The Cancellation Property of * on Integral Domains

Recall from the Integral Domains page that an integral domain is a commutative ring $(R, +, *)$ that contains no zero divisors. That is, the ring $(R, +, *)$ satisfies all of the ring axioms as well as:

• $a * b = b * a$ for all $a, b \in R$.

• If $a * b = 0$ then either $a = 0$ or $b = 0$ or both.

Now recall from the Basic Theorems Regarding Groups page that in any group $(G, *)$ we have that for all $a, b, c \in G$ that $a * b = a * c$ implies that $b = c$. Since every ring $(R, +, *)$ is a group with respect to $+$ only, i.e., $(R, +)$ is a group, then we have that $a + b = a + c$ implies that $b = c$.

We would like to formulate a similar result for the operation $*$ of a ring $(R, +, *)$. In general, if $(R, +, *)$ is a ring and $a, b, c \in R$ then $a * b = a * c$ does NOT imply that $b = c$. For example, consider the ring of $2 \times 2$ matrices with real coefficients $(M_{22}, +, *)$ with $+$ denoting standard matrix addition and $*$ denoting standard matrix multiplication. Let $A = \begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 0 & 1\\ 0 & 1 \end{bmatrix}$, and $C = \begin{bmatrix} 0 & 2\\ 0 & 0 \end{bmatrix}$. Then:

(1)

\begin{align} \quad AB = \begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 2\\ 0 & 0 \end{bmatrix} \end{align}

(2)

\begin{align} \quad AC = \begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 2\\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2\\ 0 & 0 \end{bmatrix} \end{align}

Therefore $AB = AC$ but clearly $B \neq C$. Therefore a ring need not satisfy the cancellation law for $*$. In the following theorem, we will see that the cancellation law does hold on $*$ provided that $(R, +, *)$ is an integral domain.

Theorem 1: If $(R, +, *)$ is an integral domain with $0$ being the identity of $+$. Then for all $a, b, c \in R$ where $a \neq 0$, $a * b = a * c$ implies that $b = c$.

• Proof: Let $(R, +, *)$ be an integral domain and let $a, b, c \in R$. Suppose that $a * b = a * c$. Then:

(3)

\begin{align} \quad a * b = a * c \\ \quad (a * b) - (a * c) \\ \quad a * (b + (-c)) = 0 \end{align}

• Since $(R, +, *)$ is an integral domain we have that either $a = 0$, $b + (-c) = 0$, or both. But we're given that $a \neq 0$. Therefore $b + (-c) = 0$ so $b = c$. $\blacksquare$

Sometimes we would instead like to determine whether a ring is an integral domain. The following theorem tells us that if we have a commutative ring with identity for which the cancellation property holds, then our ring is furthermore an integral domain.

Theorem 2: Let $(R, +, *)$ is a commutative ring with identity. Then $(R, +, *)$ is an integral domain if and only if for all $a, b, c \in R$ with $a \neq 0$ we have that $a * b = a * c$ implies that $b = c$.

• Proof: $\Rightarrow$ Suppose that $(R, +, *)$ is an integral domain. Then for all $a, b \in R$, if $a * b = 0$ then either $a = 0$ or $b = 0$.

• Suppose that for $a, b, c \in R$ with $a \neq 0$ we have that:

(4)

\begin{align} \quad a * b = a * c \end{align}

• Then:

(5)

\begin{align} \quad 0 &= (a * b) - (a * c) \\ &= a * (b - c) \end{align}

• So either $a = 0$ or $b - c = 0$. We assumed that $a \neq 0$, so $b - c = 0$. But then $b = c$!

• $\Leftarrow$ Suppose that for all $a, b, c \in R$ with $a \neq 0$ we have that $a * b = a * c$ implies that $b = c$. Let $a, b \in R$ and suppose that:

(6)

\begin{align} \quad a * b = 0 \end{align}

• This equation can be rewritten as:

(7)

\begin{align} \quad a * b = a * 0 \end{align}

• This implies that $b = 0$, so $(R, +, *)$ is an integral domain. $\blacksquare$