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let $R$ be a domain with field of fractions $F=Frac(R)$ and $A$ the integral closure of $R$. we have a canonical inclusion map $\varphi:R \to A$. if we assume that $\varphi$ is a flat map why does this already imply that $R=A$? in this question the OP used

Fact: Let $A$ be an integral domain and $K$ be its field of fractions. Also let $B$ be a finitely generated $A$-submodule of $K$. Then $B$ is flat iff $B$ is locally free of rank $1$.

why this Fact holds? I don't understand why the equivalence that $B$ is flat iff $B$ is locally free of rank $1$ holds. I shall be delighted if anybody could give a sketch of the proof. I know that locally every flat map is free. what I know about the rank? this means that for every prime $\mathfrak{p}$ of $R$ the $R_{\mathfrak{p}}$-module $A_{\mathfrak{p}}$ is free of some rank $r_p$. since $R$ and $A$ have same fraction field the rank is $1$ at generic point. the question is why is then the rank $r_p$ equals $1$ at every prime $\mathfrak{p}$ of $R$? guess that there exist a flatness argument but haven't found such one.

  • What do you mean by "why this fact can be applied here?" What fact and what here? – user26857 Aug 04 '19 at 16:32
  • the quoted *Fact. excuse I think my formulation was quite unprecise. basically the question is why under given conditions on $B$ and $A$ as given in the Fact* follows that $B$ is flat iff $B$ is locally free of rank $1$ –  Aug 04 '19 at 18:09
  • Are you familiar with homological algebra? This might be an overkill, but for any module $M$ over a (commutative) ring, we have that $M$ is flat over $R$ $\iff$ for every maximal ideal $m$, $M_m$ is flat over $R_m$. This follows from the fact that flatness can be characterized using $\mathrm{Tor}$ and $\mathrm{Tor}$ functor is well behaved under localization. Are you ok with this argument? – Parthiv Basu Aug 04 '19 at 21:54
  • non quite. using this characterization it suffice to consider only localisations $A_{\mathfrak{m}}$ at maximal ideals. therefore we can assume by localizing $A$ to $A_{\mathfrak{m}}$ that $A$ is local & flat now. so it is a free $R$ module. can we from here conclude that it must have rank one under assumption that $A$ and $R$ have same fraction field? –  Aug 04 '19 at 22:57
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    $B$ is finitely generated and flat over $A$, which implies $B_m$ is flat and finitely generated over the local ring $A_m$, which implies $B_m$ is free over $A_m$. Now the rank is equal to the dimension of the $K$-vector space $B_m \otimes_{A_m} K$, which has to be one since $B_m$ is contained in $K$. – Parthiv Basu Aug 05 '19 at 00:02
  • yes I see. concretely by base change argument on free modules to generic stalk, I guess –  Aug 05 '19 at 09:30

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