Prove that $R$ is an integral domain $\Leftrightarrow$ $R[x]$ is an integral domain

Solution 1:

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.

Solution 2:

This is a bit extreme, but this is also a consequence of the famed McCoy's Theorem. If $f(x)\in R[x]$ is a zero divisor, then there is a non-zero $r\in R$ such that $r\cdot f(x)=0$. Thus if $R[x]$ were not a domain, there would be an $f(x)$ which is non-zero and a zero-divisor. But that would give you an $r\in R$ that would kill all of the coefficients of $f(x)$, so $R$ would not be a domain. Bill Dubuque sketches a proof here: Zero divisor in $R[x]$