Given: domains $A$ and $B$ such that $A\subseteq B$ and such that their fraction fields are equal. Also $A$ is a unique factorization domain and if $a \in A$ is a unit in $B$, it is a unit in $A$. I want to show that if $A \subseteq B$ is flat then $B=A$. I think I need to use the fact that every element in $B$ can be written as $\frac{a_1}{a_2}$ with $a_1,a_2 \in A$. Does anyone have an idea?
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Something is flat as a module (or a module morphism). But $A$ is not, in general, a module over $B$. The only way I can attribute a meaning to “$A\subseteq B$ is flat” is that $B$ is flat as an $A$-module. – egreg Oct 22 '13 at 23:41
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@egreg: This is very common terminology. – Martin Brandenburg Oct 23 '13 at 08:00
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@MartinBrandenburg Not that $A$ is flat. – egreg Oct 23 '13 at 08:13
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Oh, you refered to an older version of the question. – Martin Brandenburg Oct 23 '13 at 22:49
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The assumption $A \cap B^* = A^*$ implies that $\mathfrak{p}B \neq B$ for all prime ideals $\mathfrak{p} \subseteq A$, i.e. that $B$ is faithfully flat over $A$. Now use SE/482908.
Martin Brandenburg
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Why $\mathfrak pB\neq B$ and why this is enough to conclude that $B$ is f.f. over $A$? – Oct 23 '13 at 08:26
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1Due to the lack of any explanation I consider this answer as being far from complete (possible wrong) and I give it a -1. – Oct 27 '13 at 06:57