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In Lang's "Algebra" (chap 16, p 614) he states the following without proof:

"Let $R$ be some ring in a finite extension $K$ of $\mathbb{Q}$, and such that $R$ is a finite module over $\mathbb{Z}$ but not integrally closed. Let $R'$ be its integral closure. Let $\mathfrak{p}$ be a maximal ideal of $R$ and suppose that $\mathfrak{p}R'$ is contained in two distinct maximal ideals $\mathfrak{P}_1$ and $\mathfrak{P}_2.$ Then it can be shown that $R'$ is not flat over $R,$ otherwise $R'$ would be free over the local ring $R_{\mathfrak{p}},$ and the rank would have to be 1, thus precluding the possibility of the two primes $\mathfrak{P}_1$ and $\mathfrak{P}_2.$"

I would appreciate either a proof of Lang's assertion, or to be guided to a relevant source.

It is shown later in the chapter that a flat, finitely generated module over a local ring is free, but I am not able to use this to prove the above statement and I suspect that other arguments are needed.

The situation described by Lang occurs naturally in algebraic number theory where we take $R= \mathbb{Z}$ and $R'= \mathcal{O}_K$ to the the ring of integers of the number field $K,$ so I am surprised I haven't been able to find anything on the internet so far.

user26857
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user142700
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    In reasonable geometric terms here's what he's saying: the cardinality of the fibers of a flat map (to a connected space) is constant. For a normalization map the map is an isomorphism almost everywhere, and so if there were some point which was non-normal the normalization map would have non-constant fiber size, so it's not constant. – Alex Youcis Oct 27 '15 at 07:23

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I don't understand Lang's assertion "$R'$ would be free over the local ring $R_{\mathfrak p}$". In fact, I can't understand the whole passage.

In fact, $R'$ is never flat (unless $R=R'$). Suppose $R\subset R'$ is flat. Then it is faithfully flat and since $R$ and $R'$ have the same field of fractions we get $R=R'$. (See Matsumura, CRT, Exercise 7.2.)

user26857
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  • He's trying to give a proof of your second paragraph. He's assuming that $R'/R$ is flat. Then, since it's finite type (since, I assume, we're working with something like number rings) this implies that $R'/R$ is projective. I think he then meant to say that $(R')\mathfrak{p}$ is free over $R\mathfrak{p}$. But, as I said in my above comment, or you mentioned here, it's just true that normalization maps are not isomorphisms unless they are isomorphisms—insert Serre $S_1$ and $R_2$ statements here. – Alex Youcis Oct 27 '15 at 10:39
  • @AlexYoucis I've tried to follow this line: $R'{\mathfrak p}$ is $R{\mathfrak p}$-flat, so we actually can assume that $R$ is local and want to show that $R'$ is also local, but need extra assumptions in order to get this; see here and here. – user26857 Oct 27 '15 at 10:49
  • I'm confused. I don't think he's making the claim that $R'_\mathfrak{p}$ is local, is he? I mean, this isn't true in any generality that I can see applying this problem. – Alex Youcis Oct 27 '15 at 10:53
  • @AlexYoucis This is the way I've thought that one can show that there is only one maximal ideal lying over $\mathfrak p$ if assume $R'/R$ flat. (Of course, this also happens only for $R=R'$.) – user26857 Oct 27 '15 at 10:54
  • I mean, that is certainly one approach, although I do not see how to use it by looking at that one prime. I don't know if you read my comment above, but I thought about just asserting the constancy of fiber size to argue that flatness (+finiteness, I suppose) should imply our map is an isomorphism, since it's birational. – Alex Youcis Oct 27 '15 at 10:57