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Let $ B $ be a faithfully flat $ A $-algebra. Let $ I \subset A $ an ideal. Shows that $ IB \cap A = I $.

This is the second item of Exercise 2.6, Chapter 1, of the Qing Liu's book Algebraic Geometry and Arithmetic Curves.

If I suppose that $ I $ is a prime ideal I can solve the problem, but in the general case...I tried to localize $ A$ at $ I $ but I can not finish.

Miguel F D
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1 Answers1

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I am sure you can find a proof in any commutative algebra book; anyway, here is one. The natural map $i:B\rightarrow B\otimes _AB$ admits a retraction (given by multiplication), hence $$i\otimes 1:B\otimes_A A/I\rightarrow (B\otimes _AB)\otimes _AA/I$$ is injective. Since $B$ is faithfully flat over $A$, this implies that $A/I\rightarrow B\otimes A/I\cong B/IB$ is injective, which is equivalent to $IB\cap A=I$.