Exercise A0.III.1.15

Proposition

$R[x]$ is an integral domain if and only if $R$ is an integral domain.

Proof

Suppose $R[x]$ is an integral domain. Let $a, b \in R[x]$ be two non-zero constant polynomials. Being constants, $a$ and $b$ live in the $R$-shaped subring of $R[x].$ Since $R[x]$ is an integral domain, $ab \neq 0,$ and in particular, this inequality holds in the $R$-subring of $R[x],$ demonstrating that $R$ is an integral domain.

Suppose conversely that $R$ is an integral domain, and let $f, g \in R[x]$ such that $fg = 0.$ Let $c_1, c_2 \in R$ be leading coefficients of $f$ and $g.$ That is, $f = c_1x^p + f'$ and $g = c_2x^q +g',$ where $p = \deg(f)$ and $q = \deg(g).$ Then

$$fg = (c_1x^p + f')(c_2 x^q + g') = c_1 c_2 x^{p + q} + \text{lower degree terms}.$$

Since $x^{p+q} \neq 0,$ it must be that $c_1 c_2 = 0.$ Hence, either $c_1$ or $c_2$ must be zero. The only polynomial whose leading coefficient is $0$ is the zero polynomial, so $f = 0$ or $g = 0.$ Thus, $R[x]$ is an integral domain.

$\blacksquare$