$operatorname{ord}_p(sum_{i=0}^{p-1}T(overline i)^{p-1-k}psi(overline i))=?$
$begingroup$
Let $Z_p$ denote the p-adic integers. Let $T:mathbb{F_p}to Z_p$ be a function with the following properties:
- $forall x in mathbb{F}_p[overline {T(x)}=x]$
- $forall x in mathbb{F}_p[T(x)^p=T(x)]$
Let $psi: mathbb{F}_p to Z_p(zeta_p)$ have the following properties:
$forall x,y in mathbb{F}_p[psi(x+y)=psi(x)psi(y)]$
$exists x in mathbb{F}_p[psi(x) neq 1]$
Moreover, assume $1 leq k leq p-2$.
By the way, the overline notation indicates reduction mod p.
I want to find the p-adic order of $sum_{i=0}^{p-1}T(overline i)^{p-1-k}psi(overline i)$
I have been able to find for the special case $p=3$ and $k=1$ using an ad-hoc method. In the special case, I use the fact that $T(overline 2)=-1$ and the sum becomes $psi(1)-psi(2)=psi(1)[1-psi(1)],$ which has p-order $frac 1 {p-1}=frac 1 2$ by this theorem. But I don't see how to solve the general case. My professor thinks that the order in the general case should be $frac k {p-1}$ but, alas, I don't see why.
abstract-algebra number-theory prime-numbers finite-fields p-adic-number-theory
edited Dec 19 '18 at 5:35
Kemono Chen
3,0871743
asked Dec 10 '18 at 23:04
Pascal's WagerPascal's Wager
362315
$endgroup$
add a comment |
$begingroup$
Let $Z_p$ denote the p-adic integers. Let $T:mathbb{F_p}to Z_p$ be a function with the following properties:
- $forall x in mathbb{F}_p[overline {T(x)}=x]$
- $forall x in mathbb{F}_p[T(x)^p=T(x)]$
Let $psi: mathbb{F}_p to Z_p(zeta_p)$ have the following properties:
$forall x,y in mathbb{F}_p[psi(x+y)=psi(x)psi(y)]$
$exists x in mathbb{F}_p[psi(x) neq 1]$
Moreover, assume $1 leq k leq p-2$.
By the way, the overline notation indicates reduction mod p.
I want to find the p-adic order of $sum_{i=0}^{p-1}T(overline i)^{p-1-k}psi(overline i)$
I have been able to find for the special case $p=3$ and $k=1$ using an ad-hoc method. In the special case, I use the fact that $T(overline 2)=-1$ and the sum becomes $psi(1)-psi(2)=psi(1)[1-psi(1)],$ which has p-order $frac 1 {p-1}=frac 1 2$ by this theorem. But I don't see how to solve the general case. My professor thinks that the order in the general case should be $frac k {p-1}$ but, alas, I don't see why.
abstract-algebra number-theory prime-numbers finite-fields p-adic-number-theory
edited Dec 19 '18 at 5:35
Kemono Chen
3,0871743
asked Dec 10 '18 at 23:04
Pascal's WagerPascal's Wager
362315
$endgroup$
1
$begingroup$
Let $f$ a function $mathbb{Z}/pmathbb{Z} to K$ such that $f(lm) = f(l)f(m)$ and $a in K$ then $G_a(k) = sum_{l=1}^{p} f(l) a^{lk} = f(k^{-1})sum_{l=1}^{p} f(lk) a^{lk}=f(k^{-1}) G_a(1)$. If $a^p = 1, a ne 1$ then $sum_{k=1}^p a^{-mk} G_a(k) =sum_{l=1}^{p} f(l) sum_{k=1}^pa^{(l-m)k} = p f(m)$ Whence $p = p f(1)=sum_{k=1}^p a^{-k} G_a(k)=sum_{k=1}^p a^{-k} f(k^{-1}) G_a(1) = G_a(1)^*G_a(1) $ where $*$ is the map sending a root of unity $zeta$ to $zeta^{-1}$. If $K = mathbb{Q}_p(zeta_p),a = zeta_p$ then $|G_a(1)^*|_p = |G_a(1)|_p =|p^{1/2}|_p = p^{-1/2}$.
$endgroup$
– reuns
Dec 11 '18 at 1:24
$begingroup$
@reuns: Why not post that as an answer?
$endgroup$
– Torsten Schoeneberg
Dec 12 '18 at 18:48
add a comment |
$begingroup$
Let $Z_p$ denote the p-adic integers. Let $T:mathbb{F_p}to Z_p$ be a function with the following properties:
- $forall x in mathbb{F}_p[overline {T(x)}=x]$
- $forall x in mathbb{F}_p[T(x)^p=T(x)]$
Let $psi: mathbb{F}_p to Z_p(zeta_p)$ have the following properties:
$forall x,y in mathbb{F}_p[psi(x+y)=psi(x)psi(y)]$
$exists x in mathbb{F}_p[psi(x) neq 1]$
Moreover, assume $1 leq k leq p-2$.
By the way, the overline notation indicates reduction mod p.
I want to find the p-adic order of $sum_{i=0}^{p-1}T(overline i)^{p-1-k}psi(overline i)$
I have been able to find for the special case $p=3$ and $k=1$ using an ad-hoc method. In the special case, I use the fact that $T(overline 2)=-1$ and the sum becomes $psi(1)-psi(2)=psi(1)[1-psi(1)],$ which has p-order $frac 1 {p-1}=frac 1 2$ by this theorem. But I don't see how to solve the general case. My professor thinks that the order in the general case should be $frac k {p-1}$ but, alas, I don't see why.
abstract-algebra number-theory prime-numbers finite-fields p-adic-number-theory
edited Dec 19 '18 at 5:35
Kemono Chen
3,0871743
asked Dec 10 '18 at 23:04
Pascal's WagerPascal's Wager
362315
$endgroup$
Let $Z_p$ denote the p-adic integers. Let $T:mathbb{F_p}to Z_p$ be a function with the following properties:
- $forall x in mathbb{F}_p[overline {T(x)}=x]$
- $forall x in mathbb{F}_p[T(x)^p=T(x)]$
Let $psi: mathbb{F}_p to Z_p(zeta_p)$ have the following properties:
$forall x,y in mathbb{F}_p[psi(x+y)=psi(x)psi(y)]$
$exists x in mathbb{F}_p[psi(x) neq 1]$
Moreover, assume $1 leq k leq p-2$.
By the way, the overline notation indicates reduction mod p.
I want to find the p-adic order of $sum_{i=0}^{p-1}T(overline i)^{p-1-k}psi(overline i)$
I have been able to find for the special case $p=3$ and $k=1$ using an ad-hoc method. In the special case, I use the fact that $T(overline 2)=-1$ and the sum becomes $psi(1)-psi(2)=psi(1)[1-psi(1)],$ which has p-order $frac 1 {p-1}=frac 1 2$ by this theorem. But I don't see how to solve the general case. My professor thinks that the order in the general case should be $frac k {p-1}$ but, alas, I don't see why.
abstract-algebra number-theory prime-numbers finite-fields p-adic-number-theory
abstract-algebra number-theory prime-numbers finite-fields p-adic-number-theory
edited Dec 19 '18 at 5:35
Kemono Chen
3,0871743
asked Dec 10 '18 at 23:04
Pascal's WagerPascal's Wager
362315
edited Dec 19 '18 at 5:35
Kemono Chen
3,0871743
asked Dec 10 '18 at 23:04
Pascal's WagerPascal's Wager
362315
edited Dec 19 '18 at 5:35
Kemono Chen
3,0871743
edited Dec 19 '18 at 5:35
Kemono Chen
3,0871743
edited Dec 19 '18 at 5:35
Kemono Chen
3,0871743
3,0871743
asked Dec 10 '18 at 23:04
Pascal's WagerPascal's Wager
362315
asked Dec 10 '18 at 23:04
Pascal's WagerPascal's Wager
362315
asked Dec 10 '18 at 23:04
Pascal's WagerPascal's Wager
362315
362315
1
$begingroup$
Let $f$ a function $mathbb{Z}/pmathbb{Z} to K$ such that $f(lm) = f(l)f(m)$ and $a in K$ then $G_a(k) = sum_{l=1}^{p} f(l) a^{lk} = f(k^{-1})sum_{l=1}^{p} f(lk) a^{lk}=f(k^{-1}) G_a(1)$. If $a^p = 1, a ne 1$ then $sum_{k=1}^p a^{-mk} G_a(k) =sum_{l=1}^{p} f(l) sum_{k=1}^pa^{(l-m)k} = p f(m)$ Whence $p = p f(1)=sum_{k=1}^p a^{-k} G_a(k)=sum_{k=1}^p a^{-k} f(k^{-1}) G_a(1) = G_a(1)^*G_a(1) $ where $*$ is the map sending a root of unity $zeta$ to $zeta^{-1}$. If $K = mathbb{Q}_p(zeta_p),a = zeta_p$ then $|G_a(1)^*|_p = |G_a(1)|_p =|p^{1/2}|_p = p^{-1/2}$.
$endgroup$
– reuns
Dec 11 '18 at 1:24
$begingroup$
@reuns: Why not post that as an answer?
$endgroup$
– Torsten Schoeneberg
Dec 12 '18 at 18:48
add a comment |
1
$begingroup$
Let $f$ a function $mathbb{Z}/pmathbb{Z} to K$ such that $f(lm) = f(l)f(m)$ and $a in K$ then $G_a(k) = sum_{l=1}^{p} f(l) a^{lk} = f(k^{-1})sum_{l=1}^{p} f(lk) a^{lk}=f(k^{-1}) G_a(1)$. If $a^p = 1, a ne 1$ then $sum_{k=1}^p a^{-mk} G_a(k) =sum_{l=1}^{p} f(l) sum_{k=1}^pa^{(l-m)k} = p f(m)$ Whence $p = p f(1)=sum_{k=1}^p a^{-k} G_a(k)=sum_{k=1}^p a^{-k} f(k^{-1}) G_a(1) = G_a(1)^*G_a(1) $ where $*$ is the map sending a root of unity $zeta$ to $zeta^{-1}$. If $K = mathbb{Q}_p(zeta_p),a = zeta_p$ then $|G_a(1)^*|_p = |G_a(1)|_p =|p^{1/2}|_p = p^{-1/2}$.
$endgroup$
– reuns
Dec 11 '18 at 1:24
$begingroup$
@reuns: Why not post that as an answer?
$endgroup$
– Torsten Schoeneberg
Dec 12 '18 at 18:48
1
1
$begingroup$
Let $f$ a function $mathbb{Z}/pmathbb{Z} to K$ such that $f(lm) = f(l)f(m)$ and $a in K$ then $G_a(k) = sum_{l=1}^{p} f(l) a^{lk} = f(k^{-1})sum_{l=1}^{p} f(lk) a^{lk}=f(k^{-1}) G_a(1)$. If $a^p = 1, a ne 1$ then $sum_{k=1}^p a^{-mk} G_a(k) =sum_{l=1}^{p} f(l) sum_{k=1}^pa^{(l-m)k} = p f(m)$ Whence $p = p f(1)=sum_{k=1}^p a^{-k} G_a(k)=sum_{k=1}^p a^{-k} f(k^{-1}) G_a(1) = G_a(1)^*G_a(1) $ where $*$ is the map sending a root of unity $zeta$ to $zeta^{-1}$. If $K = mathbb{Q}_p(zeta_p),a = zeta_p$ then $|G_a(1)^*|_p = |G_a(1)|_p =|p^{1/2}|_p = p^{-1/2}$.
$endgroup$
– reuns
Dec 11 '18 at 1:24
$begingroup$
Let $f$ a function $mathbb{Z}/pmathbb{Z} to K$ such that $f(lm) = f(l)f(m)$ and $a in K$ then $G_a(k) = sum_{l=1}^{p} f(l) a^{lk} = f(k^{-1})sum_{l=1}^{p} f(lk) a^{lk}=f(k^{-1}) G_a(1)$. If $a^p = 1, a ne 1$ then $sum_{k=1}^p a^{-mk} G_a(k) =sum_{l=1}^{p} f(l) sum_{k=1}^pa^{(l-m)k} = p f(m)$ Whence $p = p f(1)=sum_{k=1}^p a^{-k} G_a(k)=sum_{k=1}^p a^{-k} f(k^{-1}) G_a(1) = G_a(1)^*G_a(1) $ where $*$ is the map sending a root of unity $zeta$ to $zeta^{-1}$. If $K = mathbb{Q}_p(zeta_p),a = zeta_p$ then $|G_a(1)^*|_p = |G_a(1)|_p =|p^{1/2}|_p = p^{-1/2}$.
$endgroup$
– reuns
Dec 11 '18 at 1:24
$begingroup$
@reuns: Why not post that as an answer?
$endgroup$
– Torsten Schoeneberg
Dec 12 '18 at 18:48
$begingroup$
@reuns: Why not post that as an answer?
$endgroup$
– Torsten Schoeneberg
Dec 12 '18 at 18:48
add a comment |
0
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$begingroup$
Let $f$ a function $mathbb{Z}/pmathbb{Z} to K$ such that $f(lm) = f(l)f(m)$ and $a in K$ then $G_a(k) = sum_{l=1}^{p} f(l) a^{lk} = f(k^{-1})sum_{l=1}^{p} f(lk) a^{lk}=f(k^{-1}) G_a(1)$. If $a^p = 1, a ne 1$ then $sum_{k=1}^p a^{-mk} G_a(k) =sum_{l=1}^{p} f(l) sum_{k=1}^pa^{(l-m)k} = p f(m)$ Whence $p = p f(1)=sum_{k=1}^p a^{-k} G_a(k)=sum_{k=1}^p a^{-k} f(k^{-1}) G_a(1) = G_a(1)^*G_a(1) $ where $*$ is the map sending a root of unity $zeta$ to $zeta^{-1}$. If $K = mathbb{Q}_p(zeta_p),a = zeta_p$ then $|G_a(1)^*|_p = |G_a(1)|_p =|p^{1/2}|_p = p^{-1/2}$.
$endgroup$
– reuns
Dec 11 '18 at 1:24
$begingroup$
@reuns: Why not post that as an answer?
$endgroup$
– Torsten Schoeneberg
Dec 12 '18 at 18:48