User laurent berger - MathOverflow

A criterion for freeness over a local ring

2013-04-04T15:28:08Z

Let $A=K[[X_1,\dots,X_n]]$ where $K$ is a field. Let $M$ be a finitely generated torsion-free $A$-module, such that

for all $k$, the $A[1/X_k]$-module $M[1/X_k]$ is free of rank $d$;
for every $i \neq j$, we have $M = M[1/X_i] \cap M[1/X_j]$.

Does this imply that $M$ is free?

It certainly does if $n=1$ (easy), and also if $n=2$ (reduce $M$ modulo $X$), but things seem trickier if $n \geq 3$.

This question comes up when trying to prove that some $(\varphi,\Gamma)$-modules over rings in several variables are free.

Answer by Laurent Berger for Galois descent for semilinear endomorphisms

2013-03-27T08:53:37Z

If $E$ is of dimension $1$, then $\phi$ is given in some basis by a scalar $x \in L$ and you are asking whether there is an element $y \in L^\times$ such that $x \cdot \sigma(y)/y$ belongs to $K$. I don't see why there should be a simple general criteria to decide whether such a $y$ exists. If $\sigma$ fixes $K$, then the norm of $x$ should be the norm of an element of $K$, and conversely this would be enough if in addition $Gal(L/K)$ was generated by $\sigma$.

More generally, you can at least rewrite your problem in terms of cohomology: the group $\mathbb{Z}$ acts on $L$ by $n \cdot x = \sigma^n(x)$, and you're asking about the image of $H^1(\mathbb{Z},GL_d(K)) \to H^1(\mathbb{Z},GL_d(L))$. But without more specific information on $K$, $L$ and $\sigma$, I don't see what one could say in general.

Answer by Laurent Berger for Describing the ratio of uniformizers in B_dR

2012-12-23T08:56:36Z

What you're asking for is a description of $$a_1 = \theta(\frac{[\tilde{p}]-p}{t}).$$ It is an element of $C_p$ and you can't really "write it down explicitly". What you can do is let $G_{Qp}$ act on it and see what happens. If $g \in G_{Qp}$ then $g(\tilde{p})=\tilde{p} \cdot \epsilon^{c(g)}$ where $c(\cdot)$ is the Kummer cocycle and $g(t)=\chi(g) t$ where $\chi(\cdot)$ is the cyclotomic character. This should imply that $$g(a_1) = \theta(\frac{[\tilde{p}][\epsilon^{c(g)}]-p}{\chi(g) t}) = \frac{a_1}{\chi(g)} + p \frac{c(g)}{\chi(g)}.$$ Now that you see this formula, you should recognize it. Let $V$ be the semistable extension of $Q_p$ by $Q_p(1)$. It has Hodge-Tate weights $0$ and $1$ and $a_1$ is basically the period corresponding to weight $0$, ie the period that tells you that $(V \otimes_{Qp} C_p)^{G_{Qp}} \neq 0$.

Answer by Laurent Berger for Fastest way to factor integers < 2^60

2012-11-21T08:34:37Z

A long long time ago I programmed my HP48 (4 MHz! Wow!) in assembly language to factor numbers. It would do trial division by all integers congruent to $\pm 1 \bmod{6}$, up to some bound B and then use Pollard's rho method. I guess in your case you need to figure out what B should be depending on your implementation. At the time, I was using "Prime Numbers and Computer Methods for Factorization" by Hans Riesel as a reference. Since $2^{60}$ is relatively small, I'm not sure that more powerful methods would really help. Pollard's rho method is believed to be in something like $O(p^{1/2})$ where $p$ is the smallest prime factor of your integer, which is quite small in your case.

Answer by Laurent Berger for Absolute Galois group of the field of Puiseux series over $\overline{\mathbb{F}}_p$

2012-11-16T09:32:01Z

Let $E$ be the field $\overline{\mathbb{F}}_p((X))$. The field of Puiseux series whose exponents have denominators prime to $p$ is a subfield of $E^{sep}$, so the group you're asking about would then be the wild inertia subgroup of $Gal(E^{sep}/E)$. The group $Gal(E^{sep}/E)$ is quite complicated, and it comes up in arithmetic geometry, for example when studying the $\pi_1$ of curves. It also occurs as a closed subgroup of $Gal(\overline{\mathbb{Q}}_p / \mathbb{Q}_p)$ by the theory of the field of norms of Fontaine and Wintenberger. Its representations on $\mathbb{Z}_p$-modules are described by ``$\varphi$-modules'' (like $(\varphi,\Gamma)$-modules without the $\Gamma$). If you want to include Puiseux series whose exponents have denominators divisible by $p$, then you're looking at the perfection of $E$. The group does not change, as $E^{sep}$ is dense in $E^{alg}$ by a theorem of Ax.

Answer by Laurent Berger for Is there any theorem like implicit function theorem in $\mathbb{Q}$ ?

2012-10-24T16:02:45Z

Here's what I think happens over $\mathbb{Q}$. Write your polynomial $P(X,Y)$ as a product of irreducible polynomials $P_i(X,Y)$. Hilbert's irreducibility theorem ( http://en.wikipedia.org/wiki/Hilbert%27s_irreducibility_theorem ) tells you that there are infinitely many $a$'s such that $P_i(X,a)$ is irreducible for every $i$. If one of them has a solution, it is therefore of degree $1$ in $X$. Some $P_i$ is therefore of degree $1$ in $X$, which answers your question.

EDIT: it does not answer the question but rather shows that there is some polynomial $Q$ such that $P(Q(Y),Y)=0$ which is more reasonable, since then $P(Q(a)),a)=0$. This should have been the question.

Answer by Laurent Berger for Slick ways to make annoying verifications

2012-10-23T17:33:22Z

A surjective map between two modules (over a commutative ring) of the same rank is a bijection.

Answer by Laurent Berger for Lost soul: loneliness in pursing math. Advice needed.

2012-10-21T07:15:49Z

Hi Flora,

it seems that many people have answered your question assuming that you're asking about becoming a research mathematician. But you can do a PhD in math and then go on to a different career, where you use your skills but don't necessarily go on to do original research. Many companies are hiring math PhDs and it seems to me that there are opportunities in very diverse fields. I'm not in the best position to discuss them ($p$-adic analysis doesn't seem to be very much in demand in the real world these days), but you should be able to find such information easily, maybe starting with the AMS?

And don't let yourself be discouraged by the behavior of your colleagues. It's true that some mathematicians spend all their waking hours thinking and talking about maths, and a few of them are very successful but these are really a minority. It's my opinion that most of those who display this behavior would really be better off if they had a more balanced lifestyle. There may be cultural factors here: in the protestant anglo-saxon world, it's important to be seen working all the time, and bad form for people to enjoy themselves.

Just do your own thing at your own pace. It's normal that you spend hours on a proof. Maths is not about being the first, but about having a deep understanding of what is going on. It certainly helps if you have flashes of insight, but there's nothing wrong about really understanding the situation first and finding the answer afterwards.

Concerning the funding, it's true that being a PhD student does not pay much and that can be a real problem. I don't know about the country where you are but in France, some companies have programs where you work for them part-time and do a PhD in the rest of your time, the PhD being in some applied domain that is then useful for the company.

Answer by Laurent Berger for The significance of modularity for all Galois representations

2012-10-05T14:27:48Z

You would lose all the p-adic families of Galois representations, that are necessary to prove properties of the modular ones. Hida's family was necessary for the proof ofMTT, for example.

Answer by Laurent Berger for Decomposition of Matrices in Semisimple and Nilpotent Parts

2012-09-29T14:27:23Z

You don't need to conjugate if you want $R$ to be diagonalizable (as opposed to diagonal).
I assume you want $R$ and $M$ to have coefficients in $K$, otherwise just work in the algebraic closure.
The statement is then true if $K$ is perfect and possibly false otherwise, as you can see by taking $A=[[0, 1], [t, 0]]$ in $K=F_2(t)$.

EDIT : see http://en.wikipedia.org/wiki/Jordan%E2%80%93Chevalley_decomposition

Answer by Laurent Berger for Are D_dR and D_st "potentially comparable"?

2012-09-29T11:44:50Z

Actually, I think that Rebecca is right and that the answer is "no". Here's a sketch of the reason why.

Let $V$ be a $p$-adic representation. If $V$ is Hodge-Tate, then $D_{dR}(V) \neq 0$. So it's enough to find a HT representation such that $D_{st}^L (V) = 0$ for any $L$. Although I can't think of an explicit one, in my paper "Représentations potentiellement triangulines de dimension 2", with Gaëtan Chenevier, we prove the following theorem: if $X$ is the universal deformation space of some mod $p$ representation, and if $X_P$ is the subset of $X$ consisting of representations whose Sen polynomial is $P$, then the subset of $X_P$ consisting of potentially trianguline representations is a "thin subset" (i.e. most representations in $X_P$ are not potentially trianguline). It remains to observe that if $D_{st}^L (V) \neq 0$, then $V$ is potentially trianguline.

EDIT : I should have said that $V$ is of dimension 2 here.

Answer by Laurent Berger for local galois representation with higher coefficient

2012-09-06T07:06:19Z

If $F$ is not finite but rather equal to $C_p$ then this is really Sen's theory (see for instance Fontaine's course notes in Astérisque 295). If $F$ is merely a finite extension of $K$, then I'm not sure that you need to introduce a lot of machinery: restrict your representation to $G_F$ so that it's linear, do what you have to do, and then take in account the extra structure that you had. Alternatively, a semilinear representation is the same as an element of $H^1(G,GL_d(F))$, so you could use Galois-cohomological techniques, especially the inflation-restriction sequence with $G_F$ and $G_K$.

EDIT : this answered the question for $F$-semilinear representations of $G_K$ with $F$ an extension of $K$. Since then the OP has modified his question so my answer is not relevant anymore :(

Answer by Laurent Berger for (phi, Gamma) module of ordinary elliptic curve

2012-08-04T16:25:23Z

If your elliptic curve has good ordinary reduction, then the attached Galois representation is reducible : it is an extension of $\eta_2$ by $\eta_1 \chi$ where $\eta_{1,2}$ are unramified characters and $\chi$ is the cyclotomic character. The $(\phi,\Gamma)$-modules of $\eta_2$ and $\eta_1 \chi$ are easy to compute, so it remains to say something about the extension, ie the upper right star in the matrices of $\phi$ and $\gamma \in \Gamma$. This is less easy; one can't simply write general formulas, but by using the results of Cherbonnier-Colmez (JAMS) and Colmez (eg his paper on trianguline representations), you can say a number of interesting things.

Answer by Laurent Berger for Incidences of rigorous proofs used in legal proceedings

2012-07-11T09:52:19Z

In France some roadside apparatus will take a picture of you at point A and another apparatus will take a picture at point B on the same highway. They will then apply the mean value theorem to determine if you deserve a ticket for speeding.

Answer by Laurent Berger for What's the name for the analogue of divided power algebras for x^i/i?

2012-06-25T11:56:15Z

One possible way to generalize divided powers is to model sequences $x^n/a_n$ where $a_{n+m}/(a_na_m)$ is integral (an integer, or at least integral in some sense). You can model these in the obvious way, like divided powers. Choosing $a_n = n!$ gives you divided powers, and choosing $a_n=p^n$ gives you a sequence which, from the point of view of $p$-adic analysis, is much more "regular" in its growth. You can come up with other examples.

As you may know, one uses divided powers to construct Fontaine's ring of periods $B_{cris}$ and using $a_n=p^n$ instead of $a_n=n!$ gives a ring $B_{max}$ that is similar in nature to $B_{cris}$ but much better behaved (see III.2 of Colmez' 1998 paper "Théorie d'Iwasawa des représentations de de Rham d'un corps local").

More generally, tweaking divided powers to get various convergence conditions is something that happens often in $p$-adic Hodge theory. See for instance 5.2.3 of Fontaine's "Le corps des périodes $p$-adiques" for one of many examples of custom made convergence conditions.

In the $p$-adic setting, I would say that your log series belongs to the "space of holomorphic functions on the $p$-adic open unit disk, having order of growth $\leq 1$". Using precise analytic conditions rather than "divided powers" seems the right way to study the analytic series that interest you.

Answer by Laurent Berger for 1-dimensional semi-stable Galois representations with coefficients

2012-05-27T14:14:23Z

If $E \neq Q_p$ then there may be more $1$-dimensional crystalline representations that the ones you mention. By Lubin-Tate theory, every character of $G_K$ can be written as an unramified character times a character of $O_K^\times$ (after making proper choices and identifications). The algebraic characters of $O_K^\times$ are then crystalline and if $E$ contains $K$, then these provide examples of crystalline characters.

EDIT : oops sorry, I did not read the question carefully enough, I did not see that you were asking for an explicit description of all such representations. As David Loeffler pointed out, the answer is in Brian Conrad's paper (now in appendix B, the paper having grown since last time I mentioned it).

Here is what it basically says, using the same identifications as above: assume that $E$ contains $K^{gal}$ (it's usually harmless to assume that the coefficient field is large enough). If $s$ runs through the set of embeddings $s : K \to E$ and the $a_s$ are integers, then $x \mapsto \prod_s s(x)^{a_s}$ gives rise to a crystalline character of $G_K$ and they're all of this type times an unramified character.

Answer by Laurent Berger for Where to find Asterisque online?

2012-05-08T07:33:04Z

The issue of providing online access to Asterisque is a difficult one. The SMF offers electronic versions of its other journals, and the subscription rates for these has apparently significantly dropped. Asterisque is now the most profitable publication of the SMF and there is a lot of reluctance to join the 21st century and jeopardize this profitability. I agree with Obelisque that the best solution is to buy the old issues, but the ones that are most in demand are also out of print. All of these topics are currently being discussed (electronic versions, reprinting the back issues, ...). It is also not impossible that the Bourbaki seminars will soon be available online, maybe with a five year buffer, as the contract beween Bourbaki and the SMF gives total freedom to Bourbaki.

EDIT : of course, one reason why the standard model for journals does not quite apply to Asterisque is because it's part journal, part book series...

Answer by Laurent Berger for are the smooth vectors of a Frechet space dense?

2012-03-14T08:10:56Z

In full generality, the answer is "no". If you have a unitary irreducible representation of a $p$-adic Lie group on a $p$-adic Banach space, then the locally constant vectors are usually not dense. There may even be no nonzero locally algebraic vector. The locally constant vectors are dense however if you have an $\ell$-adic Lie group acting on a $p$-adic space, with $\ell \neq p$ (this is a theorem of Vigneras).

Answer by Laurent Berger for Can a p-adic representation and its twist by a non-crystalline character both have nontrivial $D_{cris}$?

2011-12-20T16:36:05Z

Let me use Colmez' article "Representations triangulines" as a reference. Let $V$ be a repn which satisfies your condition.

By proposition 4.3, $V$ is trianguline. By proposition 4.10, the HT weight of $\chi$ has to be an integer. You can then assume that $\chi$ has finite order, and this implies that $V$ is potentially crystalline on an abelian extension of $Q_p$ (aka crystabelline).

Conversely, it seems likely that one can give examples of irreducible crystabelline representations $V$ such that $D_{cris}(V)$ and $D_{cris}(V \otimes \chi)$ are both nonzero, with $\chi$ of finite order (see the examples in 2.4 of Berger-Breuil's "Sur quelques representations potentiellement cristallines de $GL_2(Q_p)$").

Banach-Mazur applied to a Hilbert space

2011-12-05T18:40:39Z

The Banach-Mazur theorem says that every separable Banach space is isometric to a subspace of $C^0([0;1],R)$, the space of continuous real valued functions on the interval $[0;1]$, with the sup norm.

If we apply this to $\ell^2(R)$, then we see that $C^0([0;1],R)$ has a subspace which is a Hilbert space for the sup norm.

My question is can one write down explicitly such a subspace of $C^0([0;1],R)$?

I'm just curious, that's all.

Answer by Laurent Berger for P-adic representations

2011-11-25T10:11:27Z

The whole thing is done with more details in Tate's original article "p-divisible groups", section 3.2. Tate proves that one can approximate a cocyle in $C_p(i)$ by cocyles with values in $Q_p^{alg}(i)$ and this is how he reduces the computation to the "discrete case".

I would suggest that it's better to prove the result by $p$-adic approximation. This way, you can basically work with cocycles with values in $O_{C_p}(i)/p^n$, also a discrete space.

Answer by Laurent Berger for Is an identity that is true for matrix Lie groups true for all Lie groups?

2011-11-22T14:44:54Z

Every finite dimensional Lie algbera over a field of characteristic $0$ is a subalgebra of some matrix Lie algebra (Ado's theorem), so every Lie group is locally isomorphic to a group of matrices.

Answer by Laurent Berger for Elementary results with p-adic numbers

2011-11-20T08:29:35Z

In other words the question is: if I had prepared something about Galois theory I would have finished with the application to resolubility of polynomial or compass and straightedge constructions; if it had been something about modular forms, it would have been for sure Fermat's last theorem; with 3-surfaces it would have been Poincaré conjecture and so on. What if it's about p-adic numbers or p-adic analysis?

These results have very different levels of sophistication. I would propose the Weil conjectures as an application : http://en.wikipedia.org/wiki/Weil_conjectures#Statement_of_the_Weil_conjectures

The first statement (rationality of the Zeta function) was originally proved by Dwork using purely $p$-adic methods. It's a beautiful application of $p$-adic analysis and $p$-adic functional analysis. In addition, it should not be too hard to state the theorem, since it's about counting solutions of polynomials in finite fields.

Finally, you can also say that Kedlaya now has given a purely $p$-adic proof of the complete conjecture (previously proved by Deligne using other methods).

Answer by Laurent Berger for Enumerating non-abelian extensions of $\mathbb{Q}_p$?

2011-10-17T11:59:17Z

This is standard stuff. Here is (in French) the solution as an exercise, copy-pasted from the final exam of a course I gave on local fields.

Soit $K$ une extension totalement ramifiée de degré $n$ de $Q_p$ et $\pi_K$ une uniformisante de $K$. On suppose pour l'instant que $p \nmid n$.

Montrer que si $w \in Q_p$ et $w^n=1$, alors $w^m=1$ où $m = n \wedge (p-1)$ (si $p \neq 2$) et $m=2$ si $p=2$.
Montrer que l'application $x \mapsto x^n$ de $1+M_K$ dans lui-même est surjective.
Montrer que dans $O_K$, on peut écrire $\pi_K^n = p w (1+z)$ où $w^{p-1} = 1$ et $z \in M_K$.
En déduire que $Q_p$ admet exactement $n$ extensions totalement ramifiées de degré $n$.

Answer by Laurent Berger for Is there a "trianguline period ring", or is one expected?

2011-07-14T06:45:05Z

The category of trianguline representation is stable under all the usual representation-theoretic operations (subs, quotients, $\oplus$, $\otimes$), so by some general tannakian formalism, there does exist a ring $B_{tri}$. The rough idea is to look at $Q_p^{alg} \otimes B_{st} \langle \langle \log(t) \rangle \rangle$ where "$\langle \langle \log(t) \rangle \rangle$" means "power series with some non zero radius of convergence" and $t$ is the usual $t$ in this business. It's interesting to note that I first heard about this ring from Fontaine (around 2003-04 maybe - I was still at Harvard) when trianguline representations had not yet been defined. Fontaine told me at the time that repns admissible for this ring should be interesting! A few comments are in order:

$B_{st}$ does not have the structure of a Banach space, so you need to figure out what "radius of convergence" means
$\exp(\log(t))=t$, so there are relations in the definition of your ring
you need to decide if you want a ring for "trianguline" or "split-trianguline" repns

I thought about this again a few weeks ago and, if I remember correctly, came to the conclusion that if you take $B = Q_p(\mu_p) \otimes \hat{Q}_p^{nr} \otimes B_e \otimes Q_p \langle \langle \log(t) \rangle \rangle \otimes Q_p[\log(\tilde{p})]$ (whew!), where $B_e=B_{cris}^{\phi=1}$, then $B$-adm reps of $G_{Q_p}$ are trianguline, and conversely split trianguline reps of $G_{Q_p}$ with integer slopes are $B$-adm. This hopefully gives an idea of the kind of ring which one should be looking for.

Answer by Laurent Berger for Longest coinciding pair of integer sequences known

2011-07-03T18:04:13Z

There is also, of course, what comes out of the answer which was given to this question:

http://mathoverflow.net/questions/11517/computer-algebra-errors

Answer by Laurent Berger for Tamagawa numbers of crystalline Galois representations

2011-05-26T08:57:48Z

I think that for $K_n=\mathbf{Q}_p$ what you're looking for is in my (unpublished) paper

http://perso.ens-lyon.fr/laurent.berger/autrestextes/tamag0919.pdf

see proposition II.2 for instance.

This paper is unpublished because it was rewritten and massively expanded with/by Denis Benois. There is some stuff in the paper with Benois about going up the cyclotomic tower which may help. I'm sorry I don't have time to look in Benois-Berger to see if there's an answer to your full question.

Answer by Laurent Berger for Resources for mathematics advising.

2011-04-26T10:56:54Z

Since the OP is asking for resources, I'll give two references that I've found to be useful.

The Survival of a Mathematician, by Steven Krantz
I Want to Be a Mathematician: An Automathography, by Paul Halmos

Both books have something to say on the subject.

And it can't hurt to add the AMS ethical guidelines:

http://www.ams.org/about-us/governance/policy-statements/sec-ethics

Answer by Laurent Berger for Crystalline Characters

2011-04-17T07:18:26Z

Hi Kevin,

Go to http://math.stanford.edu/~conrad/
Download "Grunwald--Wang for global character groups"
Read appendix A, especially prop A.3

The answer (note that $K$ and $L$ are switched in Brian's paper) is that once you've identified your character as a character of $K^\times$ via local class field theory, it should be "algebraic" on $O_K^\times$.

Answer by Laurent Berger for how irregular can a $p$-adic Galois representation be?

2011-03-09T14:29:29Z

Here are two things that can occur:

1) If V is the representation attached to an overconvergent modular form f, then V will be unramified at almost every prime but will not be de Rham at p (unless f is classical).

2) Ramakrishna has written an article "Infinitely ramified Galois representations". Here's part of the introduction: "In this paper we show how to construct [...] representations [...] that are ramified at an infinite number of primes." Under GRH, these repns are crystalline at p.

So both conditions in FM can fail independently.

Comment by Laurent Berger

2013-04-04T16:36:08Z

Great, thank you!

Comment by Laurent Berger

2012-12-17T08:41:20Z

Concerning your first question: a $G_K$-equivariant map from V to $C_p$ exists, for example, whenever $V$ is of Hodge-Tate type with one weight equal to $0$. This happens for lots of repns for which $V^{G_K} = 0$.

Comment by Laurent Berger

2012-12-04T15:49:59Z

$\mathbf{C}_p$ has plenty of immediate extensions, that is extensions that have the same value group and the same residue field. The compositum of all these is its spherical completion. Now lots of intermediate fields between $\mathbf{C}_p$ and its spherical completion would not be algebraic maximal.

Comment by Laurent Berger

2012-11-23T15:35:36Z

There is a (not serious) mention of this in "A canticle for Leibowitz", but sadly I don't remember where exactly. Can somebody find the exact quotation?

Comment by Laurent Berger

2012-11-20T12:27:22Z

@beginner If I can be allowed to advertize my own stuff: I wrote a survey on p-adic representations a few years ago, and one chapter concerns $(\varphi,\Gamma)$-modules. See [05] of perso.ens-lyon.fr/laurent.berger/publications.php

Comment by Laurent Berger

2012-11-20T08:59:50Z

@Filippo You can do it in a few words: Colmez recalls the theory of $(\varphi,\Gamma)$-modules in a footnote, footnote 86 of his Bourbaki seminar on Kato's Euler system ([21] of math.jussieu.fr/~colmez/publications.html).

Comment by Laurent Berger

2012-11-17T14:48:14Z

Ah, so the story about Gauss inventing the least squares method to compute orbital parameters of asteroids is bogus?

Comment by Laurent Berger

2012-11-16T09:25:10Z

@Spice: if you replace $\overline{\mathbb{F}}_p$ iwth $\mathbb{C}$, then the field you get is algebraically closed.

Comment by Laurent Berger

2012-10-24T07:44:51Z

@Charles : yes, sorry, I forgot to say "free", so the modules should be free of the same rank. Thanks.

Comment by Laurent Berger

2012-10-23T17:29:20Z

And the closed graph criteria also works in a compact space!

Comment by Laurent Berger

2012-09-29T11:56:59Z

By the way, these "thin subsets" are called "parties fines" in our article. I suspect that my coauthor was playing a joke on me (and the readers) since I later found out that in French, "partie fine" also means "orgy".

Comment by Laurent Berger

2012-09-06T11:50:00Z

If you're looking at linear representations of $G$ with coefficients, then everything works "the same". See for example 3.1 of Breuil-Mézard's 2002 Duke paper.

Comment by Laurent Berger

2012-09-06T07:15:18Z

You also changed the setting completely. Is $F$ an extension of $K$ or a subfield of $K$??

Comment by Laurent Berger

2012-08-16T16:01:55Z

This ring has not been studied. At some point, my student Di Matteo was interested in it, so you could ask him.

Comment by Laurent Berger

2012-08-04T16:32:32Z

Isn't this just a matter of saying $v(b \otimes c) = v(b) + v(c)$ and letting the valuation of an element of $B \otimes C$ be the sup, over all possible ways to write the element as $\sum b_i \otimes c_i$, of $\inf_i v(b_i \otimes c_i)$?