All rings are commutative with unity.

Let $\phi:A \to B$ be a faithfully flat ring homomorphism. Let $f \in A$, $g = \phi(f) \in B$, and $\psi:A_f \to B_g$ the induced homomorphism on the localizations. Assume that $f,g$ aren't zero divisors, so we may think of $A$ and $B$ as a subrings of $A_f$ and $B_g$.

Question. Is it true that $\psi^{-1}(B) \subseteq A$?


  1. Just 'flat' instead of 'faithfully flat' is not enough -- for instance, we may take $B = A_f$ and $\phi$ the usual localization map, and then the claim is (almost always) false.
  2. If mild finiteness hypotheses are necessary, please assume them by all means. In fact, for what I need, there's a huge additional assumption, namely that $\phi$ is also etale, but my feeling was that something like faithfully flat should still work. If etale (or possibly just smooth) is needed, I'm equally happy (at least as far as my specific need goes).
  3. A proof would be good, a reference even better.

1 Answer 1


In hindsight, this is trivial/well-known. The key point is that for faithfully flat extensions, contraction of extension = original ideal (for instance https://math.stackexchange.com/questions/1175630/extending-and-contracting-an-ideal-by-a-faithfully-flat-homomorphism ).

In our case, $\phi^{-1}((g)) =(f)$ and even $\phi^{-1}((g^n)) =(f^n)$. Now, if $\psi(a/f^n) \in B$, then $\phi(a) \in (g^n)$, done!

Sorry for asking a trivial question!


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