Let $R$ be a subring of a commutative ring $S$ . If $f$ is a polynomial in $R[X]$ , it defines an evaluation homomorphism from $S$ to $S$ . Any element $\alpha$ of $S$ satisfying $$f(\alpha) = 0$$ is a zero of the polynomial $f$ .

If $R$ also is equipped with a non-zero unity, then the polynomial $f$ is in $S[X]$ divisible by the binomial $x\!-\!\alpha$ (cf. the factor theorem). In this case, if $f$ is divisible by $(X\!-\!\alpha)^n$ but not by $(X\!-\!\alpha)^{n+1}$ , then $\alpha$ is a zero of the order $n$ of the polynomial $f$ . If this order is 1, then $\alpha$ is a simple zero of $f$ .

For example, the real number $\sqrt{2}$ ($\in \mathbb{R}$ ) is a zero of the polynomial $X^2\!-\!2$ of the polynomial ring $\mathbb{Q}[X]$ .