As the example above illustrates,

Theorem 16.1.18 is a modest beginning in the study of which algebraic manipulations are possible when working with rings. A fact in elementary algebra that is used frequently in problem solving is the cancellation law. We know that the cancellation laws are true under addition for any ring, based on group theory. Are the cancellation laws true under multiplication, where the group axioms can’t be counted on? More specifically, let

\([R; +, \cdot ]\) be a ring and let

\(a, b, c\in R\) with

\(a \neq 0\text{.}\) When can we cancel the

\(a\)’s in the equation

\(a \cdot b = a \cdot c\text{?}\) We can do so if

\(a^{-1}\) exists, but we cannot assume that

\(a\) has a multiplicative inverse. The answer to this question is found with the following definition and the theorem that follows.

Now, here is why zero divisors are related to cancellation.

We prove the theorem using the left cancellation axiom, namely that if \(a \neq 0\) and \(a \cdot b = a \cdot c\text{,}\) then \(b = c\) for all \(a, b, c\in R\text{.}\) The proof using the right cancellation axiom is its mirror image.

#####
(⇒)

Assume the left cancellation law holds in \(R\) and assume that \(a\) and \(b\) are two elements in \(R\) such that \(a \cdot b = 0\text{.}\) We must show that either \(a = 0\) or \(b = 0\text{.}\) To do this, assume that \(a \neq 0\) and show that \(b\) must be 0.

\begin{equation*}
\begin{split}
a\cdot b = 0 &\Rightarrow a\cdot b = a\cdot 0\\
& \Rightarrow b = 0\quad \textrm{ by the left cancellation law}\\
\end{split}\text{.}
\end{equation*}

#####
(⇐)

Conversely, assume that \(R\) has no zero divisors and we will prove that the left cancellation law must hold. To do this, assume that \(a,b, c \in R\text{,}\) \(a \neq 0\text{,}\) such that \(a \cdot b = a \cdot c\) and show that \(b = c\text{.}\)

\begin{equation*}
\begin{split}
a \cdot b = a \cdot c & \Rightarrow a \cdot b - a \cdot c=0\\
&\Rightarrow\ a\cdot (b-c) =0\\
& \Rightarrow b-c = 0\quad \textrm{ since there are no zero divisors}\\
&\Rightarrow b=c\\
\end{split}
\end{equation*}

Hence, the only time that the cancellation laws hold in a ring is when there are no zero divisors. The commutative rings with unity in which the two conditions are true are given a special name.

####
Definition 16.1.23. Integral Domain.

A commutative ring with unity containing no zero divisors is called an integral domain.

\([\mathbb{Z}; +, \cdot]\text{,}\) \(\left[\mathbb{Z}_p; +_p , \times_p \right]\) with \(p\) a prime, \([\mathbb{Q}; +, \cdot ]\text{,}\) \([\mathbb{R}; +, \cdot ]\text{,}\) and \([\mathbb{C}; +, \cdot ]\) are all integral domains. The key example of an infinite integral domain is \([\mathbb{Z}; +, \cdot ]\text{.}\) In fact, it is from \(\mathbb{Z}\) that the term integral domain is derived. Our main example of a finite integral domain is \(\left[\mathbb{Z}_p; +_p , \times_p \right]\text{,}\) when \(p\) is prime.

We close this section with the verification of an observation that was made in Chapter 11, namely that the product of two algebraic systems may not be an algebraic system of the same type.

Both \(\left[\mathbb{Z}_2; +_2 , \times_2 \right]\) and \(\left[\mathbb{Z}_3; +_3 , \times_3 \right]\) are integral domains. Consider the direct product \(\mathbb{Z}_2\times \mathbb{Z}_3\text{.}\) It’s true that \(\mathbb{Z}_2 \times \mathbb{Z}_3\) is a commutative ring with unity (see Exercise 13). However, \((1,0)\cdot (0, 2) = (0, 0)\text{,}\) so \(\mathbb{Z}_2\times \mathbb{Z}_3\) has zero divisors and is therefore not an integral domain.