Prove that if a commutative ring $R$ is integral domain, then the polynomial ring $R[x]$ is also an integral domain.

Question

I know that it is true that if a commutative ring $R$ is integral domain, then the polynomial ring $R[x]$ is also an integral domain. But I am having troubles with proving this statement. Can I ask for someone's help, please?

Answer 1

Suppose that $R[x]$ is not an integral domain. Then there are two nonzero elements, $p,q\in R[x]$ such that $pq=0$. Write $p$ and $q$ as

$$p(x) = \sum_{i=0}^n a_ix^i;\qquad q(x)=\sum_{i=0}^mb_ix^i$$

where deg$(p)=n$ and deg$(q)=m$. Consider the coefficient of the $x^{n+m}$ term, $a_nb_m$, in the product $pq$. Since $pq=0$, this implies $a_nb_m=0$. This contradicts the assumption that $R$ is an integral domain.

Answer 2

You have to show that if $f(x), g(x) \in R[x]$ are elements, then if $f(x)g(x) = 0$, it must mean that neither $f(x)$ or $g(x)$ are the zero divisors.

Such elements take the form $f(x) = \sum a_n x^n$ and $g(x) = \sum b_m x^m$, then their product $f(x)g(x) = \sum_{n,m}a_nb_mx^{n+m} =0 \iff a_nb_m = 0$. But $a_n, b_m \in R$, so neither of these are zero divisors meaning the original polynomials are not zero divisors. Since these polynomials are chosen arbitrarily, you are done.