The Integral Domain of Z/pZ

The Integral Domain of Z/pZ

Recall from the Integral Domains page that a ring $(R, +, *)$ is said to be an integral domain if it is a commutative ring and $R$ contains no zero divisors, that is, if $0$ is the identity of $+$ then $a * b = 0$ implies that $a = 0$ or $b = 0$.

We noted that the rings $( \mathbb{C}, +, *)$ and $(\mathbb{R}, +, *)$ where $+$ is standard addition and $*$ is standard multiplication are integral domains. Also, we saw earlier on the Zero Divisors of Z/nZ that if $n$ is a composite number then $(\mathbb{Z} / n \mathbb{Z}, +, *)$ always contains zero divisors and hence $(\mathbb{Z} / n \mathbb{Z}, +, *)$ is not an integral domain.

In the theorem below we will show that if instead we have a prime $p$ then the ring $(\mathbb{Z} / p \mathbb{Z}, +, *)$ will have no zero divisors and will hence be an integral domain. Recall that the set $\mathbb{Z} / p \mathbb{Z}$ is defined as:

(1)
\begin{align} \quad \mathbb{Z} / p\mathbb{Z} = \{ [0]_p, [1]_p, ..., [p-1]_p \} \end{align}

For each $k \in \{0, 1, ..., p-1 \}$ we define the set $[k]_p$ to be the set of integers that when divided by $p$ leave a remainder $k$, that is:

(2)
\begin{align} \quad [k]_p = \{ z \in \mathbb{Z} : z \equiv k \pmod p \} \end{align}

Then for each $k, j \in \{ 0, 1, ..., p-1 \}$ we define $+$ by $[k]_p + [j]_p = [k + j]_p$ and $*$ by $[k]_p * [j]_p = [kj]_p$.

We will now prove that for any prime $p$ that the ring $(\mathbb{Z}/p\mathbb{Z}, +, *)$ is an integral domain.

 Theorem 1: If $p$ is a prime number then the ring $(\mathbb{Z}/p\mathbb{Z}, +, *)$ is an integral domain.
• Proof: For any $k, j \in \{ 0, 1, 2, ..., n - 1 \}$ clearly $*$ is commutative since:
(3)
\begin{align} \quad [k]_p * [j]_p = [k \cdot j]_p = [j \cdot k]_p =[j]_p + [k]_p \end{align}
• Now since $p$ is a prime number, the only positive divisors of $p$ are $1$ and $p$ itself. Therefore $p \equiv 0 \pmod p$.
• Now consider the set $\mathbb{Z} / p\mathbb{Z}$ and assume that this set contains a zero divisor. For some $k \in \{1, 2, ..., p -1 \}$ let $[k]_p \in \mathbb{Z} / p\mathbb{Z}$ be this zero divisor. Then there exists another $j \in \{1, 2, ..., p - 1 \}$ such that $[j]_p \in \mathbb{Z} / p\mathbb{Z}$ is such that:
(4)
\begin{align} \quad [k]_p * [j]_p = [k \cdot j]_p = [0]_p \end{align}
• Therefore we must have that $k \cdot j = 0$, so $k \cdot j \equiv 0 \pmod p$. Therefore $p \mid k \cdot j$. Since $p$ is a prime number and by The Division Algorithm we have that $1 \leq k, j < p$ we have that $\gcd (p, k) = 1$ and $\gcd (p, j) = 1$. Therefore $\gcd (p, k \cdot j) = 1$ which implies that $p \not \mid k \cdot j$ which is a contradiction.
• Therefore the assumption that $\mathbb{Z} / p \mathbb{Z}$ has a zero divisor is false. Therefore the ring $(\mathbb{Z}/p\mathbb{Z}, +, *)$ is an integral domain. $\blacksquare$