Does UFD $R$ wrap around its ideal $I$ infinitely and surjectively as $\Bbb{Z}$ does $(n)$?

Question

If $R$ is a UFD, and $I$ is an ideal of $R$, then do elements of $R$ wrap around $I$ as they do in the case of $\Bbb{Z}$ and $(n)$. And by that I mean, letting $\pi : R \to R / I$ be the natural projection, is $\pi(x) = \bar{x} = y$ for infinitely many $x \in R$ when $R$ is infinite, except maybe at $y = 0$?

Answer 1

Because you assume $R$ to be an infinite integral domain, every ideal $I$ other than $\{0\}$ contains infinitely many elements. Therefore so does each and every coset of $I$ settling your claim in the affirmative. Won't need unique factorization for this.

Of course, if $I=\{0\}$, then $\pi$ injective, and the claim is false. If $R$ is a finite domain, then it is a field, and necessarily $I=\{0\}$ or $I=R$.