Is $(\mathbb{Z}/p\mathbb{Z})[x]$ an integral domain? Is $(\mathbb{Z}/p\mathbb{Z})[x]$ an integral domain?
Take two polynomials $a_nx^n+\ldots+a_1x+a_0$ and $b_mx^m+\ldots+b_1x+b_0$, and suppose their product is $0$. Then we have that either $a_n=0$ or $b_m=0$ (looking at the $n+m$-degree term of the product.) If $b_m=0$, then either $a_n=0$ or $b_{m-1}=0$ (looking at the $n+m-1$-degree term of the product), etc.
It is not clear whether this reasoning will yield that one of the polynomials must be $0$.

More generally, $R[x]$ is an integral domain iff $R$ is an  integral domain.
Indeed:


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*$R[x]$ is an integral domain $\implies$ $R$ is an integral domain: 
Easy because $R$ is a subring of $R[x]$.

*$R$ is an  integral domain $\implies$  $R[x]$ is an integral domain: 
Take $f,g\in R[x]\setminus 0$. Then $f=a_nx^n + \cdots$ and $b=b_mx^m + \cdots$ with $a_n\ne0$ and $b_m\ne0$. Since $R$ is a domain, we have $a_nb_m\ne0$ and so $fg=a_nb_mx^{n+m}+\cdots \ne 0$.
Now, $\mathbb{Z}/p\mathbb{Z}$ is a domain (assuming $p$ is a prime) and so $ \mathbb{Z}/p\mathbb{Z}[x]$ is a domain.
