Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Why is an open faithfully-flat morphism fpqc?

In other words, why must an open faithfully flat morphism $X\rightarrow Y$ have the property that around every $x\in X$, there is an open nbhd $U$ of $x\in X$ such that $f(U)$ is open, and the restriction of $f$ to $U\rightarrow f(U)$ is quasi-compact?


  • will
share|cite|improve this question
See, there Qing Liu has answered this question. – Martin Brandenburg Sep 15 '13 at 8:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.