## The Fundamental Curve of p-adic Hodge Theory, or How to Un-tilt a Tilted Field

As Quomodocumque once said concerning the most recent set of courses at Arizona Winter School, “Jared Weinstein [gives] a great lecture.” On that note, I am delighted to welcome our first guest post, by the man himself. Note that it has been converted from LaTeX into “wordpress” flavour of LaTeX, so any errors were probably introduced by me in the conversion.

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In any treatment of $p$-adic Hodge theory, one inevitably encounters a procedure for passing from fields of characteristic 0 to fields of characteristic $p$. In modern parlance, if $F$ is a perfectoid field of characteristic 0, one has the tilt $F^\flat$, a perfectoid field of characteristic $p$. (For instance, the construction of Fontaine’s period ring $B_{\text{dR}}$ has $\mathbb{C}_p^{\flat}$ as an intermediate step.) What happens when you try to un-tilt? That is, given a perfectoid field $E$, can we describe the set of all perfectoid fields $F$ with $F^{\flat}=E$? This question leads us right into the subject of our post, the remarkable “fundamental curve of $p$-adic Hodge theory”, due to Fargues and Fontaine. (See here, and also an English summary here.)

First we review the tilting procedure. Start with a field $F$ of characteristic 0 which is complete with respect to a nonarchimedean absolute value, whose residue field is perfect of characteristic $p$. (Pedantic note: The absolute value itself doesn’t come packaged with $F$, only the topology does.) Let ${\mathcal{O}}_F$ be its ring of integers. Then form

${\mathcal{O}}_{F^\flat}= \varprojlim_{x\mapsto x^p} {\mathcal{O}}_F/p.$

Then ${\mathcal{O}}_{F^\flat}$ is a ring in characteristic $p$. It’s not hard to see that ${\mathcal{O}}_{F^\flat}$ is a domain. Let $F^{\flat}$ be its fraction field.
Typically we only care about the case when $F$ is a perfectoid field, which means that it satisfies the following two properties:

1. The value group $\left\lvert{F^\times}\right\lvert$ is non-discrete.
2. The Frobenius map $x\mapsto x^p$ is surjective on ${\mathcal{O}}_F/p$.

The field ${\mathbb{Q}}_p$ satisfies the second property but not the first, and ${\mathbb{Q}}_p^\flat = {\mathbb{F}}_p$. If $\ell$ is prime to $p$, then the completion $F$ of ${\mathbb{Q}}_p(p^{1/\ell^\infty})$ satisfies the first property but not the second, and $F^{\flat}={\mathbb{F}}_p$ also. In these examples the passage from $F$ to $F^\flat$ is intolerably lossy. But if $F$ is a perfectoid field, then it turns out that $F^\flat$ is another perfectoid field.

So suppose $F$ is perfectoid. To see where the topology on $F^\flat$ comes from, one has to observe the isomorphism of multiplicative monoids (not rings)

$F^{\flat}=\varprojlim_{x\mapsto x^p} F.$

(This is a good exercise if you haven’t seen this before. Also, for this it is important that $F$ be complete.)
Given an element $e\in F^{\flat}$ which corresponds to $(f,f^{1/p},\dots)$ in the above bijection, we put $e^\sharp = f$. If $\left\lvert{\;}\right\lvert$ is an absolute value which defines the topology on $F$, we can define a corresponding absolute value on $F^\flat$ by setting $\left\lvert{e}\right\lvert=\left\lvert{e^\sharp}\right\lvert$.

For instance, let $F=\hat{{\mathbb{Q}}}_p(\mu_{p^\infty})$ be the completion of the field obtained by adjoining all $p$th power roots of unity to ${\mathbb{Q}}_p$. Then ${\mathcal{O}}_{F^{\flat}}$ contains an element $t=(0,1-\zeta_p,1-\zeta_{p^2},\dots)$ which is topologically nilpotent, as well as a system of roots $t^{1/p^n}$. This means that ${\mathcal{O}}_{F^{\flat}}$ must contain the ring ${\mathbb{F}}_p\llbracket{t^{1/p^\infty}}\rrbracket$, this being the $t$-adic completion of ${\mathbb{F}}[t^{1/p^\infty}]$. In fact $F^{\flat}={\mathbb{F}}_p(\!({t^{1/p^\infty}})\!)$, the $t$-adic completion of ${\mathbb{F}}_p(t^{1/p^\infty})$. Similarly, if $F=\hat{{\mathbb{Q}}}_p(p^{1/p^\infty})$, we may set $t=(0,p^{1/p},p^{1/p^2},\dots)$, and then once again $F^{\flat}={\mathbb{F}}_p(\!({t^{1/p^\infty}})\!)$.

Probably the most striking relationship between $F$ and $F^{\flat}$ is this:

Theorem A: The absolute Galois groups of $F$ and $F^{\flat}$ are naturally isomorphic.

The precise statement of this theorem is that there is an equivalence of categories between finite étale $F$-algebras and finite étale $F^\flat$-algebras. Applied to $F=\hat{{\mathbb{Q}}}_p(\mu_{p^\infty})$, Theorem A lies at the heart of the construction of $(\phi,\Gamma)$-modules attached to $p$-adic Galois representations. Scholze’s work on perfectoid spaces provides a version of Theorem A that works in families; using this he was able to prove Deligne’s weight-monodromy conjecture for a hypersurface over ${\mathbb{Q}}_p$ by wrestling it into characteristic $p$, where the conjecture was known previously.

Anyway, we promised to talk about how to un-tilt. Thus suppose we are given $E$, a perfectoid field in characteristic $p$. Does there exist a valued field $F$ with $F^{\flat}=E$? If so, is $F$ unique? Evidently not: We have just seen that the fields $\hat{{\mathbb{Q}}}_p(\mu_{p^\infty})$ and $\hat{{\mathbb{Q}}}_p(p^{1/p^\infty})$ both have tilt ${\mathbb{F}}_p(\!({t^{1/p^\infty}})\!)$. So there are at least two ways to un-tilt the latter field. Let’s make precise what we mean by un-tilt:

Definition: An un-tilt of $E$ is an isomorphism class of pairs $(F,\iota)$, where $F$ is a perfectoid field of characteristic 0 and $\iota\colon E\hookrightarrow F^{\flat}$ is an embedding of topological fields, such that $F^{\flat}/\iota(E)$ is a finite extension. (Two such pairs $(F,\iota)$ and $(F',\iota')$ are isomorphic if there is an isomorphism $F\cong F'$ making the obvious diagram commute.) The degree of $(F,\iota)$ is the degree of $F^{\flat}/\iota(E)$.
Let $\left\lvert{Y_E}\right\lvert$ be the set of un-tilts of $E$.

The idea behind these definitions is that there ought to be some kind of geometric object $Y_E$ (something like a rigid space) whose set of closed points $\left\lvert{Y_E}\right\lvert$ parametrize un-tilts. I should explain why I’m including un-tilts of degree $>1$, rather than using a stricter definition requiring that $\iota$ be an isomorphism. The reason is that if $E'/E$ is a Galois extension with group $G$, then we want $G$-orbits of $\left\lvert{Y_{E'}}\right\lvert$ to be in bijection with $\left\lvert{Y_E}\right\lvert$, and the definition is exactly what is necessary to make this happen. Note that a $G$-orbit of size $g$ constisting of points of $\left\lvert{Y_{E'}}\right\lvert$ of degree $d$ corresponds to a single point of $\left\lvert{Y_E}\right\lvert$ of degree $dg$.

Let $\phi\colon E\to E$ be the $p$th power Frobenius automorphism. Then there is an action of $\phi^{\mathbb{Z}}$ on $\left\lvert{Y_E}\right\lvert$, given by sending $(F,\iota)$ to $(F,\iota\circ\phi^n)$. We wanted to parametrize the un-tilts of $F$, but it seems like two un-tilts which differ by $\phi^{\mathbb{Z}}$ aren’t all that different. Let us call two un-tilts of $E$ equivalent if they differ by some power of Frobenius, so that the set of equivalence classes of un-tilts of $E$ is $\left\lvert{Y_E}\right\lvert/\phi^{\mathbb{Z}}$.

This is one of the main theorems of Fargues-Fontaine:

Theorem B: There exists a complete$\dagger$ curve* $X_E$ whose closed points are naturally in bijection with equivalence classes of un-tilts of $E$. If $x\in \left\lvert{X_E}\right\lvert$ corresponds to the class of the un-tilt $(F,\iota)$, then $F$ is the residue field of $x$.

I now have to explain the asterisk and the dagger, and in doing so I will try to get across just how strange the object $X_E$ is. First, the asterisk. A “curve” is a separated integral noetherian scheme which is regular of dimension 1. In other words, a curve is built by gluing together spectra of Dedekind rings. Thus $\mathrm{Spec}\ {\mathbb{Z}}$ is a curve, and so are the affine line $\mathbb{A}^1_K$ and the projective line $\mathbb{P}^1_K$ over any field $K$. In the latter two examples, the residue fields of closed points are finite extensions of the base field $K$. But un-tilts of $E$ don’t seem to lie over any common base field–recall that the fields $\hat{{\mathbb{Q}}}_p(\mu_{p^\infty})$ and $\hat{{\mathbb{Q}}}_p(p^{1/p^\infty})$ are both un-tilts of ${\mathbb{F}}_p(\!({t^{1/p^\infty}})\!)$, and these fields are certainly not finite over any common subfield.

So perhaps $X_E$ is more like ${\mathrm{Spec} \ } {\mathbb{Z}}$, in the sense that it doesn’t admit a finite type morphism to any ${\mathrm{Spec} \ } K$, for $K$ a field. Fine, except that $X_E$ is also complete. What does complete mean here, if not that it admits a proper morphism to some ${\mathrm{Spec} \ } K$? Fargues and Fontaine define it this way: A complete curve is a curve $X$ admitting a map $\deg\colon \left\lvert{X}\right\lvert \to {\mathbb{Z}}$, such that the degree of any principal divisor is 0. Theorem B then says that $X_E$ is complete with respect to the degree map which has already been defined on un-tilts.

Is there an analogue of Theorem B for $Y_E$? Not quite. The situation is analogous to the situation of the Tate curve in rigid-analytic geometry. Let $Y$ be the multiplicative group $\mathbb{G}_m$, considered as a rigid-analytic space over ${\mathbb{Q}}_p$, and let $q\in {\mathbb{Z}}_p$ have positive valuation. Then $q$ acts discontinously on $Y$ without fixed points, one can form the quotient $X=Y/q^{\mathbb{Z}}$, which ends up being the analytification of an elliptic curve with $j$-invariant $j(q)$. So $X$ is a complete curve and $Y$ is not.

Let’s sketch the construction of $X_E$. Since $X_E$ is supposed to parametrize un-tilts of $E$, which are in characteristic 0, perhaps it is not surprising that Witt vectors get involved. Say $x\in Y_E$ corresponds to $(F,\iota)$. Then the sharp map $\sharp\colon {\mathcal{O}}_E\to{\mathcal{O}}_F$ induces an honest ring homomorphism $W({\mathcal{O}}_E)\to {\mathcal{O}}_F$, characterized by $[e]\mapsto e^\sharp$. This extends to a surjective homomorphism $\theta_x\colon W({\mathcal{O}}_E)[1/p]\to F$, which we also write as $f\mapsto f(x)$. The kernel of $\theta_x$ is a maximal ideal, so we get a map $\left\lvert{Y_E}\right\lvert \to {\mathrm{MaxSpec} \ } W({\mathcal{O}}_E)[1/p]$. Unfortunately I highly doubt this map is a bijection–$W({\mathcal{O}}_E)[1/p]$ probably has complicated maximal ideals whose residue fields aren’t un-tilts of $E$.

It seems that $W({\mathcal{O}}_E)[1/p]$ is not the full ring of functions on $Y_E$. We’ll construct a ring $B_E$ which contains $W({\mathcal{O}}_E)[1/p]$ which has the property that closed maximal ideals of $B_E$ are in bijection with $\left\lvert{Y_E}\right\lvert$. The construction is analytic in nature. Let $\left\lvert{\;}\right\lvert$ be an absolute value on $E$ which induces its topology. For $r>0$, define a norm $\left\lvert{\;}\right\lvert_r$ on $W({\mathcal{O}}_E)[1/p]$ by

$\left\lvert{\sum_{n\gg -\infty} [a_n] p^n}\right\lvert = \sup_n \left\lvert{a_n} \right\lvert p^{-rn}.$

Now suppose $x=(F,\iota)$ is an un-tilt of $F$. Let $\left\lvert{\;}\right\lvert_F$ be the absolute value on $F$ for which $\left\lvert{p}\right\lvert_F=1/p$. This absolute value induces $\left\lvert{\;}\right\lvert_{F^{\flat}}$ on $F^{\flat}$, which is a finite extension of $E$ via $\iota$. The two absolute values on $E$ must be equivalent, in the sense that there exists $r>0$ for which $\left\lvert{e}\right\lvert^r=\left\lvert{\iota(e)}\right\lvert_{F^\flat}$ for all $e\in E$. Thus $\left\lvert{e}\right\lvert^r=\left\lvert{e^\sharp}\right\lvert_F$.

Given $f\in W({\mathcal{O}}_E)[1/p]$, we can compare $\left\lvert{f(x)}\right\lvert_E$ and $\left\lvert{f}\right\lvert_r$. If $f=\sum [a_n]p^n$, then

\begin{aligned} \left\lvert{f(x)}\right\lvert_F = & \ \left\lvert{\sum a_n^\sharp p^n}\right\lvert_F\\ \leq & \ \sup \left\lvert{a_n^{\sharp}}\right\lvert_Fp^{-n} \\ = & \ \sup \left\lvert{a_n}\right\lvert^{1/r}p^{-n}\\ = & \ \left\lvert{f}\right\lvert_r^{1/r} \end{aligned}

It follows from this inequality that if $f_i$ is a Cauchy sequence in $W({\mathcal{O}}_E)[1/p]$ with respect to $\left\lvert{\;}\right\lvert_r$, then $f_i(x)$ converges in $F$.

Definition: Let $B_E$ be the Fréchet completion of $W({\mathcal{O}}_E)[1/p]$ with respect to the norms $\left\lvert{\;}\right\lvert_r$ for $r>0$. That is, $B_E$ is the ring of sequences in $W({\mathcal{O}}_E)[1/p]$ which are Cauchy with respect to every $\left\lvert{\;}\right\lvert_r$, modulo those sequences which converge to 0 with respect to every $\left\lvert{\:}\right\lvert_r$.

In light of the foregoing discussion, if $f\in B_E$, then $f(x)$ makes sense for any $x\in \left\lvert{Y_E}\right\lvert$. Thus for every un-tilt $(F,\iota)$ of $E$, we get a continuous surjection $B_E\to F$ extending $\theta_x$, whose kernel is a closed maximal ideal of $B_E$.

Theorem C: Closed maximal ideals of $B_E$ are in bijection with $\left\lvert{Y_E}\right\lvert$.

This tempts us to define $Y_E$ as a rigid space by setting $Y_E={\mathrm{MaxSpec} \ } B_E$, except that $B_E$ isn’t anything like a Tate algebra. It turns out that $Y_E$ can be given a meaningful definition as an adic space, but this is the topic for another post.

At this point we can link $Y_E$ to classical $p$-adic Hodge theory. If $F$ is a perfectoid field in characteristic 0, and $E=F^{\flat}$, then we get a point $\infty\in \left\lvert{Y_E}\right\lvert$, and a maximal ideal $\mathfrak{m}\subset B_E$. Then the completion of $B_E$ with respect to $\mathfrak{m}$ is $B_{dR,F}^+$, the de Rham period ring associated to $F$. This is a complete DVR with residue field $F$.

Now we turn to $X_E$, which ought to be the quotient $Y_E/\phi^{\mathbb{Z}}$. This is supposed to be something like a projective curve, and I would like to motivate the construction of $X_E$ with projective curves in mind. To that end, suppose $X$ is a curve which is proper over a field, and you would like to give some kind of explicit presentation for $X$. (For intance, $X$ could be a smooth curve of genus 1, and you would like to show that $X$ is isomorphic to a plane cubic.) The usual thing to do is to find a very ample line bundle $\mathscr{L}$ on $X$, in which case

$X={\mathrm{Proj} \ }\left(\bigoplus_{n\geq 0} H^0(X,\mathcal{L}^{\otimes n})\right).$

In the case of the Fargues-Fontaine curve $X_E$, what should $\mathscr{L}$ be? Whatever it is, it must pull back to a line bundle on $Y_E$ which is $\phi$-equivariant. Since $B_E$ is the ring of analytic functions on $Y_E$, this should be the same as giving a free $B_E$-module $M$ of rank 1 together with a $\phi$-semilinear map $\phi\colon M\to M$. Let $M=B^+e$, where $\phi(e)=p^{-1}e$. This corresponds to a line bundle $\mathscr{L}$ on the (not yet defined) $X_E$. Then we ought to have, for any $n\in{\mathbb{Z}}$,

$H^0(X_E,\mathscr{L}^{\otimes n})=(Be^{\otimes n})^{\phi=1}=B^{\phi=p^n}.$

This prompts the following definition.

Definition: $X_E={\mathrm{Proj} \ } P$, where $P=\oplus_{n\geq 0} P_n$ is the graded ${\mathbb{Q}}_p$-algebra with $P_n=B_E^{\phi=p^n}$

To convince you this was the right thing to do, let me list the following facts, which hold when $E$ is algebraically closed:

1. $P$ is a graded factorial ring, whose irreducible homogeneous elements are exactly the nonzero elements of degree 1.
2. If $t\in P_1$ is nonzero, then its divisor in $X_E$ is $(\infty_t)$, for a point $\infty_t\in \left\lvert{X_E}\right\lvert$ of degree 1.
3. Conversely, if $\infty\in \left\lvert{X_E}\right\lvert$ then there exists $t\in P_1$ whose divisor is $(\infty_t)$.
4. More generally, the divisor of a nonzero element of $P_n$ has degree $n$.

(I didn’t say exactly what the divisor of an element $f\in B_E$ is, but it’s what you think: a formal sum of points in $\left\lvert{X_E}\right\lvert$, weighted with multiplicities. Since each $B_{dR,F}$ is a DVR, the multiplicities make sense.) Thus $X_E$ resembles nothing so much as the projective line over a field! In fact, Fargues and Fontaine show that (again under the assumption that $E$ is algebraically closed) $X_E$ is simply connected.

Let me close with an amusing observation. Consider the field ${\mathbb{C}}_p$, which is of course a perfectoid field. Let $E={\mathbb{C}}_p^\flat$. (I would call it $B$, but $B$ was taken!) Of course one of the un-tilts of $E$ is ${\mathbb{C}}_p$, but what are the others? Are they all isomorphic to ${\mathbb{C}}_p$?

I don’t know. But suppose instead we took $E$ to be the field of Malcev-Neumann series $k(\!({x^{\Gamma}})\!)$, where $k$ is algebraically closed and $\Gamma$ is a divisible ordered abelian group. Elements of $E$ are “power series” in $x$ with coefficients in $k$ and exponents in $\Gamma$, where the only restriction is that the support of each power series be a well-ordered subset of $\Gamma$. You get a valuation on $E$ by looking at the least exponent of $x$ that appears in such a series.

$E$ is a maximally complete field, meaning that any valued extension field $E'/E$ either has a larger residue field or else a larger value group. There is also a characteristic 0 construction, which I’d like to call $F=W(k)(\!({p^\Gamma})\!)$, and then $F^\flat=E$. A result of Bjorn Poonen is that any maximally complete field with residue field $k$ and value group $\Gamma$ has to be isomorphic $F$ or $E$, depending on its characteristic. (His proof uses the axiom of choice in an essential way.)

(Sometimes I feel that the true $p$-adic analogue of the complex numbers isn’t ${\mathbb{C}}_p$ but rather $F$. The field ${\mathbb{C}}$ is spherically complete, meaning that any nested sequence of balls has nonempty intersection. ${\mathbb{C}}_p$ doesn’t have this property, but $F$ does. Furthermore, elements of ${\mathbb{C}}$ have decimal expansions. Elements of $F$ also do, in the sense that every element is a power series in $p$. How do you write down a generic element of ${\mathbb{C}}_p$? You basically can’t.)

Let $(K,\iota)$ be any un-tilt of $E$. Then $K$ is maximally complete. (Exercise. Hint: Tilting preserves both residue field and value group.) By Poonen’s result, there is an isomorphism $f\colon K\to F$. This induces an isomorphism $f^{\flat}\colon K^{\flat}\to F^{\flat}=E$. Composing $f^{\flat}$ with $\iota\colon E\to K^\flat$ gives an automorphism of $E$ (as a topological field).

This argument shows that ${\mathrm{Aut} \ } E$ acts transitively on the set of un-tilts $\left\lvert{Y_E}\right\lvert$. The un-tilt $F$ of $E$ gives a point $\infty\in \left\lvert{Y_E}\right\lvert$, and it is almost tautological to see that the stabilizer of $\infty$ is the image of ${\mathrm{Aut} \ } F$ under the natural map ${\mathrm{Aut} \ } F\to {\mathrm{Aut} \ } E$. Thus there is a bijection

$\left\lvert{Y_E}\right\lvert \cong ({\mathrm{Aut} \ } E)/({\mathrm{Aut} \ } F).$

Similarly, you get a description of $\left\lvert{X_E}\right\lvert$ as $({\mathrm{Aut} \ } E)/({\mathrm{Aut} \ } F)\phi^{{\mathbb{Z}}}$. I find this rather amazing, since ${\mathrm{Aut} \ } E$ and ${\mathrm{Aut} \ } F$ are unimaginably huge groups with no obvious geometric structure, while $X_E$ is a proper curve. I’m not sure if this observation is useful to the study of $X_E$, but given the rising role of maximally complete fields in $p$-adic Hodge theory, it’s worth a look.

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### 10 Responses to The Fundamental Curve of p-adic Hodge Theory, or How to Un-tilt a Tilted Field

1. Kevin Buzzard says:

Whilst I’m completely happy with the equiv of cats under Thm A, I’ve always had trouble with the way people state Thm A itself. As you know, a field doesn’t have “an” absolute Galois group, and I think that you would be hard pressed to argue that if X is the algebraic closure of the rationals in C, and Y is the algebraic closure of the rationals in C_p, then Gal(X/Q) and Gal(Y/Q) were in any way at all *naturally* isomorphic, even though they are both “the” absolute Galois group of Q. As a consequence I find it hard to believe any non-tautological assertion of the form “the absolute Galois group of blah is naturally isomorphic to blah”.

• Text suitably modified!

• Kevin Buzzard says:

Somehow *something* stronger than “isomorphic” is true — because it _is_ I believe true that given an algebraic closure of $F$, one can construct in a natural way an algebraic closure of $F^\flat$ (e.g. via the equivalence of categories, which I think _is_ “natural” in some sense that I am not sure how to make precise). So somehow you _can_ do better than just “isomorphic” — but the only way I know of formalising how one can do better is via the equiv of cats. I guess one could say that the etale topoi of Spec(F) and Spec(F-flat) were naturally isomorphic — perhaps that is a statement more in line with the attempt to summarise what’s really going on in a succinct way.

• Jay says:

Without getting into an argument about what “natural” “should” mean in mathematics, I’d suggest that the Galois groups are “constructively” isomorphic. There is a direct algorithm (in my sense, and only in the computer scientist’s for a *very* long tape) for the ${}^\flat$-functor constructing extensions of $E$ from extensions of $F$, hence taking any starting algebraic closure of $E$ to one of $F$, and identifying the Galois groups.

2. PS says:

OK, I think here is what *something* is in Kevin’s comment. As is well-known, the fundamental group (resp. absolute Galois group) is not natural (i.e., depends on the choice of a base point, resp. an algebraic closure), but the fundamental groupoid (resp. “absolute Galois groupoid”(?)) is natural. In the topological context, this is the groupoid whose objects are points of the space, and morphisms are paths. In the algebraic context, objects are algebraic closures of your field, and morphisms are isomorphisms. Then these groupoids are equivalent as groupoids to actual groups, which are the fundamental group, resp. absolute Galois group, but this reduction of structure is noncanonical. The intuition here is that there is some fuzzyness about the absolute Galois group, and this fuzzyness is caught up by a space of possible choices — the topological space itself in the topological case, and the “space” of algebraic closures in the algebraic case.

Now, what is true is that the “absolute Galois groupoids” of F and F-flat are naturally equivalent. Even better, there are explicit functors in both directions, and explicit natural isomorphisms (i.e., natural transformations consisting of isomorphisms) from the identity functor to either composition. (To get these functors, use the tilting functor in one direction, and the Witt vector description of untilting in the other. The natural transformations are just the observation that their composition is “the same” ring, in the sense that there is a canonical map in one direction, which is an isomorphism.) In other words, the fuzzyness in the definition of the absolute Galois group of F and F-flat is exactly the same, and a choice of algebraic closure on one side determines an entirely canonical algebraic closure on the other side (as indicated by Kevin). In summary, I think this state of affairs may be summarized as “the absolute Galois groups are naturally isomorphic”.

3. Wholesome Breakfast says:

Kevin, regarding canonical things, isn’t there a relevant Witt vector construction here? E.g. if $R^{\flat}$ is an $\mathcal{O}_{F^{\flat}}$-algebra then

$W(R^{\flat}) \otimes_{W(\mathcal{O}_{F^{\flat}}),\theta} \mathcal{O}_{F}$

is an $\mathcal{O}_{F}$-algebra, where $\theta: W(\mathcal{O}_{F^\flat}) \rightarrow \mathcal{O}_{F}$ is the usual map.

• Kevin Buzzard says:

Sure! There is a completely canonical equivalence of categories in either direction. The only thing that “bothered” me (and I’m just repeating myself here — sorry) was that I was a little uncomfortable saying “_the_ absolute Galois group” is *anything*, because there isn’t “an” absolute Galois group. But PS’s suggestion to talk about the absolute Galois groupoid instead seems to me to be a very nice way of getting round this issue.

4. rmb says:

Do the non-classical points of the curve have any moduli-theoretic meaning?

• Jared Weinstein says:

Yes, and the meaning is straightforward if you think of X_E as an adic space over Q_p. Then points of X_E (classical or otherwise) parameterize equivalence classes of un-tilts F of E, except that the finiteness condition on F^{flat}/E is dropped. I talk about this in my next post.