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

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.

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.

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 $R[x]$

A simple observation: assume $f,g \in R[x] \setminus \{0\}$ then $$f(x) = \sum_{n=0}^N a_n x^n \ g(x) = \sum_{n=0}^M b_n x^n$$ where $a_N,b_M \in R \setminus \{0\}$.

The leading coefficient of $fg$, i.e. the coefficient of degree $N+M$ is $a_Nb_M$ which is not null since $a_N,b_M \ne 0$ and $R$ is a domain. As a consequence $fg$ as degree at least $N+M$ so it can not be the null polynomial.

If $R$ is an integral domain, but $R[x]$ is not, there must exist some minimal counter-example, that is some $f$ of minimal $\text{deg}\ n \geq 0$, for which there exists $g \neq 0$ such that $fg = 0$.
If $f(x) = a_0 + a_1x +\cdots + a_nx^n$ and: $g(x) = b_0 + b_1x +\cdots + b_mx^m$
it follows from $fg = 0$ that $a_nb_m = 0$.
But $g \neq 0 \implies b_m \neq 0 \implies a_n = 0$ (since $R$ is an integral domain), contradicting the minimality of $n$.