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How to describe all ring homomorphisms $f: A \rightarrow B$, such that corresponding affine scheme morphism $f: Spec \, B \rightarrow Spec \, A$ is open immersion?

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    $\begingroup$ $\def\Spec{\operatorname{Spec}}$Don't we have $f\colon\Spec B \to \Spec A$? $\endgroup$ – martini Oct 30 '12 at 9:54
  • $\begingroup$ Of course! I'm sorry. I've edited. $\endgroup$ – user46336 Oct 30 '12 at 10:06
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    $\begingroup$ I asked the same question here: mathoverflow.net/questions/20782/… $\endgroup$ – Manny Reyes Oct 30 '12 at 10:54
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The answers in the link given by Manny are great. Let me just add another sufficient condition which may be simpler to check than, e.g. flatness. If

  • $A$ is an integrally closed domain,
  • $B$ is contained in $\mathrm{Frac}(A)$ and finitely presented over $A$ (as $A$-algebra),
  • $f$ is quasi-finite (i.e. for all prime ideals $p$ of $A$, $B/pB$ is artinian),

then $f$ is an open immersion. This is a form of Zariski's Main Theorem.

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