Lemma 15.104.2. Let $A \to B$ be a ring map such that $B \otimes _ A B \to B$ is flat. Let $N$ be a $B$-module. If $N$ is flat as an $A$-module, then $N$ is flat as a $B$-module.

**Proof.**
Assume $N$ is a flat as an $A$-module. Then the functor

\[ \text{Mod}_ B \longrightarrow \text{Mod}_{B \otimes _ A B},\quad N' \mapsto N \otimes _ A N' \]

is exact. As $B \otimes _ A B \to B$ is flat we conclude that the functor

\[ \text{Mod}_ B \longrightarrow \text{Mod}_ B,\quad N' \mapsto (N \otimes _ A N') \otimes _{B \otimes _ A B} B = N \otimes _ B N' \]

is exact, hence $N$ is flat over $B$. $\square$

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