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Let $A$ be a Dedekind domain, $K:=\text{Frac}(A)$ and $L/K$ finite so that the integral closure $B$ of $A$ in $L$ is Dedekind. If $A$ is a PID, for example, then there exists an integral basis : $B$ is free over $A$, and this basis gives a basis of $L/K$. In particular, if $A=\mathbb{Z}$ or $\mathbb{Z}_p$ (or even the ring of integers of a finite extension of $\mathbb{Q}_p$), and $L/K$ finite then an integral basis exists.

I'm looking for a non-example ; $A$ Dedekind, $L/K$ finite for which there does not exist an integral basis. I suspect that taking $A$ to be the ring of algebraic integers of a finite extension of $\mathbb{Q}$ which is a UFD but not PID might do the trick.

Highly pathological examples (e.g. outside the realm of number fields) are very much welcome !

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A Dedekind domain is a UFD if and only if it is a PID. What you want is an example where the ring of integers has class number greater than one. Keith Conrad has a writeup called something like "A non-free relative extension" with an example of the kind you're seeking (or at least once did). –  Keenan Kidwell Apr 14 '10 at 14:23
    
I should clarify: as a candidate for an example of the kind you're after, you want a number field whose ring of integers has class number greater than one. Not being a PID is obviously a necessary condition, though not necessarily sufficient. –  Keenan Kidwell Apr 14 '10 at 14:28
    
It is true that the integral closure of a Dedekind domain in any finite extension of its fraction field is Dedekind, but if the field extension is inseparable then extra work is needed since the integral closure need not be a f.g. module over the original ring. Most books only treat the case of separable field extensions. One book which handles the general case is Jacobsons's Basic Algebra II. –  KConrad Apr 14 '10 at 16:47
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up vote 8 down vote accepted

Watch out: just because $A$ is a PID does not make $B$ a free $A$-module. You need to know that $B$ is finitely generated over $A$ to conclude $B$ has an $A$-basis when $A$ is a PID. If $L/K$ were separable then using discriminants you can stuff $B$ inside a finitely generated $A$-module so $B$ is finite free if $A$ is a PID.

If we drop the separability condition on $L/K$ in this corollary, then the whole discriminant argument breaks down and in fact $B$ need not be a finitely generated $A$-module, even if $A$ is a PID! For an example, see Exercise 11 on p. 205 of Borevich and Shafarevich's "Number Theory".

I wrote up a course handout with examples in number fields where the top ring of integers is not a free module over the bottom ring of integers: look at http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/notfree.pdf. Note that the bottom ring having class number greater than 1 is a necessary condition for such a phenomenon to occur but it's not sufficient. An example illustrating that is at the end of the handout.

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