# Number Rings and (Galois) Descent

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 $$K$$?

• "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. – Alison Miller Jan 31 '19 at 23:59
• Sorry, you're right. I forgot those have the right rank. – Dean Young Feb 1 '19 at 0:04
• 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. – F Zaldivar Feb 1 '19 at 1:09
• 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. – Dean Young Feb 1 '19 at 1:44
• 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. – F Zaldivar Feb 1 '19 at 1:48

## 1 Answer

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.

• 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! – Dean Young Feb 1 '19 at 1:03
• @DeanYoung No problem, let me know if I can add anything else to help. – Alec Rhea Feb 1 '19 at 1:05
• 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. – Dean Young Feb 1 '19 at 1:30
• I also notice that the next chapter of this book, chapter 6, provides generalizations which may be able to accomidate to this. – Dean Young Feb 1 '19 at 1:32
• @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. – Alec Rhea Feb 1 '19 at 1:35