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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 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.)

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.)

Let $B\subset A$ be an inclusion of 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 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.)

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 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.)