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$B$ is a flat $A$-algebra. For any ideals $I, J \in A$, does $IB \cap JB = (I \cap J)B$ holds? (The answer seems to be YES, how to prove it?)

Furthermore, if $f: A \to B$ is injective (maybe $B$ is not a flat $A$-module), does the equality still holds?

Lwins
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  • The answer to your first question is a "yes". See Tag 0BBY (currently Lemma 10.38.2) in the Stacks Project: http://stacks.math.columbia.edu/tag/0BBY – darij grinberg Dec 21 '16 at 03:49
  • I think the answer to your second question is a "no", and $A = \mathbb{Z}\left[x,y\right]$ and $B = \mathbb{Z}\left[x,y,z\right] / \left(xz-yz\right)$ should provide a counterexample (with $f$ being the canonical map); but I don't have the time to check everything. – darij grinberg Dec 21 '16 at 03:51

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