Integral Domains

Recall from the Zero Divisors in Rings page that if we consider a ring $(R, +, *)$ where $0$ is the identity of $+$ then a zero divisor of $R$ is an element $a \in R \setminus \{ 0 \}$ such that there exists an element $b \in R \setminus \{ 0 \}$ for which either $a * b = 0$ or $b * a = 0$.

We saw that the ring $(M_{22}, +, *)$ of $2 \times 2$ matrices with real coefficients under the operation of standard addition $+$ and standard multiplication $*$ has zero divisors. We also noted that the rings $(\mathbb{C}, +, *)$ and $(\mathbb{R}, +, *)$ of complex and real numbers respectively under standard addition $+$ and standard multiplication $*$ has no zero divisors.

The rings of complex and real numbers described above are particularly handy in being commutative rings and not having an zero divisors. In fact, we give a special name to rings that are commutative and have no zero divisors.

Definition: An Integral Domain is a commutative ring $(R, +, *)$ that has no zero divisors. That is, it satisfies the extra condition that $a * b = b * a$ and if $0$ is the identity for $+$ then for all $a, b \in R$ we have that $a * b = 0$ implies that $a = 0$, $b = 0$ or both.

We can therefore call the rings $(\mathbb{C}, +, *)$ and $(\mathbb{R}, +, *)$ integral domains.

One important property we should note of is that if $(R, +, *)$ is an integral domain, then any subring $(S, +, *)$ is also an integral domain as we prove in the following theorem.

Theorem 1: If $(S, +, *)$ is a subring of the integral domain $(R, +, *)$ then $(S, +, *)$ is an integral domain.

• Proof: Suppose that $(S, +, *)$ is a subring of the integral domain $(R, +, *)$ and assume that $(S, +, *)$ is not an integral domain.

• If $(S, +, *)$ is not commutative then there exists a pair of elements $a, b \in S$ such that $a * b \neq b * a$. But since $S \subseteq R$ then $a, b \in R$ is such that $a * b \neq b * a$ so $(R, +, *)$ is not commutative and hence is not an integral domain which is a contradiction.

• Now if $0$ is the identity for $+$ and there exists a zero divisor $a \in S \setminus \{ 0 \}$ then there also exists an element $b \in S \setminus \{ 0 \}$ such that $a * b = 0$ or $b * a =0$. But since $S \subseteq R$ we have that then $S \setminus \{ 0 \} \subseteq R \setminus \{ 0 \}$ so $a, b \in R \setminus \{ 0 \}$. Thus $a \in R$ is a zero divisor of $R$ which implies that $(R, +, *)$ is not an integral domain which is a contradiction.

• Therefore the assumption that $(S, +, *)$ is not an integral domain is false. Therefore any subring $(S, +, *)$ is an integral domain of $(R, +, *)$. $\blacksquare$