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I want to prove the following:

$A \subset B$ be integral and flat extension of rings then it is faithfully flat.

Clearly enough to show that for every ideal $I$ of $A$, $I^{ec}=I$. Since the extension is integral for every prime ideal $p$ of $A$ $p^{ec}=p$. I can’t show that it also holds for arbitrary ideal. I need some help. Thanks.

user371231
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1 Answers1

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Hint:

It is enough to prove that for any prime ideal $\mathfrak p\in\operatorname{Spec} A$, there exists a prime ideal $\mathfrak q\in\operatorname{Spec} B$ such that $\;\mathfrak q\cap A=\mathfrak p$.

Bernard
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  • Actually sir my query is why it is enough ? – user371231 Mar 17 '19 at 12:39
  • This depends on what you know about faithful flatness.. In Bourbaki, this assertion for maximal ideals is one of the equivalent properties that make up the definition. So what are you supposed to know for the definition? – Bernard Mar 17 '19 at 12:46
  • yes I missed out that maximal ideal property. It easily follows from the lying above and the maximal ideal property. Thanks to remind me. – user371231 Mar 17 '19 at 13:04