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Let $K$ be a number field, and $R$ an order of $K$. Consider the category $\mathcal{M}$ of all finitely generated $R$-submodules of $K$. If $X$ is an object of $\mathcal{M}$ such that $R=\textrm{End}_R(X)$, consider the contravariant functor $h_X:\mathcal{M}\to\mathcal{M}$ given by $T\to\textrm{Hom}_R(T,X)$, where the structure of $R$-module on $h_X(T)$ is induced from that on $X$.

I am interested in having conditions on the ring $R$ ensuring the existence of an $X$ as above for which $h_X$ is an anti-equivalence of categories.

So far I only know that in the following two cases such an $X$ exists:

  • if $R$ is the maximal order of $K$, then $X=R$ works;
  • if $K$ has a subfield $K_0$ with $[K:K_0]=2$ and such that $R\cap K_0$ is the maximal order of $K_0$ then $X=R$ works.

(The question is closely related to (and inspired by) the following: let $k$ be a finite field and $\mathcal{C}$ a $k$-isogeny class of $k$-simple ordinary abelian varieties over $k$. Does there exist an object $A$ of $\mathcal{C}$ such that any finite $k$-subgroup $H\subset A$ is equal to the intersection of the kernels of a suitable (finite) collection of isogenies $\varphi_i:A\to A$?)

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Dear Tommaso, is it obvious that $h_X(T)$ is still contained in $K$? –  Filippo Alberto Edoardo Sep 3 '13 at 17:57
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Dear Filippo: Just localize at the fraction field (which commutes with the formation of the Hom-modules in question) to turn $X$, $R$, and $T$ into $K$. –  user36938 Sep 4 '13 at 4:30
    
@user36938: nice point, I did'n think that way. –  Filippo Alberto Edoardo Sep 4 '13 at 8:56
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It seems more natural to define the category to consist of those finitely generated $R$-modules whose generic fiber is 1-dimensional (without choosing a preferred basis). I use this viewpoint below.

If $R$ is Gorenstein then you can take $X=R$ since the dualizing complex for a Gorenstein local ring is concentrated in a single degree and as such as an invertible module (so the Hom above is the same as an RHom). This conceptually explains your second case with $R \cap K_0 = O_{K_0}$ for a subfield $K_0$ over which $K$ is quadratic since in such cases $R$ is Gorenstein (as after completion over $O_{K_0}$ we are confronted with a "quadratic order" over a discrete valuation ring, which is always monogenic and hence easily checked to be Gorenstein).

In general $R$ is Cohen-Macaulay, so it is tempting to try to take $X$ to be its dualizing module (as the dualizing complex for a CM local ring is supported in a single degree), but if not projective (equivalently, $R$ not Gorenstein) then presumably the discrepancy between Hom and RHom may create some problems. No doubt the experts in commutative algebra can supply a counterexample or explain why it is a non-issue in these circumstances.

In fact, one is led to wonder (in the absence of any motivation being given for the question) whether the setup is simply "wrong": it is always true by taking $X$ to be the dualizing module (put in degree 0) that on the derived category $D^b_c(R)$ of "bounded complexes of $R$-modules with finitely generated homologies" that $T \mapsto {\rm{RHom}}(T,X)$ is an involutory auto-equivalence and ${\rm{RHom}}(X,X) = R$. That is, the dualizing module always works if you work in the appropriate derived category setting (which eliminates the 1-dimensionality restriction on the generic fiber, etc.). In the Gorenstein case we can apply ${\rm{H}}^0$'s throughout to recover the more concrete assertion with ordinary Hom's in that case. Is that not adequate for whatever motivated the question?

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Nice answer, but I don't follow completely the argument for the quadratic case being Gorenstein. I guess you want to use that if $R$ becomes Gorenstein after completion, it was already so: but you are completing at any prime of $\mathcal{O}_{K_0}$, right? Then you use that $\mathcal{O}_{K_0}$ is a DVR and hence any monogenic extension of it is still such; but in the local case not only quadratic ones are monogenic, and also I do not understand how to go back to $R$. –  Filippo Alberto Edoardo Sep 4 '13 at 8:55
    
@Filippo: Completion commutes in an evident manner with module-finite algebras, and beyond the quadratic case there will be tons of non-monogenic orders (so one cannot make a uniform claim of the Gorenstein property based just on degree beyond the quadratic case). –  Marguax Sep 4 '13 at 13:05
    
Marguaux: Ok, then I agree. I thought you were claiming something special about the quadratic case. –  Filippo Alberto Edoardo Sep 4 '13 at 15:14
    
Thanks for the nice answer and for giving my question some more appropriate context (I got the question from looking at subgroups of ordinary abelian varieties over a finite field. In the situation I had in mind R is Z[\pi], where \pi is an ordinary Weil-number). –  Tommaso Centeleghe Sep 4 '13 at 16:21
    
@ Marguaux: By the way, as you say $X$ would be (concentrated in degree $0$ and) projective iff $R$ is Gorenstein. But Tommaso insisted that $X\in\mathcal{M}$ so I think he wants it to be projective. Doesn't it somehow forces $R$ to be Gorenstein? Of course, it depends if satisfying duality for all modules in $\mathcal{M}$ is equivalent to being the dualizing module, which I ignore. –  Filippo Alberto Edoardo Sep 4 '13 at 16:31
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