Can $R/I$ be $R$-flat if $R$ is a domain?

Question

Let $R$ be a domain and $I$ a nonzero proper ideal of $R$. Can $R/I$ be flat?

For example, I can show if $R$ is Noetherian no such ideal $I$ exists. To see this, if $R/I$ is flat than by considering the injective map $I\to R$ and applying $\otimes R/I$ gives an injective map $I\otimes R/I\to R\otimes R/I$ and applying the isomorphism that $M\otimes R/I = M/I $ shows that $I/I^2\to R/I$ is injective so $I=I^2$. This cannot occur if $I$ is finitely generated.

Answer 1

Let $i \in I$ be non-zero and consider the injective map $R \to R$ given by multiplication with $i$. If we tensor this with $R/I$, we get the zero map, which is not injective.