Showing the lemma $operatorname{ord}_p(1+ζ_p)=0$ if $p>2$












2












$begingroup$


Just to give some background regarding my motivation, I'm trying to prove a lemma to help me solve How do we prove p-order of $g_k$ is $frac {k} {p-1}$?



Let $Z_p$ denote the p-adic integers, and let us adjoin a primitive p-th root of unity $ζ_p$. Assume $p>2DeclareMathOperator{ord}{ord}$.



I want to show that $ord_p(1-ζ_p)=ord_p(1-ζ_p^2)$, but I see that it is enough to prove $ord_p( frac {1-ζ_p^2} {1-ζ_p})=0$ or, equivalently, $ord_p(1+ζ_p)=0$.



I have tried applying the properties from here. (Even though they are for rational numbers, I'm going to assume for now that they hold in $Z_p(ζ_p)$ also. If not, please correct me.)



I see that, if $ord_p(ζ_p) neq ord_p(1)=0$, we will have $ord_p(1+ζ_p)=inf {ord_p(1), ord_p(ζ_p)} leq 0$ and we will get a contradiction if $<0$
.



But if instead $ord_p(ζ_p)=0,$ then I don't know what to do. We will have $ord_p(1+ζ_p) geq inf {ord_p(1), ord_p(ζ_p)}=0$, but I don't see how $>0$ would give us a contradiction.






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







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$endgroup$








  • 1




    $begingroup$
    That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
    $endgroup$
    – reuns
    Dec 7 '18 at 20:15
















2












$begingroup$


Just to give some background regarding my motivation, I'm trying to prove a lemma to help me solve How do we prove p-order of $g_k$ is $frac {k} {p-1}$?



Let $Z_p$ denote the p-adic integers, and let us adjoin a primitive p-th root of unity $ζ_p$. Assume $p>2DeclareMathOperator{ord}{ord}$.



I want to show that $ord_p(1-ζ_p)=ord_p(1-ζ_p^2)$, but I see that it is enough to prove $ord_p( frac {1-ζ_p^2} {1-ζ_p})=0$ or, equivalently, $ord_p(1+ζ_p)=0$.



I have tried applying the properties from here. (Even though they are for rational numbers, I'm going to assume for now that they hold in $Z_p(ζ_p)$ also. If not, please correct me.)



I see that, if $ord_p(ζ_p) neq ord_p(1)=0$, we will have $ord_p(1+ζ_p)=inf {ord_p(1), ord_p(ζ_p)} leq 0$ and we will get a contradiction if $<0$
.



But if instead $ord_p(ζ_p)=0,$ then I don't know what to do. We will have $ord_p(1+ζ_p) geq inf {ord_p(1), ord_p(ζ_p)}=0$, but I don't see how $>0$ would give us a contradiction.






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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
    $endgroup$
    – reuns
    Dec 7 '18 at 20:15














2












2








2


1



$begingroup$


Just to give some background regarding my motivation, I'm trying to prove a lemma to help me solve How do we prove p-order of $g_k$ is $frac {k} {p-1}$?



Let $Z_p$ denote the p-adic integers, and let us adjoin a primitive p-th root of unity $ζ_p$. Assume $p>2DeclareMathOperator{ord}{ord}$.



I want to show that $ord_p(1-ζ_p)=ord_p(1-ζ_p^2)$, but I see that it is enough to prove $ord_p( frac {1-ζ_p^2} {1-ζ_p})=0$ or, equivalently, $ord_p(1+ζ_p)=0$.



I have tried applying the properties from here. (Even though they are for rational numbers, I'm going to assume for now that they hold in $Z_p(ζ_p)$ also. If not, please correct me.)



I see that, if $ord_p(ζ_p) neq ord_p(1)=0$, we will have $ord_p(1+ζ_p)=inf {ord_p(1), ord_p(ζ_p)} leq 0$ and we will get a contradiction if $<0$
.



But if instead $ord_p(ζ_p)=0,$ then I don't know what to do. We will have $ord_p(1+ζ_p) geq inf {ord_p(1), ord_p(ζ_p)}=0$, but I don't see how $>0$ would give us a contradiction.






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







share|cite|improve this question











$endgroup$




Just to give some background regarding my motivation, I'm trying to prove a lemma to help me solve How do we prove p-order of $g_k$ is $frac {k} {p-1}$?



Let $Z_p$ denote the p-adic integers, and let us adjoin a primitive p-th root of unity $ζ_p$. Assume $p>2DeclareMathOperator{ord}{ord}$.



I want to show that $ord_p(1-ζ_p)=ord_p(1-ζ_p^2)$, but I see that it is enough to prove $ord_p( frac {1-ζ_p^2} {1-ζ_p})=0$ or, equivalently, $ord_p(1+ζ_p)=0$.



I have tried applying the properties from here. (Even though they are for rational numbers, I'm going to assume for now that they hold in $Z_p(ζ_p)$ also. If not, please correct me.)



I see that, if $ord_p(ζ_p) neq ord_p(1)=0$, we will have $ord_p(1+ζ_p)=inf {ord_p(1), ord_p(ζ_p)} leq 0$ and we will get a contradiction if $<0$
.



But if instead $ord_p(ζ_p)=0,$ then I don't know what to do. We will have $ord_p(1+ζ_p) geq inf {ord_p(1), ord_p(ζ_p)}=0$, but I don't see how $>0$ would give us a contradiction.






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




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






share|cite|improve this question















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edited Dec 7 '18 at 19:41









Bernard

119k740113




119k740113










asked Dec 7 '18 at 19:28









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357315








  • 1




    $begingroup$
    That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
    $endgroup$
    – reuns
    Dec 7 '18 at 20:15














  • 1




    $begingroup$
    That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
    $endgroup$
    – reuns
    Dec 7 '18 at 20:15








1




1




$begingroup$
That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
$endgroup$
– reuns
Dec 7 '18 at 20:15




$begingroup$
That $|zeta_p|_p = 1$ is obvious from $zeta_p^p = 1$
$endgroup$
– reuns
Dec 7 '18 at 20:15










3 Answers
3






active

oldest

votes


















0












$begingroup$

Let $$(x-1)^p-1 = f(x)g(x) in mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+zeta_p$ so $$f(x) = prod_{alpha in Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)cdot (1+zeta_p)} (x-alpha)$$



by definition $|.|_p$ is $Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)$ invariant and$$|1+zeta_p|_p=|f(0)|_p^{1/deg(f)}$$



and since $g(0) in mathbf{Z}_p$
$$1 ge |f(0)|_p ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$



whence $|f(0)|_p= 1$ and $|1+zeta_p|_p = 1$.






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$endgroup$





















    0












    $begingroup$

    Note that $frac {ζ^2-1} {ζ-1}=1+ζ$ and $frac {ζ-1} {ζ^2-1} =1+ζ^2+ζ^4+...+ζ^{p-1}$.



    In the case that $ord_p(ζ)=0$, we will have $ord_p(ζ^2)=ord_p(ζ^3)=...=0$.



    This would imply $ord_p(ζ^2-1)-οrd_p(ζ-1)=ord_p(1+ζ) geq inf {ord_p(1),ord_p(ζ)}=0 $ and also $οrd_p(ζ-1)-ord_p(ζ^2-1)=ord_p(1+ζ^2+ζ^4+...+ζ^{p-1})geq inf{ord_p(1),...,ord_p(ζ^{p-1})}=0$.



    From these two inequalities, it should be easy to see $ord_p(ζ^2-1)=οrd_p(ζ-1)$ as needed.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Here’s yet another argument:

      Start with the minimal polynomial for $zeta_p$ ,
      begin{align}
      text{Irr}(zeta_p,Bbb Q_p)&=frac{X^p-1}{X-1}\
      f=text{Irr}(zeta_p-1,Bbb Q_p)&=frac{(X+1)^p-1}X&text{($p$-Eisenstein, root $pi$)}\
      v_p(pi)=v_p(zeta_p-1)&=frac1{p-1}\
      v_p(pi+2)=v_p(zeta_p+1)&=v_p(2)=0&text{(’cause $pne2$)},.
      end{align}






      share|cite|improve this answer









      $endgroup$













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        3 Answers
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        3 Answers
        3






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        active

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        active

        oldest

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        0












        $begingroup$

        Let $$(x-1)^p-1 = f(x)g(x) in mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+zeta_p$ so $$f(x) = prod_{alpha in Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)cdot (1+zeta_p)} (x-alpha)$$



        by definition $|.|_p$ is $Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)$ invariant and$$|1+zeta_p|_p=|f(0)|_p^{1/deg(f)}$$



        and since $g(0) in mathbf{Z}_p$
        $$1 ge |f(0)|_p ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$



        whence $|f(0)|_p= 1$ and $|1+zeta_p|_p = 1$.






        share|cite|improve this answer











        $endgroup$


















          0












          $begingroup$

          Let $$(x-1)^p-1 = f(x)g(x) in mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+zeta_p$ so $$f(x) = prod_{alpha in Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)cdot (1+zeta_p)} (x-alpha)$$



          by definition $|.|_p$ is $Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)$ invariant and$$|1+zeta_p|_p=|f(0)|_p^{1/deg(f)}$$



          and since $g(0) in mathbf{Z}_p$
          $$1 ge |f(0)|_p ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$



          whence $|f(0)|_p= 1$ and $|1+zeta_p|_p = 1$.






          share|cite|improve this answer











          $endgroup$
















            0












            0








            0





            $begingroup$

            Let $$(x-1)^p-1 = f(x)g(x) in mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+zeta_p$ so $$f(x) = prod_{alpha in Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)cdot (1+zeta_p)} (x-alpha)$$



            by definition $|.|_p$ is $Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)$ invariant and$$|1+zeta_p|_p=|f(0)|_p^{1/deg(f)}$$



            and since $g(0) in mathbf{Z}_p$
            $$1 ge |f(0)|_p ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$



            whence $|f(0)|_p= 1$ and $|1+zeta_p|_p = 1$.






            share|cite|improve this answer











            $endgroup$



            Let $$(x-1)^p-1 = f(x)g(x) in mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+zeta_p$ so $$f(x) = prod_{alpha in Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)cdot (1+zeta_p)} (x-alpha)$$



            by definition $|.|_p$ is $Gal(overline{mathbf{Q}_p}/mathbf{Q}_p)$ invariant and$$|1+zeta_p|_p=|f(0)|_p^{1/deg(f)}$$



            and since $g(0) in mathbf{Z}_p$
            $$1 ge |f(0)|_p ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$



            whence $|f(0)|_p= 1$ and $|1+zeta_p|_p = 1$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 7 '18 at 20:16

























            answered Dec 7 '18 at 20:10









            reunsreuns

            19.9k21148




            19.9k21148























                0












                $begingroup$

                Note that $frac {ζ^2-1} {ζ-1}=1+ζ$ and $frac {ζ-1} {ζ^2-1} =1+ζ^2+ζ^4+...+ζ^{p-1}$.



                In the case that $ord_p(ζ)=0$, we will have $ord_p(ζ^2)=ord_p(ζ^3)=...=0$.



                This would imply $ord_p(ζ^2-1)-οrd_p(ζ-1)=ord_p(1+ζ) geq inf {ord_p(1),ord_p(ζ)}=0 $ and also $οrd_p(ζ-1)-ord_p(ζ^2-1)=ord_p(1+ζ^2+ζ^4+...+ζ^{p-1})geq inf{ord_p(1),...,ord_p(ζ^{p-1})}=0$.



                From these two inequalities, it should be easy to see $ord_p(ζ^2-1)=οrd_p(ζ-1)$ as needed.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Note that $frac {ζ^2-1} {ζ-1}=1+ζ$ and $frac {ζ-1} {ζ^2-1} =1+ζ^2+ζ^4+...+ζ^{p-1}$.



                  In the case that $ord_p(ζ)=0$, we will have $ord_p(ζ^2)=ord_p(ζ^3)=...=0$.



                  This would imply $ord_p(ζ^2-1)-οrd_p(ζ-1)=ord_p(1+ζ) geq inf {ord_p(1),ord_p(ζ)}=0 $ and also $οrd_p(ζ-1)-ord_p(ζ^2-1)=ord_p(1+ζ^2+ζ^4+...+ζ^{p-1})geq inf{ord_p(1),...,ord_p(ζ^{p-1})}=0$.



                  From these two inequalities, it should be easy to see $ord_p(ζ^2-1)=οrd_p(ζ-1)$ as needed.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Note that $frac {ζ^2-1} {ζ-1}=1+ζ$ and $frac {ζ-1} {ζ^2-1} =1+ζ^2+ζ^4+...+ζ^{p-1}$.



                    In the case that $ord_p(ζ)=0$, we will have $ord_p(ζ^2)=ord_p(ζ^3)=...=0$.



                    This would imply $ord_p(ζ^2-1)-οrd_p(ζ-1)=ord_p(1+ζ) geq inf {ord_p(1),ord_p(ζ)}=0 $ and also $οrd_p(ζ-1)-ord_p(ζ^2-1)=ord_p(1+ζ^2+ζ^4+...+ζ^{p-1})geq inf{ord_p(1),...,ord_p(ζ^{p-1})}=0$.



                    From these two inequalities, it should be easy to see $ord_p(ζ^2-1)=οrd_p(ζ-1)$ as needed.






                    share|cite|improve this answer









                    $endgroup$



                    Note that $frac {ζ^2-1} {ζ-1}=1+ζ$ and $frac {ζ-1} {ζ^2-1} =1+ζ^2+ζ^4+...+ζ^{p-1}$.



                    In the case that $ord_p(ζ)=0$, we will have $ord_p(ζ^2)=ord_p(ζ^3)=...=0$.



                    This would imply $ord_p(ζ^2-1)-οrd_p(ζ-1)=ord_p(1+ζ) geq inf {ord_p(1),ord_p(ζ)}=0 $ and also $οrd_p(ζ-1)-ord_p(ζ^2-1)=ord_p(1+ζ^2+ζ^4+...+ζ^{p-1})geq inf{ord_p(1),...,ord_p(ζ^{p-1})}=0$.



                    From these two inequalities, it should be easy to see $ord_p(ζ^2-1)=οrd_p(ζ-1)$ as needed.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 7 '18 at 21:11









                    Pascal's WagerPascal's Wager

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                    357315























                        0












                        $begingroup$

                        Here’s yet another argument:

                        Start with the minimal polynomial for $zeta_p$ ,
                        begin{align}
                        text{Irr}(zeta_p,Bbb Q_p)&=frac{X^p-1}{X-1}\
                        f=text{Irr}(zeta_p-1,Bbb Q_p)&=frac{(X+1)^p-1}X&text{($p$-Eisenstein, root $pi$)}\
                        v_p(pi)=v_p(zeta_p-1)&=frac1{p-1}\
                        v_p(pi+2)=v_p(zeta_p+1)&=v_p(2)=0&text{(’cause $pne2$)},.
                        end{align}






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Here’s yet another argument:

                          Start with the minimal polynomial for $zeta_p$ ,
                          begin{align}
                          text{Irr}(zeta_p,Bbb Q_p)&=frac{X^p-1}{X-1}\
                          f=text{Irr}(zeta_p-1,Bbb Q_p)&=frac{(X+1)^p-1}X&text{($p$-Eisenstein, root $pi$)}\
                          v_p(pi)=v_p(zeta_p-1)&=frac1{p-1}\
                          v_p(pi+2)=v_p(zeta_p+1)&=v_p(2)=0&text{(’cause $pne2$)},.
                          end{align}






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Here’s yet another argument:

                            Start with the minimal polynomial for $zeta_p$ ,
                            begin{align}
                            text{Irr}(zeta_p,Bbb Q_p)&=frac{X^p-1}{X-1}\
                            f=text{Irr}(zeta_p-1,Bbb Q_p)&=frac{(X+1)^p-1}X&text{($p$-Eisenstein, root $pi$)}\
                            v_p(pi)=v_p(zeta_p-1)&=frac1{p-1}\
                            v_p(pi+2)=v_p(zeta_p+1)&=v_p(2)=0&text{(’cause $pne2$)},.
                            end{align}






                            share|cite|improve this answer









                            $endgroup$



                            Here’s yet another argument:

                            Start with the minimal polynomial for $zeta_p$ ,
                            begin{align}
                            text{Irr}(zeta_p,Bbb Q_p)&=frac{X^p-1}{X-1}\
                            f=text{Irr}(zeta_p-1,Bbb Q_p)&=frac{(X+1)^p-1}X&text{($p$-Eisenstein, root $pi$)}\
                            v_p(pi)=v_p(zeta_p-1)&=frac1{p-1}\
                            v_p(pi+2)=v_p(zeta_p+1)&=v_p(2)=0&text{(’cause $pne2$)},.
                            end{align}







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 8 '18 at 5:45









                            LubinLubin

                            44.2k44585




                            44.2k44585






























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                                Mont Emei

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                                Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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