Prove that Zn is an Integral Domain iff n is Prime

The ring Zn 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.

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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 ℤn is an integral domain iff n is a prime number.

Table of Contents

Proof

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

(⇒) First suppose that ℤn 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 $\overline{n}=\overline{0}$

⇒ $\overline{rs}=\overline{0}$

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

As ℤn 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 ℤn is an integral domain, we need to show that it contains no zero divisors. Let us assume that

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

⇒ $\overline{rs}=\overline{0}$.

⇒ 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 ℤn contains no zero divisors which makes it an integral domain.

Therefore, ℤn 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

Q1: Is ℤ6 an integral domain?

Answer: No, ℤ6 is not an integral domain as 6 is not a prime number. This is because ℤ6 contains zero divisors as $\overline{2} \cdot \overline{3}=\overline{0}$.

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