If $(R:_{Q(R) } S)$ is non-zero, then does $(R:_{Q(R) } S)$ contain a non-zero-divisor?

Question

Let $R$ be a reduced Noetherian local ring of dimension $1$ with total ring of fractions $Q(R)$. Consider a ring extension $R \subseteq S \subseteq Q(R)$.

If $(R:_{Q(R) } S)$ is non-zero, then does $(R:_{Q(R) } S)$ contain a non-zero-divisor?

Answer 1

Consider $R=k[x,y]_{(x,y)}/(xy)$ and $S=R[\frac{x}{x+y},\frac{x}{x^2+y},\dots]$, then $(R:_{Q(R)}S)$ contains $y$, so it is non-zero. Suppose that $d \in (R:_{Q(R)}S)$. Then for all $n$, we have $d\frac{x}{x^n+y} \in R$, so $$dx \in \bigcap_{n \geq 1} (x^n+y) \subset \bigcap_{n \geq 1}(x^n,y)=(y)$$ where the last equality comes from applying Krull's intersection theorem to the ring $R/(y)\cong k[x]_{(x)}$. This implies that $dx^2 \in (xy)=(0)$, so $d$ is a zero-divisor.