Proving $ord_p(ζ_p-1)=1/(p-1)$












1












$begingroup$


After proving this, I was able to deduce an even more general result that $ord_p(ζ_p-1)=ord_p(ζ_p^2-1)=...=ord_p(ζ_p^{p-1}-1)$. Now, according to Lubin, $ord_p(ζ_p-1)$ should be $1/(p-1)$, but this isn't obvious to me so I need to prove it.



My thought was to multiply out the product $(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)$ and hope that I get something with p-order $1$ (since this would finish the proof), but I'm stuck.



I know that $ord_p(ζ_p^k)=0$ for any $k geq 0$. So if we multiply out the product, we get a sum with all of its terms having p-order $0$. I see from this that $ord_p[(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)] geq 0$.






abstract-algebra prime-numbers extension-field p-adic-number-theory roots-of-unity







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
    $endgroup$
    – Jyrki Lahtonen
    Dec 7 '18 at 22:19








  • 1




    $begingroup$
    And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
    $endgroup$
    – Jyrki Lahtonen
    Dec 7 '18 at 22:20










  • $begingroup$
    This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
    $endgroup$
    – user45765
    Dec 7 '18 at 22:32










  • $begingroup$
    @JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
    $endgroup$
    – Pascal's Wager
    Dec 7 '18 at 22:52
















1












$begingroup$


After proving this, I was able to deduce an even more general result that $ord_p(ζ_p-1)=ord_p(ζ_p^2-1)=...=ord_p(ζ_p^{p-1}-1)$. Now, according to Lubin, $ord_p(ζ_p-1)$ should be $1/(p-1)$, but this isn't obvious to me so I need to prove it.



My thought was to multiply out the product $(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)$ and hope that I get something with p-order $1$ (since this would finish the proof), but I'm stuck.



I know that $ord_p(ζ_p^k)=0$ for any $k geq 0$. So if we multiply out the product, we get a sum with all of its terms having p-order $0$. I see from this that $ord_p[(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)] geq 0$.






abstract-algebra prime-numbers extension-field p-adic-number-theory roots-of-unity







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
    $endgroup$
    – Jyrki Lahtonen
    Dec 7 '18 at 22:19








  • 1




    $begingroup$
    And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
    $endgroup$
    – Jyrki Lahtonen
    Dec 7 '18 at 22:20










  • $begingroup$
    This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
    $endgroup$
    – user45765
    Dec 7 '18 at 22:32










  • $begingroup$
    @JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
    $endgroup$
    – Pascal's Wager
    Dec 7 '18 at 22:52














1












1








1


2



$begingroup$


After proving this, I was able to deduce an even more general result that $ord_p(ζ_p-1)=ord_p(ζ_p^2-1)=...=ord_p(ζ_p^{p-1}-1)$. Now, according to Lubin, $ord_p(ζ_p-1)$ should be $1/(p-1)$, but this isn't obvious to me so I need to prove it.



My thought was to multiply out the product $(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)$ and hope that I get something with p-order $1$ (since this would finish the proof), but I'm stuck.



I know that $ord_p(ζ_p^k)=0$ for any $k geq 0$. So if we multiply out the product, we get a sum with all of its terms having p-order $0$. I see from this that $ord_p[(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)] geq 0$.






abstract-algebra prime-numbers extension-field p-adic-number-theory roots-of-unity







share|cite|improve this question









$endgroup$




After proving this, I was able to deduce an even more general result that $ord_p(ζ_p-1)=ord_p(ζ_p^2-1)=...=ord_p(ζ_p^{p-1}-1)$. Now, according to Lubin, $ord_p(ζ_p-1)$ should be $1/(p-1)$, but this isn't obvious to me so I need to prove it.



My thought was to multiply out the product $(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)$ and hope that I get something with p-order $1$ (since this would finish the proof), but I'm stuck.



I know that $ord_p(ζ_p^k)=0$ for any $k geq 0$. So if we multiply out the product, we get a sum with all of its terms having p-order $0$. I see from this that $ord_p[(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)] geq 0$.






abstract-algebra prime-numbers extension-field p-adic-number-theory roots-of-unity




abstract-algebra prime-numbers extension-field p-adic-number-theory roots-of-unity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 22:13









Pascal's WagerPascal's Wager

371315




371315








  • 3




    $begingroup$
    You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
    $endgroup$
    – Jyrki Lahtonen
    Dec 7 '18 at 22:19








  • 1




    $begingroup$
    And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
    $endgroup$
    – Jyrki Lahtonen
    Dec 7 '18 at 22:20










  • $begingroup$
    This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
    $endgroup$
    – user45765
    Dec 7 '18 at 22:32










  • $begingroup$
    @JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
    $endgroup$
    – Pascal's Wager
    Dec 7 '18 at 22:52














  • 3




    $begingroup$
    You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
    $endgroup$
    – Jyrki Lahtonen
    Dec 7 '18 at 22:19








  • 1




    $begingroup$
    And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
    $endgroup$
    – Jyrki Lahtonen
    Dec 7 '18 at 22:20










  • $begingroup$
    This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
    $endgroup$
    – user45765
    Dec 7 '18 at 22:32










  • $begingroup$
    @JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
    $endgroup$
    – Pascal's Wager
    Dec 7 '18 at 22:52








3




3




$begingroup$
You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:19






$begingroup$
You have the cyclotomic polynomial $$Phi_p(x)=x^{p-1}+x^{p-2}+cdots+x+1=prod_{k=1}^{p-1}(x-zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation?
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:19






1




1




$begingroup$
And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:20




$begingroup$
And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 22:20












$begingroup$
This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
$endgroup$
– user45765
Dec 7 '18 at 22:32




$begingroup$
This is nothing but says $p$ is totally ramified at $(1-eta_{p^r})$ with ramification index $phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$.
$endgroup$
– user45765
Dec 7 '18 at 22:32












$begingroup$
@JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
$endgroup$
– Pascal's Wager
Dec 7 '18 at 22:52




$begingroup$
@JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question.
$endgroup$
– Pascal's Wager
Dec 7 '18 at 22:52










1 Answer
1






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0












$begingroup$

$|a|_p = p^{-v_p(a)}$



Since $zeta_p^p = 1$ then $|zeta_p|_p=1$



$sum_{m=0}^{p-1} x^m =frac{x^p-1}{x-1}=prod_{k=1}^{p-1} (x-zeta_p^k)$



$prod_{k=1}^{p-1} |1-zeta_p^k| = |sum_{m=0}^{p-1} 1^m|_p = p^{-1}$ so there is some $k$ such that $|1-zeta_p^k|_p <1$.



$1-zeta_p^{kn} =1- (1+ (1-zeta_p^k))^n =1-(1+n(1-zeta_p^k)+O((1-zeta_p^k)^2)$



whence (for $p nmid n$)



$|1-zeta_p^{kn}|_p = |n|_p |1-zeta_p^{k}|_p=|1-zeta_p^{k}|_p$ and $|1-zeta_p^{k}|_p^{p-1} = p^{-1}$



letting $kn equiv 1bmod p$ you get your result $|1-zeta_p|_p =p^{-1/(p-1)}$



Note this is also a proof that $sum_{m=0}^{p-1} x^m$ is irreducible since we need to multiply all the $1-zeta_p^k, k in 1 ldots p-1$ to obtain something of integer valuation






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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    $|a|_p = p^{-v_p(a)}$



    Since $zeta_p^p = 1$ then $|zeta_p|_p=1$



    $sum_{m=0}^{p-1} x^m =frac{x^p-1}{x-1}=prod_{k=1}^{p-1} (x-zeta_p^k)$



    $prod_{k=1}^{p-1} |1-zeta_p^k| = |sum_{m=0}^{p-1} 1^m|_p = p^{-1}$ so there is some $k$ such that $|1-zeta_p^k|_p <1$.



    $1-zeta_p^{kn} =1- (1+ (1-zeta_p^k))^n =1-(1+n(1-zeta_p^k)+O((1-zeta_p^k)^2)$



    whence (for $p nmid n$)



    $|1-zeta_p^{kn}|_p = |n|_p |1-zeta_p^{k}|_p=|1-zeta_p^{k}|_p$ and $|1-zeta_p^{k}|_p^{p-1} = p^{-1}$



    letting $kn equiv 1bmod p$ you get your result $|1-zeta_p|_p =p^{-1/(p-1)}$



    Note this is also a proof that $sum_{m=0}^{p-1} x^m$ is irreducible since we need to multiply all the $1-zeta_p^k, k in 1 ldots p-1$ to obtain something of integer valuation






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $|a|_p = p^{-v_p(a)}$



      Since $zeta_p^p = 1$ then $|zeta_p|_p=1$



      $sum_{m=0}^{p-1} x^m =frac{x^p-1}{x-1}=prod_{k=1}^{p-1} (x-zeta_p^k)$



      $prod_{k=1}^{p-1} |1-zeta_p^k| = |sum_{m=0}^{p-1} 1^m|_p = p^{-1}$ so there is some $k$ such that $|1-zeta_p^k|_p <1$.



      $1-zeta_p^{kn} =1- (1+ (1-zeta_p^k))^n =1-(1+n(1-zeta_p^k)+O((1-zeta_p^k)^2)$



      whence (for $p nmid n$)



      $|1-zeta_p^{kn}|_p = |n|_p |1-zeta_p^{k}|_p=|1-zeta_p^{k}|_p$ and $|1-zeta_p^{k}|_p^{p-1} = p^{-1}$



      letting $kn equiv 1bmod p$ you get your result $|1-zeta_p|_p =p^{-1/(p-1)}$



      Note this is also a proof that $sum_{m=0}^{p-1} x^m$ is irreducible since we need to multiply all the $1-zeta_p^k, k in 1 ldots p-1$ to obtain something of integer valuation






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $|a|_p = p^{-v_p(a)}$



        Since $zeta_p^p = 1$ then $|zeta_p|_p=1$



        $sum_{m=0}^{p-1} x^m =frac{x^p-1}{x-1}=prod_{k=1}^{p-1} (x-zeta_p^k)$



        $prod_{k=1}^{p-1} |1-zeta_p^k| = |sum_{m=0}^{p-1} 1^m|_p = p^{-1}$ so there is some $k$ such that $|1-zeta_p^k|_p <1$.



        $1-zeta_p^{kn} =1- (1+ (1-zeta_p^k))^n =1-(1+n(1-zeta_p^k)+O((1-zeta_p^k)^2)$



        whence (for $p nmid n$)



        $|1-zeta_p^{kn}|_p = |n|_p |1-zeta_p^{k}|_p=|1-zeta_p^{k}|_p$ and $|1-zeta_p^{k}|_p^{p-1} = p^{-1}$



        letting $kn equiv 1bmod p$ you get your result $|1-zeta_p|_p =p^{-1/(p-1)}$



        Note this is also a proof that $sum_{m=0}^{p-1} x^m$ is irreducible since we need to multiply all the $1-zeta_p^k, k in 1 ldots p-1$ to obtain something of integer valuation






        share|cite|improve this answer









        $endgroup$



        $|a|_p = p^{-v_p(a)}$



        Since $zeta_p^p = 1$ then $|zeta_p|_p=1$



        $sum_{m=0}^{p-1} x^m =frac{x^p-1}{x-1}=prod_{k=1}^{p-1} (x-zeta_p^k)$



        $prod_{k=1}^{p-1} |1-zeta_p^k| = |sum_{m=0}^{p-1} 1^m|_p = p^{-1}$ so there is some $k$ such that $|1-zeta_p^k|_p <1$.



        $1-zeta_p^{kn} =1- (1+ (1-zeta_p^k))^n =1-(1+n(1-zeta_p^k)+O((1-zeta_p^k)^2)$



        whence (for $p nmid n$)



        $|1-zeta_p^{kn}|_p = |n|_p |1-zeta_p^{k}|_p=|1-zeta_p^{k}|_p$ and $|1-zeta_p^{k}|_p^{p-1} = p^{-1}$



        letting $kn equiv 1bmod p$ you get your result $|1-zeta_p|_p =p^{-1/(p-1)}$



        Note this is also a proof that $sum_{m=0}^{p-1} x^m$ is irreducible since we need to multiply all the $1-zeta_p^k, k in 1 ldots p-1$ to obtain something of integer valuation







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 22:46









        reunsreuns

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