Let $A\subset B$ be a finite extension of Dedekind domains. Let $0\neq b\in B$ and $0\neq a\in A$ such that $(a)=(b)\cap A$. In particular, we have $a=b\cdot c$ for some $c\in B$. Now for any $A$-linear map $\varphi:B\to A$ the map $x\mapsto\varphi(c\cdot x)$ maps the ideal $(b)$ into $(a)$. Thus we obtain an $A$-linear map $\overline{\varphi}:B/(b)\to A/(a)$. This defines a homomorphism of $A$-modules (even $B$-modules):$$\textrm{Hom}\_A(B,A)\to\textrm{Hom}\_A(B/(b),A/(a)),\, \varphi\mapsto \overline{\varphi}.$$


I have checked several examples and this map was always surjective. Is this always the case?


Moreover, if we relax the condition of $A$ and $B$ being Dedekind domains to just being some integral regular domains, can we still say something on the image of this map? Can we describe it just in terms of the finite ring extension $A/(a)\hookrightarrow B/(b)$?



Extension of Dedekind domains and their codifferent
Any $A$-module homomorphism $B/b\to A/a$ gives, by composing with $B\to B/b$, a homomorphism $B\to A/a$. Using the surjection $A\to A/a$ and the fact that $B$ is $A$-projective, the map lifts to give a homomorphism $B\to A$.


So, the only real assumption needed is that $B$ is $A$-projective.


