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**pleasehelpme1** I know that if a ring is a ED it is a PID and if it is a PID it is a UFD and if it is a UFD it is an integral domain. So this can be proven by proving ED, PID or UFD. .

You don't really have to do all of that. To show that a ring is an integral domain, you just need to show that it's commutative, have unity, and have no zero divisors.

1) Let $\displaystyle a,b \in \mathbb{Z}_p$ and $\displaystyle ab=0$. Then $\displaystyle ab=pk$ for some $\displaystyle k \in \mathbb{Z}$, which means that $\displaystyle p|a$ or $\displaystyle p|b$. Therefore, in $\displaystyle \mathbb{Z}_p, a=0$ or $\displaystyle b=0$. $\displaystyle \Box$

2) $\displaystyle \mathbb{Z}_p[x]$ is a commutative ring with unity; therefore, we need to show that there are no zero divisors.

Let $\displaystyle f(x), g(x) \in \mathbb{Z}_p[x]$, where

$\displaystyle f(x)=a_mx^m+a_{m-1}x^{m-1}+\dots+a_0$

$\displaystyle g(x)=b_nx^n+b_{n-1}x^{n-1}+\dots+b_0$,

and $\displaystyle a_m \not=0$ and $\displaystyle b_n \not=0$. Then $\displaystyle f(x)g(x)$ has leading coefficient $\displaystyle a_mb_n$, but since $\displaystyle \mathbb{Z}_p$ is an integral domain, $\displaystyle a_mb_n \not=0$. $\displaystyle \Box$