$operatorname{ord}_p(sum_{i=0}^{p-1}T(overline i)^{p-1-k}psi(overline i))=?$












1












$begingroup$


Let $Z_p$ denote the p-adic integers. Let $T:mathbb{F_p}to Z_p$ be a function with the following properties:




  1. $forall x in mathbb{F}_p[overline {T(x)}=x]$

  2. $forall x in mathbb{F}_p[T(x)^p=T(x)]$


Let $psi: mathbb{F}_p to Z_p(zeta_p)$ have the following properties:




  1. $forall x,y in mathbb{F}_p[psi(x+y)=psi(x)psi(y)]$


  2. $exists x in mathbb{F}_p[psi(x) neq 1]$



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







share|cite|improve this question











$endgroup$








  • 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
















1












$begingroup$


Let $Z_p$ denote the p-adic integers. Let $T:mathbb{F_p}to Z_p$ be a function with the following properties:




  1. $forall x in mathbb{F}_p[overline {T(x)}=x]$

  2. $forall x in mathbb{F}_p[T(x)^p=T(x)]$


Let $psi: mathbb{F}_p to Z_p(zeta_p)$ have the following properties:




  1. $forall x,y in mathbb{F}_p[psi(x+y)=psi(x)psi(y)]$


  2. $exists x in mathbb{F}_p[psi(x) neq 1]$



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







share|cite|improve this question











$endgroup$








  • 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














1












1








1





$begingroup$


Let $Z_p$ denote the p-adic integers. Let $T:mathbb{F_p}to Z_p$ be a function with the following properties:




  1. $forall x in mathbb{F}_p[overline {T(x)}=x]$

  2. $forall x in mathbb{F}_p[T(x)^p=T(x)]$


Let $psi: mathbb{F}_p to Z_p(zeta_p)$ have the following properties:




  1. $forall x,y in mathbb{F}_p[psi(x+y)=psi(x)psi(y)]$


  2. $exists x in mathbb{F}_p[psi(x) neq 1]$



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







share|cite|improve this question











$endgroup$




Let $Z_p$ denote the p-adic integers. Let $T:mathbb{F_p}to Z_p$ be a function with the following properties:




  1. $forall x in mathbb{F}_p[overline {T(x)}=x]$

  2. $forall x in mathbb{F}_p[T(x)^p=T(x)]$


Let $psi: mathbb{F}_p to Z_p(zeta_p)$ have the following properties:




  1. $forall x,y in mathbb{F}_p[psi(x+y)=psi(x)psi(y)]$


  2. $exists x in mathbb{F}_p[psi(x) neq 1]$



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 5:35









Kemono Chen

3,0521743




3,0521743










asked Dec 10 '18 at 23:04









Pascal's WagerPascal's Wager

357315




357315








  • 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




    $begingroup$
    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










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