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34 votes
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Injective ring homomorphism from $\mathbb{Z}_p[[x,y]]$ to $\mathbb{Z}_p[[x]]$

Surprisingly, at least to me, yes. Let $f$ and $g$ be two power series in $\mathbb Z_p[[x]]$ that, modulo $p$, are algebraically independent. Send $x$ to $pf$ and $y$ to $pg$. To check the map $\...
Will Sawin's user avatar
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15 votes
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Translation between formal geometry and rigid geometry

No, these are not the same thing. Formal schemes are to rigid-analytic spaces as $\mathbf{Z}_p$-schemes are to $\mathbf{Q}_p$-schemes. The book Lectures in Formal and Rigid Geometry by Bosch is an ...
dorebell's user avatar
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14 votes

Intuition about $\mathrm{Spec}\mathbb{C}[[t]]$ versus $\mathrm{Spf}\mathbb{C}[[t]]$ versus $\mathrm{Specan}\mathbb{C}[[t]]$ (and similar objects)

In a nutshell: $\mathrm{Spec}~\mathbb{C}[[t]]$ Is a “trait” i.e. the spectrum of a discrete valuation domain, with a generic (open) point and a closed point. It might be imagined as a refinement of ...
Leo Alonso's user avatar
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12 votes
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Connectedness, loops and formal moduli problems

The presentation of the formal moduli problems story in Gaitsgory-Rosenblyum A Study in Derived Algebraic Geometry, Vol 2 may be what you are looking for. We review it here (in the case over $\mathrm{...
A Rock and a Hard Place's user avatar
9 votes
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Symmetric powers of curves and completion along the diagonal

$\operatorname{Sym}^{d-1}(C) $ is ample, and $\Delta $ is not unless $C\cong \mathbb{P}^1$. To see this, consider the finite map $\pi :C^d\rightarrow \operatorname{Sym}^{d}(C) $, and the projections $\...
abx's user avatar
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7 votes
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Basic questions about formal schemes

For 1) The functorial interpretation is developed by Strickland in Formal schemes and formal groups. Homotopy invariant algebraic structures (Baltimore, MD, 1998), 263–352, Contemp. Math., 239, Amer....
Leo Alonso's user avatar
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7 votes
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Basic example of a formal affine scheme, functorial point of view

It might be illuminating to first work the example of (ordinary) affine space $\mathbb{A}^1_\mathbb{Z}$ over the integers. As a functor, $\mathbb{A}^1_\mathbb{Z}$ is the forgetful functor $\mathit{...
Emily's user avatar
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6 votes
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Vector bundles on complete rings

I will assume that $A$ is Noetherian (if you wish to work in the non-Noetherian setting, I'll see what I can do to modify my answer). Without loss of generality, we may assume that $A$ is in fact $I$-...
David Benjamin Lim's user avatar
6 votes
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Motivation for Henselian rings in algebraic geometry

First about the ralation between henselian and formal smoothness property, I think a good idea is to look at what was the first version of Hensel lemma: it says that if $f(\bar{a})=0,f'(\bar{a})\, ...
ali's user avatar
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6 votes
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GAGA for henselian schemes

OK, it turns out that $H^1((P^1_A)^h, O^h)$ is nonzero in general. A counter example can be found in a blog post on the very blog you mention in your post. Here is a link.
darx's user avatar
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6 votes

Intuition about $\mathrm{Spec}\mathbb{C}[[t]]$ versus $\mathrm{Spf}\mathbb{C}[[t]]$ versus $\mathrm{Specan}\mathbb{C}[[t]]$ (and similar objects)

One good way of seeing the difference between $\mathrm{Spec}\,\mathbb{C}[[t]]$ and $\mathrm{Spf}\,\mathbb{C}[[t]]$ is to look at what functors they represent on affine schemes. In fact we have $$\...
Denis Nardin's user avatar
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6 votes
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about morphisms of affine formal schemes $\mathrm{Spf}(B)\to \mathrm{Spf}(A)$

See EGA I, sec. 10. For the specific question you mention, see paragraph 10.2. There is further issues on the cohomology of formal schemes in EGA III, sect 3.4. Also, the existence theorem is treated ...
Leo Alonso's user avatar
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5 votes
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References on topological rings

Here are 3 references that haven't been mentioned yet. I am not sure if the latter two would be of any use to you, but probably are worth a look. Arnautov, Glavatsky, Mikhalev - Introduction to the ...
M.G.'s user avatar
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5 votes

Intuition about $\mathrm{Spec}\mathbb{C}[[t]]$ versus $\mathrm{Spf}\mathbb{C}[[t]]$ versus $\mathrm{Specan}\mathbb{C}[[t]]$ (and similar objects)

To complement the other answers, I would like to add a word on the analytic spectrum $\mathrm{Specan}(\mathbb{C}[[t]])$. First, let me say that I am not sure what $\mathrm{Specan}$ means and have no ...
Jérôme Poineau's user avatar
5 votes

Functorial point of view for formal schemes

Although this doesn't cover every example, formal completions $\hat{X}_Y$ of a scheme $X$ along a closed subscheme $Y$ have a particularly nice functorial presentation. We have $$\hat{X}_Y \simeq Y_{...
Exit path's user avatar
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5 votes
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Does $R\hat{f}_*\mathcal{F}=\hat{f}_*\mathcal{F}$ hold for affine adic morphisms?

I never liked the definition of formal completion (as defined in EGA or Hartshorne), so I'll use the following definition from Brian's formal GAGA paper. Definition: Let $X$ be a locally Noetherian ...
David Benjamin Lim's user avatar
5 votes
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Is the formal completion of an affine group necessarily a formal group?

The universal map is not a map of formal groups without some extra condition. An easy class of counterexamples comes from completions of an affine group along a closed subscheme that does not contain ...
S. Carnahan's user avatar
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4 votes
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What's the relation between pseudo-compact and admissible rings?

Neither of these properties implies the other: There is an admissible algebra that is not pseudo-compact If $A$ is a ring with the discrete topology that is not artinian then it is admissible but not ...
Simon Wadsley's user avatar
3 votes

Formal Schemes Methods: Applications

Instead of trying to answer the question in full, let me give some further appearances of the notion of formal scheme. There are several situations when a consideration of a topology in certain rings ...
Leo Alonso's user avatar
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3 votes
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Smooth loci and formal neighborhoods

No. Take for $R$ a discrete valuation ring with fraction field $K$, and for $f$ any non-smooth morphism of smooth $K$-algebras (viewed as a morphism of $R$-algebras).
Laurent Moret-Bailly's user avatar
2 votes

What is the essential image of $AbVar$ in $p-div$?

In short, I still don't know over a general ring, but for $\mathbb{F}_q$, we have an answer: The idempotent completion of the essential image of $AbVar_{/\mathbb{F}_q}$ in $FG_{/\mathbb{F}_q}$ is ...
Catherine Ray's user avatar
2 votes

Does there exists a "local slice" for an action $ \widehat{\mathbb{G}}_a $ on $ \operatorname{Spf}(\widehat{A}) $ (char zero)?

The answer is no, such a ``local slice'' does not exist in general even for algebraic actions. Let $ \delta \in T_{k[x_{1},x_{2}]/k} $ be the derivation $ x_{1} \frac{\partial}{\partial x_{2}}+x_{2}\...
Schemer1's user avatar
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2 votes
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Formal neighbourhood of a closed subscheme

If you look at a family of elliptic curves like $y^2 = x^3 - x-t^n$ as a surface $X$ mapping to a curve $C$ with parameter $t$, and $Y$ the fiber over $0$, then the neighborhood modulo $\mathcal I_Y^n ...
Will Sawin's user avatar
  • 148k
1 vote

Meaning of dagger cohomology $H^{1 \dagger}(G^\dagger)$ in "Frobenius and Monodromy Operators" by Coleman and Iovita

I think $X^{\dagger}$ must mean the dagger space associated to the weak formal scheme you get by taking the weak completion of the model $\mathcal{X}$ along the special fibre $\mathcal{X}_{k}$. Then $...
Oli Gregory's user avatar
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1 vote
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On the stability of having a normal formal model under finite extensions of the base field

As for your first question, $X_L$ indeed admits a normal formal model by virtue of normalisation. Whether $A^{\circ}\otimes_R R_L$ is already normal (which is equivalent to $A^{\circ}\otimes_R R_L=(...
ArtW's user avatar
  • 125
1 vote
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When is a formal group smooth?

I think an adaptation of Schlessinger's argument from Functors of Artin rings should work. Let $\Lambda$ be a Noetherian ring, and suppose we have a connected formal group $\mathcal{G}$ formally ...
xir's user avatar
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1 vote
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A translation between formal and rigid geometry

Using your notation: it corresponds, like in the "classic" case of polynomials over a field, to the ideal generated by $\zeta_1-x_1,\dots,\zeta_n-x_n$, where $x=(x_1,\dots,x_n)\in \mathbb B^n(K)$. In ...
Xarles's user avatar
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1 vote

Coherent modules over complete adic rings: counterexamples

Suggestion: read the paper "On Hausdorff completions of commutative rings in rigid geometry", by Fujiwara, Gabber, and Kato.
Dragon's user avatar
  • 11
1 vote

generic fibre functor for relative rigid spaces

If you look at 1.1.12 in Huber's book "Étale cohomology of rigid analytic varieties and adic spaces" he notes that if you embed admissible Noetherian formal schemes into adic spaces, then ...
Ashwin Iyengar's user avatar
1 vote

generic fibre functor for relative rigid spaces

Yes. You will find what you are looking for in Ahmed Abbes' 'Elements de Geometrie Rigide $I$', where rigid geometry a la Raynaud is developed in a relative setting. It is written in French, but it ...
js21's user avatar
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