In algebraic number theory, one chooses for each finite étale $\mathbb{Q}$-algebra $K$ a finite $\mathbb{Z}$-algebra $\mathcal{O}_K$. Usually one simply speaks of the finite $\mathbb{Q}$-algebras which are fields, but the étale $\mathbb{Q}$-algebras are just products of these.

Now $- \otimes \mathbb{Q}$ takes the ring of integers $\mathcal{O}_K$ to $K$. Moreover, any integral basis for $\mathcal{O}_K$ is a $\mathbb{Q}$-basis for $K$. In terms of descent theory, we might think of $\mathcal{O}_K$ as a $\mathbb{Z}$-form of $K$. Indeed, this seems similar to the situation for Galois descent, or descent in general. However, as far as I know, descent theory applies to Galois extensions $L/K$ of fields. In that case, a descent datum for an $L$-algebra $A$ is the structure of an $G$-module, where $G$ is the Galois group of the extension, but here, any map of $K$ fixing $\mathcal{O}_K$ seems to also fix $K$.

I guess I'll still ask: is it possible to do descent theory in a way that applies to this context? More precisely, what kind of descent data would we need to recover $\mathcal{O}_K$ from $K$?

  • $\begingroup$ "the number rings O_K are precisely the finite Z-algebras such that O_K⊗Z Q_sep splits into the product of rank_Z(O_K) many Q_sep's. " I don't think this is right: the property you've stated also holds for any order in a number field (or more generally any order in an etale Q-algebra) since it only depends on O_K ⊗Z Q. $\endgroup$ – Alison Miller Jan 31 '19 at 23:59
  • $\begingroup$ Sorry, you're right. I forgot those have the right rank. $\endgroup$ – Dean Young Feb 1 '19 at 0:04
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    $\begingroup$ I think all these is in Waterhouse "Affine Group Schemes" (Springer) and it seems to me that the descent data that generalizes Galois descent in you case is faithfully flat descent. $\endgroup$ – F Zaldivar Feb 1 '19 at 1:09
  • $\begingroup$ I thought about this for a while, but I think $\mathbb{Q} / \mathbb{Z}$ is not faithfully flat, which can be seen from the fact that $\mathbb{Z} / n \mathbb{Z}$ collapses. Something close may be the case though. I will look into the reference you provided, thanks very much. $\endgroup$ – Dean Young Feb 1 '19 at 1:44
  • $\begingroup$ You are right. I think I can see that that the étale condition gives covering data, but I don't see how to define the cocycle condition to get descent data in a non faithfully flat situation. $\endgroup$ – F Zaldivar Feb 1 '19 at 1:48

The categorical Galois theory of Borceux and Janelidze given in chapter 4 for commutative rings applies to your situation.

In particular, it applies to any 'effective Galois descent morphism' defined as follows:

Definition 1. Let $\mathcal{C}$ be a category. An arrow $f:X\to Y\in \mathsf{Hom}_\mathcal{C}$ is an effective descent morphism iff the pullback along $f$ functor $-_f:\mathcal{C}/Y\to\mathcal{C}/X$ is monadic.

Definition 2. Let $\sigma:R\to S$ be a morphism of rings. Write $\eta$ for the unit of the adjunction $$\mathsf{Sp}_S:(\mathsf{S}\text{-Alg})^{op}\rightleftarrows\mathsf{Prof}/\mathsf{Sp}(S):\mathcal{C}_S,$$ where $\mathsf{Sp}_S$ and $\mathcal{C}_S$ are defined by $$\mathsf{Sp}_S(A)=\big(\mathsf{Sp}(A)\longrightarrow\mathsf{Sp}(S)\big),$$ $$\mathcal{C}_S(X,f)=\mathsf{Hom}\big((X,f),(\coprod_MS/M,p)\big),$$ with $p:\coprod_MS/M\to\mathsf{Sp}(S)$ the projection of the Pierce structural space of the ring $S$. An $R$-algebra $A$ is split by $\sigma$ when the morphism $$\eta_{S\otimes_RA}:\mathcal{C}_S\mathsf{Sp}_S(S\otimes_R A)\longrightarrow S\otimes_RA$$ is an isomorphism.

Definition 3. A morphism of rings $\sigma:R\to S$ is of effective Galois descent iff

  1. $\sigma$ is an effective descent morphism in $\mathsf{Ring}^{op}$, and

  2. For every object $(X,\psi)\in\mathsf{Prof}/\mathsf{Sp}(S)$, the $R$-algebra $\mathcal{C}_S(X,\psi)$ is split by $\sigma$.

They then give the following Galois theorem for commutative rings.

Theorem 1. Let $\sigma:R\to S$ be an effective Galois descent morphism, with $Gal[\sigma]$ the corresponding Galois groupoid in the category of profinite spaces. Then there exists an equivalence of categories $$\big(\mathsf{Split}_R(\sigma)\big)^{op}\simeq\mathsf{Prof}^{Gal[\sigma]}$$ between the dual of the category of $R$-algebras split by $\sigma$ and the category of internal covariant presheaves on $Gal[\sigma]$ in the category of profinite topological spaces.

The reference to étale stuff was for étale morphisms of topological spaces, but your situation should fall under the scope of above theorem.

  • $\begingroup$ Thanks, I'll check that out. Maybe you can say if they give conditions for when that morphism $f : X \rightarrow Y$ defines some sort of categorical equivalence, and if they characterize the one's that split. You see, $\mathcal{O}_K$ splits when we tensor up with $\mathbb{Q}^{sep}$, but so does any order of $K$. Edit: just saw your modification- wow, thanks so much! $\endgroup$ – Dean Young Feb 1 '19 at 1:03
  • $\begingroup$ @DeanYoung No problem, let me know if I can add anything else to help. $\endgroup$ – Alec Rhea Feb 1 '19 at 1:05
  • $\begingroup$ Hmm... $\mathbb{Q} / \mathbb{Z}$ is not faithfully flat, which (maybe?) is the right condition for monadicity of the pushout functor. E.g. $\mathbb{Z} / n \mathbb{Z}$ collapses. However, we can probably modify this by restricting $\mathbb{Z}$-algebras. The key thing to notice is that any two distinct orders with the same number field would also go to the same structure in $\text{Prof}^{\text{Gal} [\sigma]}$, so there must be some further restriction besides characteristic zero to make this work. I hope (!) that we can just restrict to the integrally closed orders here to make this work. $\endgroup$ – Dean Young Feb 1 '19 at 1:30
  • $\begingroup$ I also notice that the next chapter of this book, chapter 6, provides generalizations which may be able to accomidate to this. $\endgroup$ – Dean Young Feb 1 '19 at 1:32
  • $\begingroup$ @DeanYoung Yes, they ultimately culminate with a Non-Galoisian galois theorem for arbitrary extensions of fields in chapter $7$ which is what I was after, so I'm admittedly a bit fuzzy on the details of chapter $4$. I'll see if I can address your comments after some thought tonight, but the book is excellent and I recommend it highly. $\endgroup$ – Alec Rhea Feb 1 '19 at 1:35

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