Let $R$ and $S$ be two commutative rings and $f$ and $g$ be two homomorphisms from $R$ to $S$. Also, let $E$ be their equalizer, i.e. the subring $\{x \in R \vert f(x)=g(x)\}$.
If $R$ is a flat $E$-module, does this imply that it is faithfully flat?
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Geoffrey Trang
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Here is a possible suggestion for a counter example which I haven't checked carefully, so may be wrong. Let $R=S=k[x_1,x_2,y_1,y_2]/(x_1y_1+x_2y_2-1)$ where $k$ is any field. Let $f$ be identity and $g(x_i)=x_i, g(y_1)=y_1-x_2, g(y_2)=y_2+x_1$. Then, it looks as though $E=k[x_1,y_1]$ and $E\subset R$ is flat, but not faithfully flat. – Mohan Jul 22 '19 at 01:38