$(varphi,Gamma)$-modules and valuation in Zp
up vote
7
down vote
favorite
The theory of $(varphi,Gamma)$-modules and a reciprocity law due to Cherbonnier and Colmez involve the ring $A_K$, that is described below.
The ring $A_K$:
Set $E$ to be the set of sequences $(x^{(0)},x^{(1)}, . . . )$ of elements of $C_p$ satisfying $(x^{(n+1)})^p=x^{(n)}$, with addition given by $(x + y)^{(n)} = lim_m (x^{(n+m)} + y^{(n+m)})^{p^m}$ and $(xy)^{(n)} = x^{(n)}y^{(n)}$. Then $E$ is a complete, algebraically closed field of characteristic $p$.
number-theory p-adic-number-theory valuation-theory
|
show 1 more comment
up vote
7
down vote
favorite
The theory of $(varphi,Gamma)$-modules and a reciprocity law due to Cherbonnier and Colmez involve the ring $A_K$, that is described below.
The ring $A_K$:
Set $E$ to be the set of sequences $(x^{(0)},x^{(1)}, . . . )$ of elements of $C_p$ satisfying $(x^{(n+1)})^p=x^{(n)}$, with addition given by $(x + y)^{(n)} = lim_m (x^{(n+m)} + y^{(n+m)})^{p^m}$ and $(xy)^{(n)} = x^{(n)}y^{(n)}$. Then $E$ is a complete, algebraically closed field of characteristic $p$.
number-theory p-adic-number-theory valuation-theory
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Sep 20 at 9:26
May I ask what is the reciprocity law you mentioned that is related to the theory of $(varphi,Gamma)$-modules? And what made you think that $l_v$ behaves like this?
– awllower
Sep 21 at 9:33
1
The meaning of "uniformizer" here seems to be not the usual one (as a DVR, $A_{Bbb Q_p}$ has as uniformizer $p$, not $pi$). Maybe in the context they mean "lift of a uniformizer of the residue field", and I presume that for every such lift $pi'$, every element of $A$ can also be written as a Laurent series in $pi'$. If they mean that, then it seems plausible that $l_1$ is independent of the choice of $pi'$, but $l_2, ...$ are clearly not (take $pi' := p^nu + pi$ as a different choice and see how $l_nu$ of that depends on whether you write it as series in $pi$ or in $pi'$).
– Torsten Schoeneberg
Sep 21 at 16:55
Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{I}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i*p*pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 9:32
@TorstenSchoeneberg: Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{i}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i∗p∗pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 12:01
|
show 1 more comment
up vote
7
down vote
favorite
up vote
7
down vote
favorite
The theory of $(varphi,Gamma)$-modules and a reciprocity law due to Cherbonnier and Colmez involve the ring $A_K$, that is described below.
The ring $A_K$:
Set $E$ to be the set of sequences $(x^{(0)},x^{(1)}, . . . )$ of elements of $C_p$ satisfying $(x^{(n+1)})^p=x^{(n)}$, with addition given by $(x + y)^{(n)} = lim_m (x^{(n+m)} + y^{(n+m)})^{p^m}$ and $(xy)^{(n)} = x^{(n)}y^{(n)}$. Then $E$ is a complete, algebraically closed field of characteristic $p$.
number-theory p-adic-number-theory valuation-theory
The theory of $(varphi,Gamma)$-modules and a reciprocity law due to Cherbonnier and Colmez involve the ring $A_K$, that is described below.
The ring $A_K$:
Set $E$ to be the set of sequences $(x^{(0)},x^{(1)}, . . . )$ of elements of $C_p$ satisfying $(x^{(n+1)})^p=x^{(n)}$, with addition given by $(x + y)^{(n)} = lim_m (x^{(n+m)} + y^{(n+m)})^{p^m}$ and $(xy)^{(n)} = x^{(n)}y^{(n)}$. Then $E$ is a complete, algebraically closed field of characteristic $p$.
number-theory p-adic-number-theory valuation-theory
number-theory p-adic-number-theory valuation-theory
edited Nov 25 at 22:17
asked Sep 20 at 9:24
lartom
362
362
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Sep 20 at 9:26
May I ask what is the reciprocity law you mentioned that is related to the theory of $(varphi,Gamma)$-modules? And what made you think that $l_v$ behaves like this?
– awllower
Sep 21 at 9:33
1
The meaning of "uniformizer" here seems to be not the usual one (as a DVR, $A_{Bbb Q_p}$ has as uniformizer $p$, not $pi$). Maybe in the context they mean "lift of a uniformizer of the residue field", and I presume that for every such lift $pi'$, every element of $A$ can also be written as a Laurent series in $pi'$. If they mean that, then it seems plausible that $l_1$ is independent of the choice of $pi'$, but $l_2, ...$ are clearly not (take $pi' := p^nu + pi$ as a different choice and see how $l_nu$ of that depends on whether you write it as series in $pi$ or in $pi'$).
– Torsten Schoeneberg
Sep 21 at 16:55
Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{I}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i*p*pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 9:32
@TorstenSchoeneberg: Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{i}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i∗p∗pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 12:01
|
show 1 more comment
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Sep 20 at 9:26
May I ask what is the reciprocity law you mentioned that is related to the theory of $(varphi,Gamma)$-modules? And what made you think that $l_v$ behaves like this?
– awllower
Sep 21 at 9:33
1
The meaning of "uniformizer" here seems to be not the usual one (as a DVR, $A_{Bbb Q_p}$ has as uniformizer $p$, not $pi$). Maybe in the context they mean "lift of a uniformizer of the residue field", and I presume that for every such lift $pi'$, every element of $A$ can also be written as a Laurent series in $pi'$. If they mean that, then it seems plausible that $l_1$ is independent of the choice of $pi'$, but $l_2, ...$ are clearly not (take $pi' := p^nu + pi$ as a different choice and see how $l_nu$ of that depends on whether you write it as series in $pi$ or in $pi'$).
– Torsten Schoeneberg
Sep 21 at 16:55
Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{I}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i*p*pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 9:32
@TorstenSchoeneberg: Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{i}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i∗p∗pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 12:01
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Sep 20 at 9:26
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Sep 20 at 9:26
May I ask what is the reciprocity law you mentioned that is related to the theory of $(varphi,Gamma)$-modules? And what made you think that $l_v$ behaves like this?
– awllower
Sep 21 at 9:33
May I ask what is the reciprocity law you mentioned that is related to the theory of $(varphi,Gamma)$-modules? And what made you think that $l_v$ behaves like this?
– awllower
Sep 21 at 9:33
1
1
The meaning of "uniformizer" here seems to be not the usual one (as a DVR, $A_{Bbb Q_p}$ has as uniformizer $p$, not $pi$). Maybe in the context they mean "lift of a uniformizer of the residue field", and I presume that for every such lift $pi'$, every element of $A$ can also be written as a Laurent series in $pi'$. If they mean that, then it seems plausible that $l_1$ is independent of the choice of $pi'$, but $l_2, ...$ are clearly not (take $pi' := p^nu + pi$ as a different choice and see how $l_nu$ of that depends on whether you write it as series in $pi$ or in $pi'$).
– Torsten Schoeneberg
Sep 21 at 16:55
The meaning of "uniformizer" here seems to be not the usual one (as a DVR, $A_{Bbb Q_p}$ has as uniformizer $p$, not $pi$). Maybe in the context they mean "lift of a uniformizer of the residue field", and I presume that for every such lift $pi'$, every element of $A$ can also be written as a Laurent series in $pi'$. If they mean that, then it seems plausible that $l_1$ is independent of the choice of $pi'$, but $l_2, ...$ are clearly not (take $pi' := p^nu + pi$ as a different choice and see how $l_nu$ of that depends on whether you write it as series in $pi$ or in $pi'$).
– Torsten Schoeneberg
Sep 21 at 16:55
Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{I}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i*p*pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 9:32
Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{I}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i*p*pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 9:32
@TorstenSchoeneberg: Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{i}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i∗p∗pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 12:01
@TorstenSchoeneberg: Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{i}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i∗p∗pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 12:01
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
After clarifying terminology in the comments, I think this is what's going on here:
Let $F$ be a local field with ring of integers $O_F$ whose maximal ideal is $(p)$, and residue field $k := O_F/(p)$. Let $A$ be the set of Laurent series
$displaystyle sum_{iin Bbb Z} a_i pi^i: a_i in O_F$ such that $lim_{ito -infty} a_i=0$
(i.e. the series can be infinite on both sides, but to the left, the coefficients go to $0$). Then one can show that $A$ is a complete DVR with maximal ideal $(p)$, whose residue field $A/p$ (that's $E_K$ in your question) is of the form $k((t))$ (usual Laurent series, with finite negative part, over the field $k$). Notice that $k((t))$ is itself a complete field with a discrete valuation, whose ring of integers is $k[[t]]$ (power series over $k$). The uniformisers of this are precisely the elements of the form $bar a_1 t + sum_{ige 2}bar a_i t^i$, where $bar a_i in k$ and $bar a_1 neq 0$.
Next, let $varpi in A$ be any lift of any such uniformiser. It is of the form
$varpi = displaystyle sum_{iin Bbb Z} a_i pi^i$ with $a_i in (p)$ for $ile 0$, and $a_1in O_F^*$
(a special case being $varpi = pi$ itself).
Claim: Every element of $A$ can also be written uniquely in the form $$displaystyle sum_{iin Bbb Z} b_i varpi^i: b_i in O_F text{ such that } lim_{ito -infty} b_i=0.$$
Now in the article they define functions $l_nu: A rightarrow Bbb Z$ by
$l_nu(sum_{iin Bbb Z} a_i pi^i) := minlbrace i: p^nu text{does not divide } a_irbrace$.
but one could also define, for each $varpi$, the variant
$l_{nu, varpi}(sum_{iin Bbb Z} b_i varpi^i) :=minlbrace i: p^nu text{does not divide } b_irbrace$
(so that $l_nu$ is the choice $l_{nu, pi}$).
Now:
For $nu ge 2$, in the examples $varpi = p^{nu-1} + pi$ we have $l_nu(varpi) = 0$ but $l_{nu, varpi}(varpi) = 1$.
For $nu=1$ however, a little consideration shows that both $l_{nu}$ and $l_{nu, varpi}$ can actually be described as
$l_nu(a)= l_{nu, varpi}(a) = text{ the } t-text{adic valuation of } bar a in A/(p) = k((t))$
which does not depend on the choice of $varpi$.
I think you mean $limlimits_{nrightarrow-infty}a_n=0$ in the description of $A$ and in the claim?
– awllower
Sep 23 at 5:54
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 at 14:51
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2923822%2fvarphi-gamma-modules-and-valuation-in-zp%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
After clarifying terminology in the comments, I think this is what's going on here:
Let $F$ be a local field with ring of integers $O_F$ whose maximal ideal is $(p)$, and residue field $k := O_F/(p)$. Let $A$ be the set of Laurent series
$displaystyle sum_{iin Bbb Z} a_i pi^i: a_i in O_F$ such that $lim_{ito -infty} a_i=0$
(i.e. the series can be infinite on both sides, but to the left, the coefficients go to $0$). Then one can show that $A$ is a complete DVR with maximal ideal $(p)$, whose residue field $A/p$ (that's $E_K$ in your question) is of the form $k((t))$ (usual Laurent series, with finite negative part, over the field $k$). Notice that $k((t))$ is itself a complete field with a discrete valuation, whose ring of integers is $k[[t]]$ (power series over $k$). The uniformisers of this are precisely the elements of the form $bar a_1 t + sum_{ige 2}bar a_i t^i$, where $bar a_i in k$ and $bar a_1 neq 0$.
Next, let $varpi in A$ be any lift of any such uniformiser. It is of the form
$varpi = displaystyle sum_{iin Bbb Z} a_i pi^i$ with $a_i in (p)$ for $ile 0$, and $a_1in O_F^*$
(a special case being $varpi = pi$ itself).
Claim: Every element of $A$ can also be written uniquely in the form $$displaystyle sum_{iin Bbb Z} b_i varpi^i: b_i in O_F text{ such that } lim_{ito -infty} b_i=0.$$
Now in the article they define functions $l_nu: A rightarrow Bbb Z$ by
$l_nu(sum_{iin Bbb Z} a_i pi^i) := minlbrace i: p^nu text{does not divide } a_irbrace$.
but one could also define, for each $varpi$, the variant
$l_{nu, varpi}(sum_{iin Bbb Z} b_i varpi^i) :=minlbrace i: p^nu text{does not divide } b_irbrace$
(so that $l_nu$ is the choice $l_{nu, pi}$).
Now:
For $nu ge 2$, in the examples $varpi = p^{nu-1} + pi$ we have $l_nu(varpi) = 0$ but $l_{nu, varpi}(varpi) = 1$.
For $nu=1$ however, a little consideration shows that both $l_{nu}$ and $l_{nu, varpi}$ can actually be described as
$l_nu(a)= l_{nu, varpi}(a) = text{ the } t-text{adic valuation of } bar a in A/(p) = k((t))$
which does not depend on the choice of $varpi$.
I think you mean $limlimits_{nrightarrow-infty}a_n=0$ in the description of $A$ and in the claim?
– awllower
Sep 23 at 5:54
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 at 14:51
add a comment |
up vote
1
down vote
After clarifying terminology in the comments, I think this is what's going on here:
Let $F$ be a local field with ring of integers $O_F$ whose maximal ideal is $(p)$, and residue field $k := O_F/(p)$. Let $A$ be the set of Laurent series
$displaystyle sum_{iin Bbb Z} a_i pi^i: a_i in O_F$ such that $lim_{ito -infty} a_i=0$
(i.e. the series can be infinite on both sides, but to the left, the coefficients go to $0$). Then one can show that $A$ is a complete DVR with maximal ideal $(p)$, whose residue field $A/p$ (that's $E_K$ in your question) is of the form $k((t))$ (usual Laurent series, with finite negative part, over the field $k$). Notice that $k((t))$ is itself a complete field with a discrete valuation, whose ring of integers is $k[[t]]$ (power series over $k$). The uniformisers of this are precisely the elements of the form $bar a_1 t + sum_{ige 2}bar a_i t^i$, where $bar a_i in k$ and $bar a_1 neq 0$.
Next, let $varpi in A$ be any lift of any such uniformiser. It is of the form
$varpi = displaystyle sum_{iin Bbb Z} a_i pi^i$ with $a_i in (p)$ for $ile 0$, and $a_1in O_F^*$
(a special case being $varpi = pi$ itself).
Claim: Every element of $A$ can also be written uniquely in the form $$displaystyle sum_{iin Bbb Z} b_i varpi^i: b_i in O_F text{ such that } lim_{ito -infty} b_i=0.$$
Now in the article they define functions $l_nu: A rightarrow Bbb Z$ by
$l_nu(sum_{iin Bbb Z} a_i pi^i) := minlbrace i: p^nu text{does not divide } a_irbrace$.
but one could also define, for each $varpi$, the variant
$l_{nu, varpi}(sum_{iin Bbb Z} b_i varpi^i) :=minlbrace i: p^nu text{does not divide } b_irbrace$
(so that $l_nu$ is the choice $l_{nu, pi}$).
Now:
For $nu ge 2$, in the examples $varpi = p^{nu-1} + pi$ we have $l_nu(varpi) = 0$ but $l_{nu, varpi}(varpi) = 1$.
For $nu=1$ however, a little consideration shows that both $l_{nu}$ and $l_{nu, varpi}$ can actually be described as
$l_nu(a)= l_{nu, varpi}(a) = text{ the } t-text{adic valuation of } bar a in A/(p) = k((t))$
which does not depend on the choice of $varpi$.
I think you mean $limlimits_{nrightarrow-infty}a_n=0$ in the description of $A$ and in the claim?
– awllower
Sep 23 at 5:54
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 at 14:51
add a comment |
up vote
1
down vote
up vote
1
down vote
After clarifying terminology in the comments, I think this is what's going on here:
Let $F$ be a local field with ring of integers $O_F$ whose maximal ideal is $(p)$, and residue field $k := O_F/(p)$. Let $A$ be the set of Laurent series
$displaystyle sum_{iin Bbb Z} a_i pi^i: a_i in O_F$ such that $lim_{ito -infty} a_i=0$
(i.e. the series can be infinite on both sides, but to the left, the coefficients go to $0$). Then one can show that $A$ is a complete DVR with maximal ideal $(p)$, whose residue field $A/p$ (that's $E_K$ in your question) is of the form $k((t))$ (usual Laurent series, with finite negative part, over the field $k$). Notice that $k((t))$ is itself a complete field with a discrete valuation, whose ring of integers is $k[[t]]$ (power series over $k$). The uniformisers of this are precisely the elements of the form $bar a_1 t + sum_{ige 2}bar a_i t^i$, where $bar a_i in k$ and $bar a_1 neq 0$.
Next, let $varpi in A$ be any lift of any such uniformiser. It is of the form
$varpi = displaystyle sum_{iin Bbb Z} a_i pi^i$ with $a_i in (p)$ for $ile 0$, and $a_1in O_F^*$
(a special case being $varpi = pi$ itself).
Claim: Every element of $A$ can also be written uniquely in the form $$displaystyle sum_{iin Bbb Z} b_i varpi^i: b_i in O_F text{ such that } lim_{ito -infty} b_i=0.$$
Now in the article they define functions $l_nu: A rightarrow Bbb Z$ by
$l_nu(sum_{iin Bbb Z} a_i pi^i) := minlbrace i: p^nu text{does not divide } a_irbrace$.
but one could also define, for each $varpi$, the variant
$l_{nu, varpi}(sum_{iin Bbb Z} b_i varpi^i) :=minlbrace i: p^nu text{does not divide } b_irbrace$
(so that $l_nu$ is the choice $l_{nu, pi}$).
Now:
For $nu ge 2$, in the examples $varpi = p^{nu-1} + pi$ we have $l_nu(varpi) = 0$ but $l_{nu, varpi}(varpi) = 1$.
For $nu=1$ however, a little consideration shows that both $l_{nu}$ and $l_{nu, varpi}$ can actually be described as
$l_nu(a)= l_{nu, varpi}(a) = text{ the } t-text{adic valuation of } bar a in A/(p) = k((t))$
which does not depend on the choice of $varpi$.
After clarifying terminology in the comments, I think this is what's going on here:
Let $F$ be a local field with ring of integers $O_F$ whose maximal ideal is $(p)$, and residue field $k := O_F/(p)$. Let $A$ be the set of Laurent series
$displaystyle sum_{iin Bbb Z} a_i pi^i: a_i in O_F$ such that $lim_{ito -infty} a_i=0$
(i.e. the series can be infinite on both sides, but to the left, the coefficients go to $0$). Then one can show that $A$ is a complete DVR with maximal ideal $(p)$, whose residue field $A/p$ (that's $E_K$ in your question) is of the form $k((t))$ (usual Laurent series, with finite negative part, over the field $k$). Notice that $k((t))$ is itself a complete field with a discrete valuation, whose ring of integers is $k[[t]]$ (power series over $k$). The uniformisers of this are precisely the elements of the form $bar a_1 t + sum_{ige 2}bar a_i t^i$, where $bar a_i in k$ and $bar a_1 neq 0$.
Next, let $varpi in A$ be any lift of any such uniformiser. It is of the form
$varpi = displaystyle sum_{iin Bbb Z} a_i pi^i$ with $a_i in (p)$ for $ile 0$, and $a_1in O_F^*$
(a special case being $varpi = pi$ itself).
Claim: Every element of $A$ can also be written uniquely in the form $$displaystyle sum_{iin Bbb Z} b_i varpi^i: b_i in O_F text{ such that } lim_{ito -infty} b_i=0.$$
Now in the article they define functions $l_nu: A rightarrow Bbb Z$ by
$l_nu(sum_{iin Bbb Z} a_i pi^i) := minlbrace i: p^nu text{does not divide } a_irbrace$.
but one could also define, for each $varpi$, the variant
$l_{nu, varpi}(sum_{iin Bbb Z} b_i varpi^i) :=minlbrace i: p^nu text{does not divide } b_irbrace$
(so that $l_nu$ is the choice $l_{nu, pi}$).
Now:
For $nu ge 2$, in the examples $varpi = p^{nu-1} + pi$ we have $l_nu(varpi) = 0$ but $l_{nu, varpi}(varpi) = 1$.
For $nu=1$ however, a little consideration shows that both $l_{nu}$ and $l_{nu, varpi}$ can actually be described as
$l_nu(a)= l_{nu, varpi}(a) = text{ the } t-text{adic valuation of } bar a in A/(p) = k((t))$
which does not depend on the choice of $varpi$.
edited Sep 23 at 6:01
answered Sep 22 at 20:44
Torsten Schoeneberg
3,6362832
3,6362832
I think you mean $limlimits_{nrightarrow-infty}a_n=0$ in the description of $A$ and in the claim?
– awllower
Sep 23 at 5:54
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 at 14:51
add a comment |
I think you mean $limlimits_{nrightarrow-infty}a_n=0$ in the description of $A$ and in the claim?
– awllower
Sep 23 at 5:54
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 at 14:51
I think you mean $limlimits_{nrightarrow-infty}a_n=0$ in the description of $A$ and in the claim?
– awllower
Sep 23 at 5:54
I think you mean $limlimits_{nrightarrow-infty}a_n=0$ in the description of $A$ and in the claim?
– awllower
Sep 23 at 5:54
1
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 at 5:57
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 at 14:51
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 at 14:51
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2923822%2fvarphi-gamma-modules-and-valuation-in-zp%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Sep 20 at 9:26
May I ask what is the reciprocity law you mentioned that is related to the theory of $(varphi,Gamma)$-modules? And what made you think that $l_v$ behaves like this?
– awllower
Sep 21 at 9:33
1
The meaning of "uniformizer" here seems to be not the usual one (as a DVR, $A_{Bbb Q_p}$ has as uniformizer $p$, not $pi$). Maybe in the context they mean "lift of a uniformizer of the residue field", and I presume that for every such lift $pi'$, every element of $A$ can also be written as a Laurent series in $pi'$. If they mean that, then it seems plausible that $l_1$ is independent of the choice of $pi'$, but $l_2, ...$ are clearly not (take $pi' := p^nu + pi$ as a different choice and see how $l_nu$ of that depends on whether you write it as series in $pi$ or in $pi'$).
– Torsten Schoeneberg
Sep 21 at 16:55
Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{I}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i*p*pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 9:32
@TorstenSchoeneberg: Many thanks. $pi$ is really the "lift of a uniformizer of the residue field". But why is the statement plausible? With $pi^j:=p+pi$ it is $sum_{i}a_i(pi+p)^i$ and with binomial formula $sum_{i}a_i(pi^i+i∗p∗pi^{i−1}+...+p^i)$. In this case only coefficients in the $sum_{i}a_ipi^i$ can be independent from p and it is the same as series in $pi$?
– lartom
Sep 22 at 12:01