end{align}

Note that this equality holds if and only if:

(4)

\begin{align} \quad ac = bd \quad \mathrm{and} \quad ad + bc = 0 \end{align}

Without loss of generality, assume $x \neq 0$. Then either $a \neq 0$ or $b \neq 0$ or both. Assume $a \neq 0$. Then we can divide both equations by $a$ to get:

(5)

\begin{align} \quad c = \frac{bd}{a} \: (*) \quad \mathrm{and} \quad d + \frac{bc}{a} = 0 \: (**) \end{align}

Substituting the first equation into the second yields:

(6)

\begin{align} \quad d + c \frac{b}{a} = 0 (*)\\ \quad d + \frac{bd}{a} \cdot \frac{b}{a} = 0 \\ \quad d + \frac{b^2d}{a^2} = 0 (*) \\ \quad d \left ( 1 + \frac{b^2}{a^2} \right ) = 0 \end{align}

There are two possibilities in the equation above. Either $d = 0$ or $1 + \frac{b^2}{a^2} = 0$. Clearly $1 + \frac{b^2}{a^2} = 0$ since this would imply that $\left ( \frac{b}{a} \right )^2 = - 1$. Therefore $d = 0$.

Looking at $(**)$ we see that then $\frac{bc}{a} = 0$ so either $b = 0$ or $c = 0$. If $c = 0$ we have that then $y = c + di = 0$. Meanwhile, if $c \neq 0$ then $b = 0$ and by $(*)$ this implies that $c = 0$ so then $y = c + di = 0$ again. In either case, we see that if $x = a + bi \neq 0 + 0i$ then $y = c + di = 0 + 0i$. Therefore, there exists no zero divisors in the ring $(\mathbb{C}, +, *)$.