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The problem 1.2.10 from Qing Liu's book "Algebraic Geometry and Arithmetic Curves" is the following:

Let $A$ be an integral domain, and $B$ its integral closure in the field of fractions $\operatorname{Frac}(A)$. Suppose that $B$ is a finitely generated $A$-module. Show that $B$ is flat over $A$ if and only if $B=A$. One can show that this result is true without the assumption of finiteness of $B$ over $A$.

Is that result easier to prove if one assumes that $B$ is finitely generated $A$-module than without it? I just saw a sketch that the general case might be proved by taking taking polynomial of degree $n$ and show somehow that one can find a polynomial with smaller degree to get a contradiction but I didn't understand it and how to complete it to a valid proof.

Jaakko Seppälä
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  • Solved here. (If need more details please let me know.) – user26857 Jan 26 '15 at 11:17
  • @user26857: your answer in the link indeed shows elegantly how to prove the general result. The OP however knows that the result holds without any finiteness asumption. In order to make the case that such an assumption does not really simplify the proof, you might want to add a few more hints/references in your answer, in particular since it refers to an exercise in Matsumura. Even if the OP doesn't need such complements, other users might welcome them. That said, you are the only master on board of your answer ... – Georges Elencwajg Jan 26 '15 at 12:10
  • Dear @GeorgesElencwajg, the OP posted a problem from Liu's book which can be (or not) a homework. Beside this, I think the two steps I've hinted in my old answer are not hard to prove, both being easy exercises. – user26857 Jan 26 '15 at 21:09
  • Dear @user26857, yes I understand your point of view, even if I have to confess that I didn't find the second step easy at all. Anyway, and more importantly, I'll take this opportunity to tell you how much I enjoy your posts which always very insightfully go straight to the heart of the matter. I'm looking forward to your next ones! – Georges Elencwajg Jan 26 '15 at 23:21
  • Dear @GeorgesElencwajg, if $x\in B$, then $x$ is in the field of fractions of $A$, so there is $a\in A$ such that $ax\in A$. But faithfully flatness gives $aB\cap A=aA$, so $ax\in aA$. (I don't think this can be called otherwise than easy.) Thanks for your nice words. – user26857 Jan 26 '15 at 23:27
  • @user26857 I can say that questions I post are not homeworks. I'm an unemployed mathematician an I like to improve my algebra skills by studying commutative algebra. But for me the problems are difficult so it is nice that there are kind people who helps me. – Jaakko Seppälä Jan 26 '15 at 23:35
  • Dear @user2219896: +1 (on the linked post) for your explanation/calculation and your kind offer to the OP. I personally knew the solution to step 2 (it's even at the back of the book!) but I didn't post it because , even if I argued with you in my first comment, I considered the decision was yours. – Georges Elencwajg Jan 27 '15 at 07:19
  • Dear user 2219896: +1 for your question too, and all my wishes for your future personal situation. – Georges Elencwajg Jan 27 '15 at 07:23

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