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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 $A/(a)\hookrightarrow B/(b)$?

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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.

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  • $\begingroup$ Thanks! So this shows that any homomorphism $B/b\to A/a$ comes from a homomorphismm $B\to A$ which maps $(b)$ to $(a)$. Is it clear that every such can be obtained by multiplication with $c$ as above? $\endgroup$
    – Hans
    Commented May 11, 2020 at 14:43
  • $\begingroup$ okay, I got it! $\endgroup$
    – Hans
    Commented May 11, 2020 at 15:50

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