Zero divisors in a ring of functions

Detailed Solution

$R$ is a commutative ring with pointwise operations, and the constant function $1(x)=1$ is a multiplicative identity, so (A) is true. Let $f=\chi_{[0,1]}$ and $g=\chi_{(1,2]}$; both are nonzero (e.g. $f(0)=1$, $g(1.5)=1$), but since $[0,1]$ and $(1,2]$ are disjoint, $f(x)g(x)=0$ for every $x\in\mathbb R$, so $fg=0$. Hence $f$ and $g$ are nonzero zero divisors, so (B) is true. If $f(x)\neq 0$ for every $x$, then $g(x)=1/f(x)$ defines $g\in R$ with $fg=1$, so $f$ is a unit; conversely if $fg=1$ then $f(x)g(x)=1$ forces $f(x)\neq 0$ for every $x$. So (C) is true. Since (B) exhibits an explicit pair of nonzero zero divisors, statement (D) is FALSE. Thus the correct option is $\boxed{\mathrm{D}}$.