Here is an exercise(p.129, ex.1.15) from Algebra: Chapter 0 by P.Aluffi.
Prove that $R[x]$ is an integral domain if and only if $R$ is an integral domain.
The implication part makes no problems, because $R$ is a subring of $R[x]$.
For the ''if'' part, however..
Let $R$ be an integral domain.
Now let's review pairs $f(x)=\sum \nolimits a_ix^i,g(x)= \sum \nolimits b_ix_i \in R[x]$ such that $f(x)g(x) = 0$
So, we have $\sum \limits_{k=0}^{\infty} \sum \nolimits_{i+j=k} a_ib_jx^{i+j} = 0$.
Now, I'm not sure how to deduce that $f(x) = 0 \vee g(x) = 0$
If I look at $f(x),g(x)$ such that, for example, $deg((f(x))=3, deg((g(x))=2)$, it makes sense. I begin by something like that "if $f(x)g(x)=0$ then $a_0b_0=0$ then $a_0 = 0 \vee b_0 = 0$. If $a_0 =0 $, then $a_0b_1 = a_0b_2 = 0$" and so on.
As I understand, it comes down to proving the following implication:
$(\forall k \in \mathbb{N} \sum \nolimits_{i+j=k} a_ib_j = 0)(1) \Rightarrow ((\forall n \in \mathbb{N} \ \ a_n = 0) \vee (\forall m \in \mathbb{N} \ \ b_m = 0))(2)$
We can say that $(1)$ is a system of equations in $R$. And $(2)$ is it solution.
Best Answer
Suppose that neither $f$ nor $g$ is the zero polynomial. Then there exist non-negative integers $k$ and $l$ and ring elements $a_0,a_1,\dots, a_k$, with $a_k\ne 0$, and $b_0,b_1,\dots,b_l$, with $b_l\ne 0$, such that $$f=a_0+a_1x+\cdots+a_kx^k \quad\text{and}\quad g=b_0+b_1x+\cdots+b_lx^l.$$
The coefficient of $x^{k+l}$ in the product $fg$ is $a_kb_l$. Since $R$ is an integral domain, we have $a_kb_l\ne 0$, and therefore $fg$ is not the zero polynomial.