Proving $ord_p(ζ_p-1)=1/(p-1)$












1














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







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  • 3




    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?
    – Jyrki Lahtonen
    Dec 7 at 22:19








  • 1




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










  • 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 $1/(p-1)$.
    – user45765
    Dec 7 at 22:32










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
















1














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


















  • 3




    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?
    – Jyrki Lahtonen
    Dec 7 at 22:19








  • 1




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










  • 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 $1/(p-1)$.
    – user45765
    Dec 7 at 22:32










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














1












1








1


2





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













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 at 22:13









Pascal's Wager

325115




325115








  • 3




    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?
    – Jyrki Lahtonen
    Dec 7 at 22:19








  • 1




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










  • 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 $1/(p-1)$.
    – user45765
    Dec 7 at 22:32










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














  • 3




    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?
    – Jyrki Lahtonen
    Dec 7 at 22:19








  • 1




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










  • 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 $1/(p-1)$.
    – user45765
    Dec 7 at 22:32










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








3




3




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?
– Jyrki Lahtonen
Dec 7 at 22:19






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?
– Jyrki Lahtonen
Dec 7 at 22:19






1




1




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




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












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 $1/(p-1)$.
– user45765
Dec 7 at 22:32




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 $1/(p-1)$.
– user45765
Dec 7 at 22:32












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




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










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



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



    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


























      0














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



      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
























        0












        0








        0






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



        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












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



        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 at 22:46









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