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

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?

• If you have two nonzero polynomials, when you multiply them, can the product of their highest degree terms be zero? – Ben West Mar 15 '17 at 5:35

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.

• I have a question. Taking $R:=\mathbb{F}_p$, which is a field (therefore an integral domain) we have by Fermat's little theorem that $a^{p}=a$ for all $a\in\mathbb{F}_p$, so $(x^{p-1}-1)x=0$ as polynomials in $\mathbb{F}_p[x]$. This means $\mathbb{F}_p[x]$ is not an integral domain... What am I missing? – rmdmc89 Feb 3 '18 at 13:10
• @rentatodias The problem is that $x$ is not an element of $\mathbb{F}_p$, just a free variable, and so Fermat’s little theorem doesn’t apply. That is, $x^p\ne x$ as polynomials in $\mathbb{F}_p[x]$. That said, this does mean that $x^p-x=0$ in $\mathbb{F}_p[x]/(x-a)$. – Santana Afton Feb 3 '18 at 16:50

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.