Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 3 down vote accepted

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).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.