We have previously shown gone through the process of showing that we have a ring in A Ring of Matrices. We have also shown that the relation, Congruence Modulo n, is a congruence relation on the set of integers for any integer \(n\). Today, we will look at the set of integers modulo this congruence relation when we are using a prime number for \(n\). In this case, we will not only have a ring, but an integral domain.
An Integral Domain
Let \(p \in \mathbb{Z}\) be a prime number. Then \(\mathbb{Z}_{p}\) is an integral domain.
A Ring
In order to show that \(\mathbb{Z}_{p}\) is an integral domain under addition and multiplication mod \(p\), we must first show that it is a ring. However, we have already shown that congruence modulo \(n\) is a congruence on \(Z\) for both the multiplication and addition for any integer \(n\). Then, since we defined \(\mathbb{Z}_{p}\) as the integers modulo the congruence relation modulo \(p\), we have that \(\mathbb{Z}_{p}\) is indeed a homomorphic image of \(\mathbb{Z}\). Since we already know that homomorphic image of a ring is again a ring, we have that \(\mathbb{Z}_{p}\) is indeed a ring.
That is, we will be using the definition of \(\mathbb{Z}_{p}\) along with information we have shown about homomorphisms to show that we have a ring. This is indeed less work than showing that \(\mathbb{Z}_{p}\) satisfies all the requirements of a ring individually.
Identity
In addition to being a ring, an integral domain must also have a multiplicative identity. Now, we know that \(1\) is the multiplicative identity for \(\mathbb{Z}\). Hence, the image of \(1\) under the above defined homomorphism will be the multiplicative identity of \(\mathbb{Z}_{p}\). That is \(1 \text{ mod } p\) will be the identity. To show this, let \(a \in \mathbb{Z}_{p}\). Then, since \(a*1=a=1*a\), we have that
\begin{align*}
(a \text{ mod } p)*(1 \text{ mod } p) &=
a*1 \text{ mod } p \\
&=a \text{ mod } p \\
&=1 *a \text{ mod } p \\
&=(1 \text{ mod } p)*(1 \text{ mod } p).
\end{align*}
Hence, \(1 \text{ mod } p\) is the identity on \(\mathbb{Z}_{p}\).
Zero Divisors
The last step is to show that \(\mathbb{Z}_{p}\) has no zero divisors. That is, we need to show that if \(a,b \in \mathbb{Z}_{p}\) and \((a \text{ mod } p )*(b \text{ mod } p)=0 \text{ mod } p\), then \(a = 0 \text{ mod } p\) for \(b= 0\text{ mod } p\). In order to do this, let \(a,b \in \mathbb{Z}\). Then suppose that
\begin{align*}
ab=0 \text{ mod } p
\end{align*}
Then, we must have that \(p | ab\). However, since \(p\) is prime, this implies that \(p|a\) or \(p|b\). Hence \(a =0 \text{ mod } p\) or \(b=0 \text{ mod } p\) as required.
Proof
Let \(p \in \mathbb{Z}\) be a prime number. We note that congruence modulo \(p\) is a congruence relation for both the addition and multiplication on \(\mathbb{Z}\) for any integer \(p\) by the proof in Congruence Modulo n. Since we \(\mathbb{Z}_{p}\) as the integers modulo the congruence relation for \(p\), we get that \(\mathbb{Z}_{p}\) is a homomorphic image of \(\mathbb{Z}\). Hence, \(\mathbb{Z}_{p}\) is a ring.
Next, let \(a \in \mathbb{Z}\). Then \(a \text{ mod } p \in \mathbb{Z}_{p}\) and
\begin{align*}
(a \text{ mod } p)*(1 \text{ mod } p) &=a*1 \text{ mod } p \\
&=a \text{ mod } p \\
&=1*a \text{ mod } p \\
&=(1 \text{ mod } p)*(a \text{ mod } p).
\end{align*}
Hence, \(1 \text{ mod } p\) is the multiplicative identity for \(\mathbb{Z}_{p}\).
Now, suppose that \(a,b \in \mathbb{Z}\) and that \((a \text{ mod } p)*(b \text{ mod } p)=0 \text{ mod } p\). Then, we have that
\begin{align*}
0 \text{ mod } p &= (a \text{ mod } p)*(b \text{ mod } p) \\
ab \text{ mod } p.
\end{align*}
Therefore, \(p |ab\). Since \(p\) is prime, by definition this implies that \(p|a\) or \(p|b\). That is \(a = 0 \text{ mod }p\) or \(b =0 \text{ mod } p\). Hence, \(\mathbb{Z}_{p}\) has no zero divisors.
Because \(\mathbb{Z}_{p}\) is a ring with identity and no zero divisors, it is an integral domain.
Conclusion
I hope this helps as you work through proving that sets are integral domains. If it did, make sure to share it with anyone else that may need the help.
