Some easy questions about multiplicative characters and Jacobi sums.












2












$begingroup$


First I want to give you some context. Then I will ask my questions. I think that my questions are easy and fast to answer, so I've decided to put them together in one question here.



Context



Consider the equations $ a_1x_1^{l_1} + dots + a_rx_r^{l_r} = b $ with $a_1, dots , a_r in F^{*}$, where $F$ is a finite field. Let us say that $F$ has $m$ elements. $N$ is the number of solutions. Now I know that if $b neq 0$ :



$ N = m^{r-1} + sum chi_1chi_2cdotschi_r(b) chi_1(a_1^{-1})chi_2(a_2^{-1})cdotschi_r(a_r^{-1})J(chi_1,dots,chi_r)$.
The summation is over $r$-tuples of characters $chi_1, dots, chi_r$, where
$ chi_i^{l_i} = varepsilon $ and $chi_i neq epsilon$ for $i = 1, dots, r$.



I want to specializing this to $x^2 + y^4 = 1$. Obviously $r=2$. So I want to sum over all $2$-tuples of characters $chi_1, chi_2$ where $ chi_i^{l_i} = varepsilon $ and $chi_i neq epsilon$ for $i = 1,2$.



So the $N = m + J(rho,chi) + J(rho,chi^2) + J(rho,chi^3)$, where $rho$ is a character of order $2$ and $chi$ is a character of order $4$.



Questions



1) Why $chi^2 = rho$ ? I need that to say $J(rho,chi^2) = -1 $.



2) We know $chi^4 = varepsilon$, but why this is enough to say that $chi^3 =
bar{chi}$ ?



3) Why $J(rho,bar{chi}) = overline{J(rho,chi)} $ ?



4) Now let us say that $pi = -J(rho,chi)$. We know that $rho$ takes the values $pm1$ and $chi pm1$, $pm i$. Why is this enougth to say that $pi = a+bi$?



5)If my four questions are answered I can say that $N = m - 1 - pi - bar{pi}$ . I know that $ a^2 + b^2 = pi bar{pi} = m $. Why can I say that $N = m - 1 - 2a$ ?



I hope that my question are clear. Sorry that they might be trivial for you. I'm an absolute beginner.






number-theory characters







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    First I want to give you some context. Then I will ask my questions. I think that my questions are easy and fast to answer, so I've decided to put them together in one question here.



    Context



    Consider the equations $ a_1x_1^{l_1} + dots + a_rx_r^{l_r} = b $ with $a_1, dots , a_r in F^{*}$, where $F$ is a finite field. Let us say that $F$ has $m$ elements. $N$ is the number of solutions. Now I know that if $b neq 0$ :



    $ N = m^{r-1} + sum chi_1chi_2cdotschi_r(b) chi_1(a_1^{-1})chi_2(a_2^{-1})cdotschi_r(a_r^{-1})J(chi_1,dots,chi_r)$.
    The summation is over $r$-tuples of characters $chi_1, dots, chi_r$, where
    $ chi_i^{l_i} = varepsilon $ and $chi_i neq epsilon$ for $i = 1, dots, r$.



    I want to specializing this to $x^2 + y^4 = 1$. Obviously $r=2$. So I want to sum over all $2$-tuples of characters $chi_1, chi_2$ where $ chi_i^{l_i} = varepsilon $ and $chi_i neq epsilon$ for $i = 1,2$.



    So the $N = m + J(rho,chi) + J(rho,chi^2) + J(rho,chi^3)$, where $rho$ is a character of order $2$ and $chi$ is a character of order $4$.



    Questions



    1) Why $chi^2 = rho$ ? I need that to say $J(rho,chi^2) = -1 $.



    2) We know $chi^4 = varepsilon$, but why this is enough to say that $chi^3 =
    bar{chi}$     ?



    3) Why $J(rho,bar{chi}) = overline{J(rho,chi)} $ ?



    4) Now let us say that $pi = -J(rho,chi)$. We know that $rho$ takes the values $pm1$ and $chi pm1$, $pm i$. Why is this enougth to say that $pi = a+bi$?



    5)If my four questions are answered I can say that $N = m - 1 - pi - bar{pi}$ . I know that $ a^2 + b^2 = pi bar{pi} = m $. Why can I say that $N = m - 1 - 2a$ ?



    I hope that my question are clear. Sorry that they might be trivial for you. I'm an absolute beginner.






    number-theory characters







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      First I want to give you some context. Then I will ask my questions. I think that my questions are easy and fast to answer, so I've decided to put them together in one question here.



      Context



      Consider the equations $ a_1x_1^{l_1} + dots + a_rx_r^{l_r} = b $ with $a_1, dots , a_r in F^{*}$, where $F$ is a finite field. Let us say that $F$ has $m$ elements. $N$ is the number of solutions. Now I know that if $b neq 0$ :



      $ N = m^{r-1} + sum chi_1chi_2cdotschi_r(b) chi_1(a_1^{-1})chi_2(a_2^{-1})cdotschi_r(a_r^{-1})J(chi_1,dots,chi_r)$.
      The summation is over $r$-tuples of characters $chi_1, dots, chi_r$, where
      $ chi_i^{l_i} = varepsilon $ and $chi_i neq epsilon$ for $i = 1, dots, r$.



      I want to specializing this to $x^2 + y^4 = 1$. Obviously $r=2$. So I want to sum over all $2$-tuples of characters $chi_1, chi_2$ where $ chi_i^{l_i} = varepsilon $ and $chi_i neq epsilon$ for $i = 1,2$.



      So the $N = m + J(rho,chi) + J(rho,chi^2) + J(rho,chi^3)$, where $rho$ is a character of order $2$ and $chi$ is a character of order $4$.



      Questions



      1) Why $chi^2 = rho$ ? I need that to say $J(rho,chi^2) = -1 $.



      2) We know $chi^4 = varepsilon$, but why this is enough to say that $chi^3 =
      bar{chi}$       ?



      3) Why $J(rho,bar{chi}) = overline{J(rho,chi)} $ ?



      4) Now let us say that $pi = -J(rho,chi)$. We know that $rho$ takes the values $pm1$ and $chi pm1$, $pm i$. Why is this enougth to say that $pi = a+bi$?



      5)If my four questions are answered I can say that $N = m - 1 - pi - bar{pi}$ . I know that $ a^2 + b^2 = pi bar{pi} = m $. Why can I say that $N = m - 1 - 2a$ ?



      I hope that my question are clear. Sorry that they might be trivial for you. I'm an absolute beginner.






      number-theory characters







      share|cite|improve this question









      $endgroup$




      First I want to give you some context. Then I will ask my questions. I think that my questions are easy and fast to answer, so I've decided to put them together in one question here.



      Context



      Consider the equations $ a_1x_1^{l_1} + dots + a_rx_r^{l_r} = b $ with $a_1, dots , a_r in F^{*}$, where $F$ is a finite field. Let us say that $F$ has $m$ elements. $N$ is the number of solutions. Now I know that if $b neq 0$ :



      $ N = m^{r-1} + sum chi_1chi_2cdotschi_r(b) chi_1(a_1^{-1})chi_2(a_2^{-1})cdotschi_r(a_r^{-1})J(chi_1,dots,chi_r)$.
      The summation is over $r$-tuples of characters $chi_1, dots, chi_r$, where
      $ chi_i^{l_i} = varepsilon $ and $chi_i neq epsilon$ for $i = 1, dots, r$.



      I want to specializing this to $x^2 + y^4 = 1$. Obviously $r=2$. So I want to sum over all $2$-tuples of characters $chi_1, chi_2$ where $ chi_i^{l_i} = varepsilon $ and $chi_i neq epsilon$ for $i = 1,2$.



      So the $N = m + J(rho,chi) + J(rho,chi^2) + J(rho,chi^3)$, where $rho$ is a character of order $2$ and $chi$ is a character of order $4$.



      Questions



      1) Why $chi^2 = rho$ ? I need that to say $J(rho,chi^2) = -1 $.



      2) We know $chi^4 = varepsilon$, but why this is enough to say that $chi^3 =
      bar{chi}$       ?



      3) Why $J(rho,bar{chi}) = overline{J(rho,chi)} $ ?



      4) Now let us say that $pi = -J(rho,chi)$. We know that $rho$ takes the values $pm1$ and $chi pm1$, $pm i$. Why is this enougth to say that $pi = a+bi$?



      5)If my four questions are answered I can say that $N = m - 1 - pi - bar{pi}$ . I know that $ a^2 + b^2 = pi bar{pi} = m $. Why can I say that $N = m - 1 - 2a$ ?



      I hope that my question are clear. Sorry that they might be trivial for you. I'm an absolute beginner.






      number-theory characters




      number-theory characters






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 31 '18 at 12:47









      MemoriesMemories

      9811




      9811






















          1 Answer
          1






          active

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          2












          $begingroup$

          $chi^2$ has order $2$ and $rho$ is the only character of order $2$.



          $chi^3=chi^{-1}=overlinechi$.



          $overline{J(rho,chi)}=J(overlinerho,overlinechi)=J(rho,overlinechi)$.



          $J(rho,chi)$ is the sum of terms all of which have the form $pm1$ or $pm i$.



          $pi+overlinepi=(a+bi)+(a-bi)=2a.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer. I get it now. Happy new year :).
            $endgroup$
            – Memories
            Jan 1 at 18:34










          • $begingroup$
            Sorry for disturbing you again, but can I ask you one final question? My question seems so simple that it wouldnt be necessary to use " ask Question". I showed now $ rho(a) -1 equiv 0 (2) $ and $chi(a) -1 equiv 0(1+i)$. Combine this to get $(rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. Thus $sum_{a+b=1} (rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. I have expanded this term and got $sum_a rho(a)$ and $sum_b chi(b)$. Now I need that these two sum are zero. But I don't see why.
            $endgroup$
            – Memories
            Jan 2 at 16:46








          • 1




            $begingroup$
            @Memories It's a general fact about characters that the values of a non-trivial character sum to zero.
            $endgroup$
            – Lord Shark the Unknown
            Jan 2 at 16:51











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          $chi^2$ has order $2$ and $rho$ is the only character of order $2$.



          $chi^3=chi^{-1}=overlinechi$.



          $overline{J(rho,chi)}=J(overlinerho,overlinechi)=J(rho,overlinechi)$.



          $J(rho,chi)$ is the sum of terms all of which have the form $pm1$ or $pm i$.



          $pi+overlinepi=(a+bi)+(a-bi)=2a.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer. I get it now. Happy new year :).
            $endgroup$
            – Memories
            Jan 1 at 18:34










          • $begingroup$
            Sorry for disturbing you again, but can I ask you one final question? My question seems so simple that it wouldnt be necessary to use " ask Question". I showed now $ rho(a) -1 equiv 0 (2) $ and $chi(a) -1 equiv 0(1+i)$. Combine this to get $(rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. Thus $sum_{a+b=1} (rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. I have expanded this term and got $sum_a rho(a)$ and $sum_b chi(b)$. Now I need that these two sum are zero. But I don't see why.
            $endgroup$
            – Memories
            Jan 2 at 16:46








          • 1




            $begingroup$
            @Memories It's a general fact about characters that the values of a non-trivial character sum to zero.
            $endgroup$
            – Lord Shark the Unknown
            Jan 2 at 16:51
















          2












          $begingroup$

          $chi^2$ has order $2$ and $rho$ is the only character of order $2$.



          $chi^3=chi^{-1}=overlinechi$.



          $overline{J(rho,chi)}=J(overlinerho,overlinechi)=J(rho,overlinechi)$.



          $J(rho,chi)$ is the sum of terms all of which have the form $pm1$ or $pm i$.



          $pi+overlinepi=(a+bi)+(a-bi)=2a.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for your answer. I get it now. Happy new year :).
            $endgroup$
            – Memories
            Jan 1 at 18:34










          • $begingroup$
            Sorry for disturbing you again, but can I ask you one final question? My question seems so simple that it wouldnt be necessary to use " ask Question". I showed now $ rho(a) -1 equiv 0 (2) $ and $chi(a) -1 equiv 0(1+i)$. Combine this to get $(rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. Thus $sum_{a+b=1} (rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. I have expanded this term and got $sum_a rho(a)$ and $sum_b chi(b)$. Now I need that these two sum are zero. But I don't see why.
            $endgroup$
            – Memories
            Jan 2 at 16:46








          • 1




            $begingroup$
            @Memories It's a general fact about characters that the values of a non-trivial character sum to zero.
            $endgroup$
            – Lord Shark the Unknown
            Jan 2 at 16:51














          2












          2








          2





          $begingroup$

          $chi^2$ has order $2$ and $rho$ is the only character of order $2$.



          $chi^3=chi^{-1}=overlinechi$.



          $overline{J(rho,chi)}=J(overlinerho,overlinechi)=J(rho,overlinechi)$.



          $J(rho,chi)$ is the sum of terms all of which have the form $pm1$ or $pm i$.



          $pi+overlinepi=(a+bi)+(a-bi)=2a.$






          share|cite|improve this answer









          $endgroup$



          $chi^2$ has order $2$ and $rho$ is the only character of order $2$.



          $chi^3=chi^{-1}=overlinechi$.



          $overline{J(rho,chi)}=J(overlinerho,overlinechi)=J(rho,overlinechi)$.



          $J(rho,chi)$ is the sum of terms all of which have the form $pm1$ or $pm i$.



          $pi+overlinepi=(a+bi)+(a-bi)=2a.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 13:37









          Lord Shark the UnknownLord Shark the Unknown

          102k1059132




          102k1059132












          • $begingroup$
            Thank you for your answer. I get it now. Happy new year :).
            $endgroup$
            – Memories
            Jan 1 at 18:34










          • $begingroup$
            Sorry for disturbing you again, but can I ask you one final question? My question seems so simple that it wouldnt be necessary to use " ask Question". I showed now $ rho(a) -1 equiv 0 (2) $ and $chi(a) -1 equiv 0(1+i)$. Combine this to get $(rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. Thus $sum_{a+b=1} (rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. I have expanded this term and got $sum_a rho(a)$ and $sum_b chi(b)$. Now I need that these two sum are zero. But I don't see why.
            $endgroup$
            – Memories
            Jan 2 at 16:46








          • 1




            $begingroup$
            @Memories It's a general fact about characters that the values of a non-trivial character sum to zero.
            $endgroup$
            – Lord Shark the Unknown
            Jan 2 at 16:51


















          • $begingroup$
            Thank you for your answer. I get it now. Happy new year :).
            $endgroup$
            – Memories
            Jan 1 at 18:34










          • $begingroup$
            Sorry for disturbing you again, but can I ask you one final question? My question seems so simple that it wouldnt be necessary to use " ask Question". I showed now $ rho(a) -1 equiv 0 (2) $ and $chi(a) -1 equiv 0(1+i)$. Combine this to get $(rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. Thus $sum_{a+b=1} (rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. I have expanded this term and got $sum_a rho(a)$ and $sum_b chi(b)$. Now I need that these two sum are zero. But I don't see why.
            $endgroup$
            – Memories
            Jan 2 at 16:46








          • 1




            $begingroup$
            @Memories It's a general fact about characters that the values of a non-trivial character sum to zero.
            $endgroup$
            – Lord Shark the Unknown
            Jan 2 at 16:51
















          $begingroup$
          Thank you for your answer. I get it now. Happy new year :).
          $endgroup$
          – Memories
          Jan 1 at 18:34




          $begingroup$
          Thank you for your answer. I get it now. Happy new year :).
          $endgroup$
          – Memories
          Jan 1 at 18:34












          $begingroup$
          Sorry for disturbing you again, but can I ask you one final question? My question seems so simple that it wouldnt be necessary to use " ask Question". I showed now $ rho(a) -1 equiv 0 (2) $ and $chi(a) -1 equiv 0(1+i)$. Combine this to get $(rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. Thus $sum_{a+b=1} (rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. I have expanded this term and got $sum_a rho(a)$ and $sum_b chi(b)$. Now I need that these two sum are zero. But I don't see why.
          $endgroup$
          – Memories
          Jan 2 at 16:46






          $begingroup$
          Sorry for disturbing you again, but can I ask you one final question? My question seems so simple that it wouldnt be necessary to use " ask Question". I showed now $ rho(a) -1 equiv 0 (2) $ and $chi(a) -1 equiv 0(1+i)$. Combine this to get $(rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. Thus $sum_{a+b=1} (rho(a) -1)(chi(b) -1) equiv 0(2+2i)$. I have expanded this term and got $sum_a rho(a)$ and $sum_b chi(b)$. Now I need that these two sum are zero. But I don't see why.
          $endgroup$
          – Memories
          Jan 2 at 16:46






          1




          1




          $begingroup$
          @Memories It's a general fact about characters that the values of a non-trivial character sum to zero.
          $endgroup$
          – Lord Shark the Unknown
          Jan 2 at 16:51




          $begingroup$
          @Memories It's a general fact about characters that the values of a non-trivial character sum to zero.
          $endgroup$
          – Lord Shark the Unknown
          Jan 2 at 16:51


















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