$L(s,chi)=1$ for non trivial Dirichlet character modulo a polynomial of degree $1$












0












$begingroup$


Let $m(x)in Bbb F_q[X]$ be a monic polynomial of degree $1$.



Show that for every non-trivial Dirichlet character modulo $m$ we have:
$L(s,chi)=1$.



I have seen a theorem that states that if $chi$ is non trivial, then $L(s,chi)$ is a polynomial in $q^{-s}$ of degree at most $deg(m)-1$. In my case, it means the degree is $0$, so $L(s,chi) in Bbb F_q$. I still can't understand why it equals $1$.






number-theory polynomials finite-fields dirichlet-series







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$endgroup$

















    0












    $begingroup$


    Let $m(x)in Bbb F_q[X]$ be a monic polynomial of degree $1$.



    Show that for every non-trivial Dirichlet character modulo $m$ we have:
    $L(s,chi)=1$.



    I have seen a theorem that states that if $chi$ is non trivial, then $L(s,chi)$ is a polynomial in $q^{-s}$ of degree at most $deg(m)-1$. In my case, it means the degree is $0$, so $L(s,chi) in Bbb F_q$. I still can't understand why it equals $1$.






    number-theory polynomials finite-fields dirichlet-series







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $m(x)in Bbb F_q[X]$ be a monic polynomial of degree $1$.



      Show that for every non-trivial Dirichlet character modulo $m$ we have:
      $L(s,chi)=1$.



      I have seen a theorem that states that if $chi$ is non trivial, then $L(s,chi)$ is a polynomial in $q^{-s}$ of degree at most $deg(m)-1$. In my case, it means the degree is $0$, so $L(s,chi) in Bbb F_q$. I still can't understand why it equals $1$.






      number-theory polynomials finite-fields dirichlet-series







      share|cite|improve this question











      $endgroup$




      Let $m(x)in Bbb F_q[X]$ be a monic polynomial of degree $1$.



      Show that for every non-trivial Dirichlet character modulo $m$ we have:
      $L(s,chi)=1$.



      I have seen a theorem that states that if $chi$ is non trivial, then $L(s,chi)$ is a polynomial in $q^{-s}$ of degree at most $deg(m)-1$. In my case, it means the degree is $0$, so $L(s,chi) in Bbb F_q$. I still can't understand why it equals $1$.






      number-theory polynomials finite-fields dirichlet-series




      number-theory polynomials finite-fields dirichlet-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 25 '18 at 17:29







      user401516

















      asked Dec 25 '18 at 15:29









      user401516user401516

      1,004311




      1,004311






















          1 Answer
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          $begingroup$

          $$L(T,chi) = sum_{f in F_q[X]_{monic}} chi(f) T^{deg(f)} = \
          T^{deg(1)} + sum_{a in F_q[X]/(m)} chi(a) sum_{f in F_q[X]_{monic}} T^{deg(f m +a)}$$           Can you finish from there ?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
            $endgroup$
            – user401516
            Dec 25 '18 at 17:01






          • 1




            $begingroup$
            @user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
            $endgroup$
            – reuns
            Dec 25 '18 at 17:04












          • $begingroup$
            Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
            $endgroup$
            – user401516
            Dec 25 '18 at 17:15












          • $begingroup$
            $m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
            $endgroup$
            – reuns
            Dec 25 '18 at 17:47














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          1 Answer
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          active

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          $begingroup$

          $$L(T,chi) = sum_{f in F_q[X]_{monic}} chi(f) T^{deg(f)} = \
          T^{deg(1)} + sum_{a in F_q[X]/(m)} chi(a) sum_{f in F_q[X]_{monic}} T^{deg(f m +a)}$$           Can you finish from there ?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
            $endgroup$
            – user401516
            Dec 25 '18 at 17:01






          • 1




            $begingroup$
            @user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
            $endgroup$
            – reuns
            Dec 25 '18 at 17:04












          • $begingroup$
            Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
            $endgroup$
            – user401516
            Dec 25 '18 at 17:15












          • $begingroup$
            $m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
            $endgroup$
            – reuns
            Dec 25 '18 at 17:47


















          2












          $begingroup$

          $$L(T,chi) = sum_{f in F_q[X]_{monic}} chi(f) T^{deg(f)} = \
          T^{deg(1)} + sum_{a in F_q[X]/(m)} chi(a) sum_{f in F_q[X]_{monic}} T^{deg(f m +a)}$$           Can you finish from there ?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
            $endgroup$
            – user401516
            Dec 25 '18 at 17:01






          • 1




            $begingroup$
            @user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
            $endgroup$
            – reuns
            Dec 25 '18 at 17:04












          • $begingroup$
            Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
            $endgroup$
            – user401516
            Dec 25 '18 at 17:15












          • $begingroup$
            $m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
            $endgroup$
            – reuns
            Dec 25 '18 at 17:47
















          2












          2








          2





          $begingroup$

          $$L(T,chi) = sum_{f in F_q[X]_{monic}} chi(f) T^{deg(f)} = \
          T^{deg(1)} + sum_{a in F_q[X]/(m)} chi(a) sum_{f in F_q[X]_{monic}} T^{deg(f m +a)}$$           Can you finish from there ?






          share|cite|improve this answer











          $endgroup$



          $$L(T,chi) = sum_{f in F_q[X]_{monic}} chi(f) T^{deg(f)} = \
          T^{deg(1)} + sum_{a in F_q[X]/(m)} chi(a) sum_{f in F_q[X]_{monic}} T^{deg(f m +a)}$$           Can you finish from there ?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 25 '18 at 17:07

























          answered Dec 25 '18 at 16:51









          reunsreuns

          20.7k21353




          20.7k21353












          • $begingroup$
            Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
            $endgroup$
            – user401516
            Dec 25 '18 at 17:01






          • 1




            $begingroup$
            @user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
            $endgroup$
            – reuns
            Dec 25 '18 at 17:04












          • $begingroup$
            Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
            $endgroup$
            – user401516
            Dec 25 '18 at 17:15












          • $begingroup$
            $m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
            $endgroup$
            – reuns
            Dec 25 '18 at 17:47




















          • $begingroup$
            Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
            $endgroup$
            – user401516
            Dec 25 '18 at 17:01






          • 1




            $begingroup$
            @user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
            $endgroup$
            – reuns
            Dec 25 '18 at 17:04












          • $begingroup$
            Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
            $endgroup$
            – user401516
            Dec 25 '18 at 17:15












          • $begingroup$
            $m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
            $endgroup$
            – reuns
            Dec 25 '18 at 17:47


















          $begingroup$
          Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
          $endgroup$
          – user401516
          Dec 25 '18 at 17:01




          $begingroup$
          Why is $L(T,chi)=sum_{f ; monic}chi(f)T^{deg f}$ ? according to the definition I know: $L(T,chi)=sum_{f ; monic}frac{chi(f)}{|f|^T}$
          $endgroup$
          – user401516
          Dec 25 '18 at 17:01




          1




          1




          $begingroup$
          @user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
          $endgroup$
          – reuns
          Dec 25 '18 at 17:04






          $begingroup$
          @user401516 With $|f| = q^{deg(f)}$ then $|f|^{-s} = T^{deg(f)}$ where $T =q^{-s}$. We use the $q^{-s}$ notation only in the theory of global zeta functions. For local zeta functions we prefer to make clear they are power series in $T =q^{-s}$.
          $endgroup$
          – reuns
          Dec 25 '18 at 17:04














          $begingroup$
          Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
          $endgroup$
          – user401516
          Dec 25 '18 at 17:15






          $begingroup$
          Thank you, I understand. Why is $T^{deg(a)}=sum_{f ; monic}T^{deg(fm+a)}$ ? sorry for being slow, pretty new to this subject.
          $endgroup$
          – user401516
          Dec 25 '18 at 17:15














          $begingroup$
          $m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
          $endgroup$
          – reuns
          Dec 25 '18 at 17:47






          $begingroup$
          $m$ is monic so any $h$ monic of degree $ge m$ is of the form $h = fm+a$ with $deg(a) < deg(m)$ and $f$ monic. That leaves us with $L(T,chi)= sum_{h in F_q[X]_{monic}, deg(h) < deg(m)} chi(h) T^{deg(h)}+ sum_{f in F_q[X]_{monic}} sum_{a in F_q[X]/(m)} chi(fm+a) T^{deg(f m +a)}$. Here $deg(m) = 1$ so the first sum is easy. The second sum simplifies as $chi(fm+a) = chi(a)$. @user401516
          $endgroup$
          – reuns
          Dec 25 '18 at 17:47




















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