# Integral Domain of Integers

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.

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