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Let $R$ be an integral domain and $A$ it's integral closure. Futhermore $A$ is a finitely generated $R$-module.

I want to show that if $A$ is flat over $R$ then already $R = A$ holds.

My attempts: I tried (without success) following: Assume $R \neq A$. Let $a \in A \backslash R$. Since $A$ integral closure there exist a minimal $n$ with

(*) $$a^n +r_{n-1}a^{n-1} +... +r_0 =0$$

Since $R$ integral then $R \subset Frac(R)$ and the multiplication map $m: R \to A, r \mapsto ar$ is injective Since $A$ flat it should be stay injective after tensoring with $-\otimes A$.

Now I can multipy $a$ also with all elements from $A$. My hope was to deduce the contradiction using (*) by multiplying with a appropiate polynomial in $a$. The problem is the "constant" term r_0.

I think that I'm very close with this proof to the solution but I don't see the final step. Can anybody help. BTW: Up to now I don't used the finitely generated property.

Source: Liu's "Algebraic Geometry". Futhermore: Could the statement be true without the finiteness assumtion. Liu's says yes but how to show?

user267839
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  • See https://math.stackexchange.com/q/482908/660 – Pierre-Yves Gaillard Dec 06 '18 at 14:09
  • and https://math.stackexchange.com/questions/385364/integral-closure-tildea-is-flat-over-a-then-a-is-integrally-closed?rq=1 – Badam Baplan Dec 06 '18 at 23:44
  • @BadamBaplan: Hi. Thank you for the reference. You refering to Hagen Knaf's answer, don't you? The only not satisfying point there is that for localizations wrt a prime ideal $\mathfrak p$ he assumes that there exist always a $x_{\mathfrak p}\in B_{\mathfrak p}$ with $B_{\mathfrak p}=x_{\mathfrak p}A_{\mathfrak p}$. Could you explain why this is true? – user267839 Dec 07 '18 at 00:06
  • Actually I much prefer the other answer there, which does not use the assumption that the integral closure is an f.g. $R$-module. – Badam Baplan Dec 07 '18 at 01:11
  • The answer you mention is using the fact that f.g. flat modules are locally cyclic. My feeling is that that is not the right answer to this question, since it has nothing to do with the integral closure. That approach shows that a domain has no proper f.g. flat overrings, but the second approach shows that a domain has no proper integral flat overrings. The second result implies the first. – Badam Baplan Dec 07 '18 at 01:21
  • @BadamBaplan: Do you have maybe spontaneously are good reference or a sketch of the proof for the statement that f.g. flat modules are locally cyclic? – user267839 Dec 07 '18 at 01:38
  • let me clarify first, what i wrote before was misleading.In general, f.g. flat modules are just locally free. But supposing the module is additionally a (fractional) ideal of a domain, as in our case, you can show the local ranks are $1$ because ideals in domains are free if they are principal, since any two nonzerodivisors are linearly dependent. – Badam Baplan Dec 07 '18 at 02:35
  • @BadamBaplan: That's true. The problem is to show that after localization $B_p$ is principal over $A_p$. – user267839 Dec 07 '18 at 03:07
  • ah I guess following argument works: Two elements $a_1/a_2, a_3/a_4$ cannot be linear independent just because $a_2 a_4 a_3 a_1/a_2 = a_1 a_2 a_4 a_3/a_4$. Do you mean this argument? – user267839 Dec 07 '18 at 03:14
  • yep that is exactly the argument. but again, note that this requires the elements not to be zero divisors.

    Again I'll say I think the other proof is much better, both because it answers Liu's question about finiteness and because it immediately generalizes to rings with zero divisors.

     – Badam Baplan Dec 07 '18 at 03:42

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