Let $R$ be a Noetherian domain, $t\in R$ be a non-zero, non-unit element, then is it true that
$$\bigcap_{n \ge 1} t^nR=\{0\} \text{?} $$
It almost feels like the nilradical (which is zero for any domain) and intersection of prime ideals, but I can't quite crack it, possibly because I can't see where Noetherianness comes into play.
Please help. Thanks in advance.
Let $I = \bigcap_{n \geqslant 1} t^{n}R$, and suppose $x \in I$. Then $x = t^{i}r_{i}$ for some $r_{i} \in R$ for each $i \in \mathbb{N}$; it is clear that if $x$ is nonzero, then each $r_{i}$ is nonzero too. In particular, we have $t^{i}r_{i} = x = t^{i+1}r_{i+1}$, and since $R$ is a domain, this implies $r_{i} = tr_{i+1}$, so $r_{i+i} \mid r_{i}$ for each $i \in \mathbb{N}$. Thus, we have a chain of principal ideals $(r_{1}) \subset (r_{2}) \subset \cdots$. Since $R$ is Noetherian, this chain must stabilize eventually, i.e. we must eventually have $(r_{k}) = (r_{k+1}) = \cdots$ for some $k \in \mathbb{N}$. Then in particular, $r_{k}$ and $r_{k+1}$ are associate, so $r_{k} = tr_{k+1} = \alpha t r_{k}$ for some $\alpha \in R^{\times}$. As noted above, if $x$ is nonzero, then (in particular) $r_{k}$ is nonzero, so $\alpha t = 1$ by cancellation, which contradicts the fact that $t$ is not a unit. Hence, we must have $x = 0$, so $I = \{0\}$.