# are restrictions of a finite flat morphism of local rings to irreducible components flat?

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 $$S$$?

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 $$Y$$.