Are restrictions of a finite flat morphism of local rings to irreducible components flat

Question

Let $R \to S$ be a morphism of local Noetherian rings, and assume that $R$ is integral and $S$ is a finite flat module over $R$. Is it true that $S/I$ is flat over $R$ for $I$ minimal prime ideal of $S$?

Answer 1

This is not true. The counterexample is as follows: take the nodal cubic $Y$ and let $X=\mathbb{A}^1$ be its normalization with points $a,b$ above the node. Take two copies $X_0, X_1$ of $X$ and glue $a_0$ to $b_1$ and $a_1$ to $b_0$, getting $X'$. The morphism $X' \to Y$ is étale, but each irreducible component is clearly not flat over $Y$.