Prove that Zn is an Integral Domain iff n is Prime

The ring Z_n of integers modulo n is an integral domain when n is a prime number. In this article, we will prove that Zn is an integral domain if and only if n is a prime number.

What is an Integral Domain?

A non-trivial ring R is said to be an integral domain if

• R is a commutative ring with unity
• R has no divisors of zero.

A zero divisor is an element a such that ab=0 for some non-zero element b in R. Let us now prove that ℤ_n is an integral domain iff n is a prime number.

Proof

The ring ℤ_n is a commutative ring with unity $\overline{1}$.

(⇒) First suppose that ℤ_n is an integral domain. We need to show that n is prime. If possible assume that n is a composite number.

So n = rs for two integers r and s with 1< r, s < n.

⇒ $\overline{r} \neq \overline{0}$ and $\overline{s} \neq \overline{0}$.

Now $\overline{n}=\overline{0}$

⇒ $\overline{rs}=\overline{0}$

⇒ $\overline{r} \cdot \overline{s}=\overline{0}$.

As ℤ_n is an integral domain by assumption, it contains no zero divisors. This implies either $\overline{r}=\overline{0}$ or $\overline{s}=\overline{0}$. Thus we arrive at a contradiction. So n must be a prime number.

(⇐) Next suppose that n is a prime number.

To prove ℤ_n is an integral domain, we need to show that it contains no zero divisors. Let us assume that

$\overline{r} \cdot \overline{s}=\overline{0}$

⇒ $\overline{rs}=\overline{0}$.

⇒ n divides rs. As n is a prime number, it follows that either n divides r or n divides s. In other words, either $\overline{r}=\overline{0}$ or $\overline{s}=\overline{0}$. This proves that ℤ_n contains no zero divisors which makes it an integral domain.

Therefore, ℤ_n is an integral domain if and only if n is a prime number.

Have You Read These?

Ring Theory: Definition, Examples, Properties

Zero Divisors in a Ring

Every Finite Integral Domain is a Field

A Field is an Integral Domain

FAQs