# necessary separability condition?

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.

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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The standard example of an inseparable extension is $\mathbb{F}_p(\sqrt[p]{T})/\mathbb{F}_p(T)$, so I would check that one first (I'm not sure how to compute the integral closure of $\mathbb{F}_p[T]$ in $\mathbb{F}_p(\sqrt[p]{T})$ however). – Zev Chonoles Jun 29 '11 at 19:35
I checked that after you mentioned it in the other thread. $\sqrt[p]{T}$ itself is integral over $F_p[T]$ by its equation $X^p-T$. So the integral closure contains $F_p[\sqrt[p]{T}]$ which is free of dimension $p$, so the integral closure is certainly too. (I suspect this is the integral closure but thats not necessarz to prove here ;) ) – Peter Patzt Jun 29 '11 at 19:39
Dear Peter, If $A$ is a finite type domain over a field, then the integral closure of $A$ in any finite extension $L$ of its fraction field will be finitely generated as an $A$-module, even if $L$ is not assumed separable. This is a special case of the theory of Japanese rings referred to in Georges's answer below. (I would guess that it is in Matsumura's commutative algebra book.) Regards, – Matt E Jun 29 '11 at 22:26
Matt, what do you mean by "finite type domain"? – Peter Patzt Jun 30 '11 at 10:57

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. – Peter Patzt Jun 30 '11 at 11:04