$S^{−1} I = 0$ if and only if there exists an $s ∈ S$ such that $s I = 0$

Question

Let $R$ be a commutative ring with unity, $S \subset R$ a multiplicative set, and $I \subset R$ a finitely generated ideal. It's quite straightforward to show that $S^{−1} I = 0$ if and only if there exists an $s ∈ S$ such that $s I = 0$.

But now I'm wondering whether the same claim is true without the assumption that $I$ is finitely generated. I'm trying to find a counterexample by thinking of non Noetherian, non ID rings, such as $(\mathbb{Z} /12\mathbb{Z})[x^{\frac{a}{b}}]$, the ring of polynomial with rational powers, and coefficents in $\mathbb{Z} /12\mathbb{Z}$.

But I can't really find the right multiplicative set $S$ or the right ideal $I$. Does anyone have a counterexample or knows how to prove it?

Thanks for the help!

Answer 1

Hint : How about $R= k[x_i, y_i, i\in \mathbb N]/(x_iy_i, i\in \mathbb N)$, with obvious choices of $S,I$ ?