TITLE: Faithfully flatness of $A\subseteq B$ implies $aB\cap A = aA$
QUESTION [2 upvotes]: When $A\subseteq B$ is faithfully flat, why does $aB\cap A = aA$ hold, where $a\in A$?

REPLY [0 votes]: Since the quotient $B/A$ is flat, $M \to B \otimes_A M$ is injective for every $A$-module $M$. For $M = A/I$ this gives $IB \cap A = I$.
{"set_name": "stack_exchange", "score": 2, "question_id": 524681}