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Let $A \subseteq B \subseteq C$ be rings with $B$ a faithfully flat $A$-module and $C$ a faithfully flat $B$-module which is an integral extension of $B$. Given a maximal ideal $I \subseteq C$, is it true that $$(I \cap A)B = I \cap B$$ I know that faithful flatness is transitive and also gives $(I \cap B)C = I$. I feel like this is true, but have been unable to prove it.

Austin Shiner
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Faithful flatness implies that contraction is left inverse to extension of ideals, I think you had that backwards. In other words $IC \cap B = I$ for faithfully flat extensions, but not $(I\cap B)C = I$.

For a counterexample to your question, consider $k \subseteq k[x^2] \subseteq k[x]$ and the maximal ideal $xk[x]$. This also shows that even for the finite faithfully flat extension $k[x^2] \subseteq k[x]$, contraction of ideals is not left inverse to extension.

Badam Baplan
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  • Ah I see. Thank you. Do you know of an equivalent condition for which $(I \cap B)C = I$? I know this is equivalent to the algebra morphism $B \to C$ being surjective but do not know any others. – Austin Shiner Jun 07 '20 at 15:04
  • @AustinShiner No, it's not equivalent to the morphism being surjective. As far as I know it is a delicate property, typical of epimorphisms. For example if $B \rightarrow C$ is an epimorphism, then extension is left inverse to contraction if additionally either (1) $B \rightarrow C$ is finite (i.e. surjective); or (2) $B \rightarrow C$ is flat. – Badam Baplan Jun 07 '20 at 15:56
  • @ Badam Baplan Why is it not equivalent to the homomorphism $f: B \to C$ being surjective? The contraction ideal is defined to be $f^{-1}(I)$ and the extension ideal would then be $f(f^{-1}(I))$. If $f$ is surjective then $f(f^{-1}(S)) = S$ for any set $S$, and in particular, $f(f^{-1}(I)) = I$. Where did I go wrong in this? – Austin Shiner Jun 07 '20 at 17:37
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    @Austin Shiner You have correctly observed that surjective morphisms have the property in question. However, that doesn't establish that the property in question is equivalent to being surjective, and indeed, it is not. For example, a consequence of (2) in my previous comment is that localizations have the property in question, and localizations are typically not surjective. – Badam Baplan Jun 07 '20 at 18:35