$(varphi,Gamma)$-modules and valuation in Zp
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
asked Sep 20 '18 at 9:24
lartom
362
|
show 1 more comment
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
asked Sep 20 '18 at 9:24
lartom
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 '18 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 '18 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 '18 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 '18 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 '18 at 12:01
|
show 1 more comment
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
asked Sep 20 '18 at 9:24
lartom
362
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
asked Sep 20 '18 at 9:24
lartom
362
asked Sep 20 '18 at 9:24
lartom
362
edited Nov 25 '18 at 22:17
asked Sep 20 '18 at 9:24
lartom
362
asked Sep 20 '18 at 9:24
lartom
362
asked Sep 20 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 12:01
|
show 1 more comment
1 Answer
1
active
oldest
votes
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 $varpi$.
answered Sep 22 '18 at 20:44
Torsten Schoeneberg
3,7412833
I think you mean $limlimits_{nrightarrow-infty}a_n=0$ in the description of $A$ and in the claim?
– awllower
Sep 23 '18 at 5:54
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 '18 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 '18 at 14:51
add a comment |
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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 $varpi$.
answered Sep 22 '18 at 20:44
Torsten Schoeneberg
3,7412833
I think you mean $limlimits_{nrightarrow-infty}a_n=0$ in the description of $A$ and in the claim?
– awllower
Sep 23 '18 at 5:54
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 '18 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 '18 at 14:51
add a comment |
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 $varpi$.
answered Sep 22 '18 at 20:44
Torsten Schoeneberg
3,7412833
I think you mean $limlimits_{nrightarrow-infty}a_n=0$ in the description of $A$ and in the claim?
– awllower
Sep 23 '18 at 5:54
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 '18 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 '18 at 14:51
add a comment |
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 $varpi$.
answered Sep 22 '18 at 20:44
Torsten Schoeneberg
3,7412833
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 $varpi$.
answered Sep 22 '18 at 20:44
Torsten Schoeneberg
3,7412833
edited Sep 23 '18 at 6:01
answered Sep 22 '18 at 20:44
Torsten Schoeneberg
3,7412833
answered Sep 22 '18 at 20:44
Torsten Schoeneberg
3,7412833
answered Sep 22 '18 at 20:44
Torsten Schoeneberg
3,7412833
3,7412833
I think you mean $limlimits_{nrightarrow-infty}a_n=0$ in the description of $A$ and in the claim?
– awllower
Sep 23 '18 at 5:54
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 '18 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 '18 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 '18 at 5:54
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 '18 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 '18 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 '18 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 '18 at 5:54
1
1
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 '18 at 5:57
@awllower Yes, thank you; I will edit that.
– Torsten Schoeneberg
Sep 23 '18 at 5:57
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 '18 at 14:51
@TorstenSchoeneberg: Many, many thanks. I think I understand it.
– lartom
Sep 23 '18 at 14:51
add a comment |
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– José Carlos Santos
Sep 20 '18 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 '18 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 '18 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 '18 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 '18 at 12:01