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Let $B\subset A$ be an inclusion of Krull dimension 1 rings, and assume $A$ a Dedekind domain. I think that $\mathfrak{p}\in Spec B$ is in the image of the contraction map $\mathfrak{P} \mapsto \mathfrak{P}\cap B$ if and only if $\mathfrak{p}$ contains no units of $A$, but I can not find a proof neither a counterexample.

In other words, is it true that $\mathfrak{p}A=A$ if and only if $\mathfrak{p}\cap A^{*}\neq \emptyset$?

(In K. Conrad's notes "The conductor ideal of an order" - https://kconrad.math.uconn.edu/blurbs/gradnumthy/conductor.pdf I found a lot of useful facts if I take $B$ an order in $K$ fraction field of $A$ and $B$, $A$ maximal order. In that situation we seem to have a bijective correspondence between the primes not dividing the conductor, hence the problem would reduce to the finitely many primes of $B$ which divide the conductor. But still I do not see an easy proof or counterexample.)

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  • $\begingroup$ You have a typographical error or missing hypothesis, since there is an easy counterexample: let $B = \mathbf Z$ and $A = \mathbf Z_{(p)}$ (localization of $\mathbf Z$ at the ideal $(p)$) or $\mathbf Z_p$ ($p$-adic integers). The only prime ideals in $A$ are $(0)$ and $pA$, which contract to $(0)$ and $p\mathbf Z$, neither of which contain a unit in $A$. $\endgroup$
    – KConrad
    Commented Jan 28, 2022 at 15:57
  • $\begingroup$ Yes thank you, I meant actually $\mathfrak{p}$ does not contain units of $A$. I have edited the question. I hope now is more clear $\endgroup$
    – Hair80
    Commented Jan 28, 2022 at 16:20
  • $\begingroup$ The implication $(\Leftarrow)$ is true for all ring extensions. When $A$ is integral over $B$, without Dedekind or Krull dimension conditions, the equivalence is true since ${\mathfrak p}A \not= A$ for all prime ideals $\mathfrak p$ in $B$ (Lang, Algebraic Number Theory, Proposition 9, Chapter I), so also ${\mathfrak p} \cap A^\times = \emptyset$, as a nonempty intersection would make ${\mathfrak p}A = A$. Thus you may as well assume $A$ is not integral over $B$. $\endgroup$
    – KConrad
    Commented Jan 28, 2022 at 17:10
  • $\begingroup$ Thank you very much for your response, actually the example I am looking at consists in $B$ an order and $A$ a Dedekind domain not integral on $B$. Would you have any idea on what happens in this case? $\endgroup$
    – Hair80
    Commented Jan 29, 2022 at 5:27

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