Consider an order $R$ in a number field $L$. Let $C_R$ be the set of $R$-fractional ideals modulo $L^\times$. Let $O$ be the maximal order in $L$, and $C_O$ be the class group of $O$.

My question: Is there a formula that relates $\# C_R$ with $\# C_O$, that involves perhaps the conductor of $R$?

A note: Beware that in the definition of $C_R$ I really consider fractional ideals: I do not ask that they are non-singular, so $C_R$ is not the Picard group of the order $R$ (In fact, it is not even a group!).

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    $\begingroup$ I do not think there is a (simple) relation known or that it even should exist (I might be wrong though). There is however various work on these (semi)groups in this and more general contexts. Would you be interest in that general information, too, or do you know about this anyway? $\endgroup$ – user9072 Jul 24 '13 at 12:45
  • $\begingroup$ I wanted to find a nice bound for #C_R , preferably in terms of the conductor and data associated to the quotient field of L. In my arguments I have essentially no control on what R can be. In particular I am not very interested in the quadratic case $\endgroup$ – mnr Jul 30 '13 at 6:54
  • $\begingroup$ I did remember to have found an argument of KConrad on this site (somewhere?) , which basically says that one can still do the argument using Minkowski's theorem to bound #C_R $\endgroup$ – mnr Jul 30 '13 at 6:56
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    $\begingroup$ As said I do not think this is really known (depending on the quality and what one means by nice); the proof of the finiteness in Zanardo-Zannier I mention below is rather explcit though I think so you might extract something from the proof. Also, it is known that every high enough power of an ideal will be regular and degree of the extension minus one is 'high enough' I think. IIRC this is in the Dade et al. paper I mention, but this is possibly difficult to read as terminology progressed. But there are more recent more streamlined and generalized presentations of this. $\endgroup$ – user9072 Jul 30 '13 at 7:04
  • $\begingroup$ Also Dade et al. proof finiteness (I think), but I do not have the paper handy. Yet I think it is also somewhat constructive. As you mention KConrad, perhaps you find what you remember behind the link David Speyer gave (but I cannot know). $\endgroup$ – user9072 Jul 30 '13 at 7:08

The ideal class semigroups mentioned in the question got studied in this setting (orders of number fields) and in other and more general ones by various authors in recent years.

A starting references is:

Zanardo, P.; Zannier, U. The class semigroup of orders in number fields. Math. Proc. Cambridge Philos. Soc. 115 (1994), 379–391.

They show that this semigroup is a Clifford semigroup for orders of quadratic fields, but for degree greater 2, there always is some order such that the respective class semigroup is not a Clifford semigroup.

There is also an older paper on this theme (that it appears was overlooked in recent literature for some time but reapperas in still more recent one)

Dade, E. C.; Taussky, O.; Zassenhaus, H. On the theory of orders, in paricular on the semigroup of ideal classes and genera of an order in an algebraic number field. Math. Ann. 148 (1962) 31–64.

Starting from these two papers and looking for papers that quote them in MathSciNet for example one will find several more recent contributions. Investigating these semiclass groups; but it seems (but I donot have a good overview) the emphasis is more to generalize to more general structures (say, other domains than just orders) than more details in the number-theoretic setting.

Yet, in particular the early papers (I do not know for the recent ones) contain also some explicit information on these semigroups, but as commented earlier I would not know of in some sense simple descriptions (but again my knowledge is superficial here).


$\DeclareMathOperator{\Pic}{Pic} \DeclareMathOperator{\Spec}{Spec}$ EDIT: This answer is wrong; please un-accept it. I erroneously believed that a fact about quadratic fields held in general. See David Speyer's counterexample in the comment section.

Your "class monoid" $C_R$ may [EDIT: in the quadratic case] be described as the disjoint union of the Picard groups of all intermediate orders between $R$ and $O$, plus the additional data of how to multiply classes in different Picard groups. Specifically, for a fractional ideal $I$ of $R$, let $R_I$ be the maximum order for which $I$ (as a set) remains a fractional ideal. In other words, $R_I$ is the multiplier ring $$ R_I = \{x \in L : xI \subseteq I\}. $$ With a little work, you can show that $I$ is non-singular (= invertible = locally principal) in $R_I$ [EDIT: $I$ may not be invertible in $R_I$ when $[L:\mathbb{Q}]>2$], and in no proper suborder. Therefore, $$\#C_R = \sum_{R \subseteq R' \subseteq O} \#{\Pic{R'}}.$$

As you may already know, there are standard theorems relating $\Pic{R'}$ to $C_O$ and some other data. The idea is to look at the long exact sequence in sheaf cohomology associated to the short exact sequence of sheaves on $\Spec{R'}$ $$1 \to \mathcal{O}_{R'}^\times \to i_*\mathcal{O}_O^\times \to i_*\mathcal{O}_O^\times/\mathcal{O}_{R'}^\times \to 1,$$ or see Neukirch's Algebraic Number Theory, Ch. 1 Sec. 12 for non-sheafy details. For the purposes of counting, the upshot is that $$ \#\Pic{R'} = \frac{h_L}{(O^\times : R'^\times)} \frac{\#(O/\mathfrak{f}_{R'})^\times}{\#(R'/\mathfrak{f}_{R'})^\times}, $$ where $\mathfrak{f}_{R'}$ is the conductor of $R'$, and $h_L = \#C_O$.

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    $\begingroup$ Keith Conrad comments here mathoverflow.net/questions/17555/… that, in fields of degree higher than $2$, there are $\mathbb{Z}$ lattices which are not invertible ideals with respect to any order. He doesn't say that they are ideals in some order, but it sounds like he's implying it. Could you give a reference or proof for the fact that every ideal is invertible in it's multiplier ring? Thanks! $\endgroup$ – David E Speyer Jul 25 '13 at 13:00
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    $\begingroup$ Counterexample: Set $\pi = \sqrt[3]{2}$. Let $R = \mathbb{Z}[\pi^4, \pi^5]$ and $I = \langle 2, \pi^4 \rangle$. I get that $R$ is the multliplier ring of $I$; that $R:I = \langle 1, \pi, \pi^2 \rangle$ and $I (R:I) = \langle 2, \pi^4, \pi^5 \rangle$. I find it easier to think in terms of the images of these ideals under the $\pi$-adic valuation: $R \leadsto (0,3,4,5,6,\ldots)$, $I \leadsto (3,4,6,7,8,\ldots)$, $R:I \leadsto (0,1,2,3,4,\ldots)$, $I (R:I) \leadsto (3,4,5,6,\ldots)$. $\endgroup$ – David E Speyer Jul 25 '13 at 13:20
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    $\begingroup$ I believe (but I could be wrong) the argument in the answer (if it were true) would imply that the semigroup is a Clifford semigroup. This is however known to be not true for degree greater 2 (Zanardo-Zannier). $\endgroup$ – user9072 Jul 25 '13 at 14:00
  • $\begingroup$ Updated the answer. Thanks for pointing that out, David. $\endgroup$ – Gene S. Kopp Jul 25 '13 at 18:15
  • $\begingroup$ @DavidSpeyer: Any (full) $\mathbf Z$-lattice in a number field is a fractional ideal for its multiplier ring, so you can always interpret the lattice as a fractional ideal for an order. $\endgroup$ – KConrad Jul 26 '13 at 10:45

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