Why does the ring of entire functions have no zero divisors?

Question

Why does the ring of entire functions have no zero divisors, while the ring of infinitely differentiable functions on the real line does?

Answer 1

Hint An entire function whose zero set has an accumulation point is the zero function.

Answer 2

Let $fg\equiv0$ which $f$ and $g$ are entire function. If $f(z)\neq 0$ for some $z\in \mathbb C$, then $f$ for some $\delta \gt0$ be non-zero in $D(\delta,z)$. every point of $D(\delta,z)$ is a limit point of $D(\delta,z)$ and from the identity $fg\equiv0$ we shod have $g(z)=0$ for every $z\in D(\delta,z)$. It's mean that the zero point of $g$ in $D(\delta,z)$ has limit point. So by the identity theorem, $g\equiv0$.