6
$\begingroup$

It is known that the following are equivalent for an epimorphism $A \to B$ in $\mathbf{CRing}$:

  1. Let $S$ be the set of elements $a \in A$ such that $A [a^{-1}] \to B [a^{-1}]$ is an isomorphism. Then $S$ generates the unit ideal in $B$.

  2. $A \to B$ is a flat homomorphism of finite presentation.

  3. $A \to B$ is an étale homomorphism.

What I'd like to know is if there is an algebraic proof that 2 implies 1.

Here is a cheat proof: it is well known that flat morphisms locally of finite presentation are open maps, so $\operatorname{Spec} B → \operatorname{Spec} A$ factors as an fppf monomorphism $\operatorname{Spec} B → U$ followed by an open immersion $U → \operatorname{Spec} B$; but fppf monomorphisms are isomorphisms, so we are done. This proof is unsatisfactory to me because (a) it involves a scheme that is not known to be affine a priori and (b) the well-known fact about flat morphisms locally of finite presentation being open is (to me, at least) a deep result in scheme theory.

(Incidentally, is there a good name for ring homomorphisms corresponding to open immersions? I want to say "open localisation", but that's misleading because they are not necessarily localisations...)

$\endgroup$
  • 3
    $\begingroup$ For each prime ideal $p$ of $B$ and its contraction $q$ in $A$, I claim $A_q \rightarrow B_q$ is an isomorphism. Then the finite presentation hypothesis would provide $a \in A - q$ such that $A[1/a]\rightarrow B[1/a]$ is an isomorphism. The image of $a$ in $B$ lies outside the prime ideal $p$, so such $a$'s would generate 1 in $B$, as desired. So we can assume $A$ is local with maximal ideal $m$ hit by a prime of $B$ (hence $mB \ne B$), so $A \rightarrow B$ is faithfully flat (Theorem 7.2 in Matsumura's "Comm. Ring Theory"). You know the rest. QED (But the "cheat proof" is more illuminating!) $\endgroup$ – nfdc23 Feb 8 '16 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.