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I wonder if and where the separability condition on $L$ is used in the following theorem from Lang's Algebraic Number Theory p. 7. I suspect it is necessary to see that a subring of the finite extension $L$ is a finitely generated $A$-module, but have problems imagining a counterexample to this theorem without separability. Even if this exists, a lot of the inseparable extension still suffice the assertion.

From Lang's Algebraic Number Theory

Let me post a more elaborate version of the proof:

$B$ is a torsion-free $A$-module, because the multiplication $xa = 0$ for $x\in B$ and $a\in A$ does not allow zero-divisors as it happens in $L$. (Now we would need to deduce that $B$ is finitely generated over $A$.) From the theory of PIDs $B$ is free. Assuming $B$'s dimension is smaller than $n$, and its basis shall be $\alpha_1,\dots,\alpha_k$. Then we can find a $\beta \in L$ that is linearly independent from this basis and a $c\in A-\{0\}$ such that $c\beta$ is integral over $A$ but still $c\beta\in B$ is linearly independent from the basis of $B$. Contradiction.

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    Matt, what do you mean by "finite type domain"?2011-06-30

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A domain R is called Japanese in EGA IV (Première partie, §23, page 309), if for every finite-dimensional extension $L$ of its fraction field $K=Frac(R)$, the ring of elements in $L$ integral over $R$ is a finitely generated $R$ module. So a discrete valuation ring that is not Japanese will certainly give your required counter-example. Such a non-Japanese discrete valuation ring exists and is described in Nagata's Local Rings and in Yu's article here (pages 7,8).

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    The ring discribed in Yu's article in example 2.22 is not Japanese, but I couldnt tell whether it is principal or not. Is there a connection between principal and Japanese? I understand though, that taking only separable fields here simplifies the theorem.2011-06-30