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.
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 $\overline{n}=\overline{0}$
⇒ $\overline{rs}=\overline{0}$
⇒ $\overline{r} \cdot \overline{s}=\overline{0}$.
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
$\overline{r} \cdot \overline{s}=\overline{0}$
⇒ $\overline{rs}=\overline{0}$.
⇒ 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 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
Every Finite Integral Domain is a Field
FAQs
Answer: No, ℤ6 is not an integral domain as 6 is not a prime number. This is because ℤ6 contains zero divisors as $\overline{2} \cdot \overline{3}=\overline{0}$.
