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
Comments (0)