Prove that if $p notequiv 1 hspace{0.2 cm} (5)$ then $f(x) = x^{5} - 2$ has a unique solution in...












2












$begingroup$


To prove the statetament, i thought to define a linear application $$ phi : mathbb{F}_{p}^{*} longmapsto mathbb{F}_{p}^{*}$$



Define by : $f(x) = x^{5}$, studying the kernel of $phi$ I noticed that from $x^{5} equiv 1 hspace{0.2 cm}(5)$ $phi$ was injective (Because from the hp. we have $5 nmid p - 1 $, which translates in $p notequiv 1 hspace{0.2 cm} (5)$).



It follows that this $phi$ is an automorphism of $mathbb{F}_{p}^{*}$, in particular it is surjective,



From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ it exists $x in mathbb{F}_{p}^{*}$ such that $x^{5} = 2$.



Now i know my $f(x) = (x - alpha)p(x)$ splits into one linear factor, and one of $deg p = 4$.



I'd like to conclude saying that $p$ doesn't split and so that the solution is unique, but don't see how,



Any help or tip would be appreciated,



Thanks.






group-theory field-theory finite-fields splitting-field automorphism-group







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$endgroup$








  • 1




    $begingroup$
    Just use that $phi$ is injective.
    $endgroup$
    – Mark Bennet
    Jan 5 at 20:34










  • $begingroup$
    Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
    $endgroup$
    – Mindlack
    Jan 5 at 20:39










  • $begingroup$
    So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
    $endgroup$
    – greedoid
    Jan 5 at 20:44
















2












$begingroup$


To prove the statetament, i thought to define a linear application $$ phi : mathbb{F}_{p}^{*} longmapsto mathbb{F}_{p}^{*}$$



Define by : $f(x) = x^{5}$, studying the kernel of $phi$ I noticed that from $x^{5} equiv 1 hspace{0.2 cm}(5)$ $phi$ was injective (Because from the hp. we have $5 nmid p - 1 $, which translates in $p notequiv 1 hspace{0.2 cm} (5)$).



It follows that this $phi$ is an automorphism of $mathbb{F}_{p}^{*}$, in particular it is surjective,



From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ it exists $x in mathbb{F}_{p}^{*}$ such that $x^{5} = 2$.



Now i know my $f(x) = (x - alpha)p(x)$ splits into one linear factor, and one of $deg p = 4$.



I'd like to conclude saying that $p$ doesn't split and so that the solution is unique, but don't see how,



Any help or tip would be appreciated,



Thanks.






group-theory field-theory finite-fields splitting-field automorphism-group







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Just use that $phi$ is injective.
    $endgroup$
    – Mark Bennet
    Jan 5 at 20:34










  • $begingroup$
    Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
    $endgroup$
    – Mindlack
    Jan 5 at 20:39










  • $begingroup$
    So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
    $endgroup$
    – greedoid
    Jan 5 at 20:44














2












2








2





$begingroup$


To prove the statetament, i thought to define a linear application $$ phi : mathbb{F}_{p}^{*} longmapsto mathbb{F}_{p}^{*}$$



Define by : $f(x) = x^{5}$, studying the kernel of $phi$ I noticed that from $x^{5} equiv 1 hspace{0.2 cm}(5)$ $phi$ was injective (Because from the hp. we have $5 nmid p - 1 $, which translates in $p notequiv 1 hspace{0.2 cm} (5)$).



It follows that this $phi$ is an automorphism of $mathbb{F}_{p}^{*}$, in particular it is surjective,



From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ it exists $x in mathbb{F}_{p}^{*}$ such that $x^{5} = 2$.



Now i know my $f(x) = (x - alpha)p(x)$ splits into one linear factor, and one of $deg p = 4$.



I'd like to conclude saying that $p$ doesn't split and so that the solution is unique, but don't see how,



Any help or tip would be appreciated,



Thanks.






group-theory field-theory finite-fields splitting-field automorphism-group







share|cite|improve this question









$endgroup$




To prove the statetament, i thought to define a linear application $$ phi : mathbb{F}_{p}^{*} longmapsto mathbb{F}_{p}^{*}$$



Define by : $f(x) = x^{5}$, studying the kernel of $phi$ I noticed that from $x^{5} equiv 1 hspace{0.2 cm}(5)$ $phi$ was injective (Because from the hp. we have $5 nmid p - 1 $, which translates in $p notequiv 1 hspace{0.2 cm} (5)$).



It follows that this $phi$ is an automorphism of $mathbb{F}_{p}^{*}$, in particular it is surjective,



From here i know that for every $p notequiv 1 hspace{0.2 cm} (5)$ it exists $x in mathbb{F}_{p}^{*}$ such that $x^{5} = 2$.



Now i know my $f(x) = (x - alpha)p(x)$ splits into one linear factor, and one of $deg p = 4$.



I'd like to conclude saying that $p$ doesn't split and so that the solution is unique, but don't see how,



Any help or tip would be appreciated,



Thanks.






group-theory field-theory finite-fields splitting-field automorphism-group




group-theory field-theory finite-fields splitting-field automorphism-group






share|cite|improve this question













share|cite|improve this question











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asked Jan 5 at 20:29









jacopoburellijacopoburelli

1707




1707








  • 1




    $begingroup$
    Just use that $phi$ is injective.
    $endgroup$
    – Mark Bennet
    Jan 5 at 20:34










  • $begingroup$
    Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
    $endgroup$
    – Mindlack
    Jan 5 at 20:39










  • $begingroup$
    So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
    $endgroup$
    – greedoid
    Jan 5 at 20:44














  • 1




    $begingroup$
    Just use that $phi$ is injective.
    $endgroup$
    – Mark Bennet
    Jan 5 at 20:34










  • $begingroup$
    Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
    $endgroup$
    – Mindlack
    Jan 5 at 20:39










  • $begingroup$
    So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
    $endgroup$
    – greedoid
    Jan 5 at 20:44








1




1




$begingroup$
Just use that $phi$ is injective.
$endgroup$
– Mark Bennet
Jan 5 at 20:34




$begingroup$
Just use that $phi$ is injective.
$endgroup$
– Mark Bennet
Jan 5 at 20:34












$begingroup$
Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
$endgroup$
– Mindlack
Jan 5 at 20:39




$begingroup$
Since $x in mathbb{F}_p^* longmapsto x^5 in mathbb{F}_p^*$ is bijective, $f$ is a bijection from $mathbb{F}_p$ to itself. So it has a unique root. Compare $f$ with its derivative to prove that the root is simple iff $p neq 5$.
$endgroup$
– Mindlack
Jan 5 at 20:39












$begingroup$
So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
$endgroup$
– greedoid
Jan 5 at 20:44




$begingroup$
So you want to prove that equation $$x^5equiv _p2$$ has only one solution if $pne 1pmod 5$?
$endgroup$
– greedoid
Jan 5 at 20:44










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint If $alpha$ is a multiple root, then $f'(alpha)=0$.



Alternate solution Use the fact that there exists a primitive root $b$ modulo $p$.
Write $x=b^k, 2=b^alpha$ and solve for $k$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Let $p= 5k+r$ where $rin{0,2,3,4}$. Let us prove that $${0^5-2,1^5-2,...,(p-1)^5-2}=_{pmod p} {0,1,2,...,p-1}$$



    Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
    Since by Fermat theorem we have $a^{5k+r-1}equiv _p1$ we deduce:$$a^{5k}equiv _pb^{5k}implies a^{r-1}equiv_pb^{r-1} $$



    Case 1: $r=0$ (so $p=5$) then $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.



    Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.



    Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.



    Case 4: $r=4$ then $a^3equiv_p b^3$, then $a^6equiv_5 b^6 equiv a^5b implies pmid a^5(a-b) implies pmid a implies pmid b implies a=b$ a contradiction.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Hint If $alpha$ is a multiple root, then $f'(alpha)=0$.



      Alternate solution Use the fact that there exists a primitive root $b$ modulo $p$.
      Write $x=b^k, 2=b^alpha$ and solve for $k$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint If $alpha$ is a multiple root, then $f'(alpha)=0$.



        Alternate solution Use the fact that there exists a primitive root $b$ modulo $p$.
        Write $x=b^k, 2=b^alpha$ and solve for $k$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint If $alpha$ is a multiple root, then $f'(alpha)=0$.



          Alternate solution Use the fact that there exists a primitive root $b$ modulo $p$.
          Write $x=b^k, 2=b^alpha$ and solve for $k$.






          share|cite|improve this answer









          $endgroup$



          Hint If $alpha$ is a multiple root, then $f'(alpha)=0$.



          Alternate solution Use the fact that there exists a primitive root $b$ modulo $p$.
          Write $x=b^k, 2=b^alpha$ and solve for $k$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 at 20:32









          N. S.N. S.

          104k7114209




          104k7114209























              1












              $begingroup$

              Let $p= 5k+r$ where $rin{0,2,3,4}$. Let us prove that $${0^5-2,1^5-2,...,(p-1)^5-2}=_{pmod p} {0,1,2,...,p-1}$$



              Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
              Since by Fermat theorem we have $a^{5k+r-1}equiv _p1$ we deduce:$$a^{5k}equiv _pb^{5k}implies a^{r-1}equiv_pb^{r-1} $$



              Case 1: $r=0$ (so $p=5$) then $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.



              Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.



              Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.



              Case 4: $r=4$ then $a^3equiv_p b^3$, then $a^6equiv_5 b^6 equiv a^5b implies pmid a^5(a-b) implies pmid a implies pmid b implies a=b$ a contradiction.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let $p= 5k+r$ where $rin{0,2,3,4}$. Let us prove that $${0^5-2,1^5-2,...,(p-1)^5-2}=_{pmod p} {0,1,2,...,p-1}$$



                Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
                Since by Fermat theorem we have $a^{5k+r-1}equiv _p1$ we deduce:$$a^{5k}equiv _pb^{5k}implies a^{r-1}equiv_pb^{r-1} $$



                Case 1: $r=0$ (so $p=5$) then $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.



                Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.



                Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.



                Case 4: $r=4$ then $a^3equiv_p b^3$, then $a^6equiv_5 b^6 equiv a^5b implies pmid a^5(a-b) implies pmid a implies pmid b implies a=b$ a contradiction.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $p= 5k+r$ where $rin{0,2,3,4}$. Let us prove that $${0^5-2,1^5-2,...,(p-1)^5-2}=_{pmod p} {0,1,2,...,p-1}$$



                  Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
                  Since by Fermat theorem we have $a^{5k+r-1}equiv _p1$ we deduce:$$a^{5k}equiv _pb^{5k}implies a^{r-1}equiv_pb^{r-1} $$



                  Case 1: $r=0$ (so $p=5$) then $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.



                  Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.



                  Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.



                  Case 4: $r=4$ then $a^3equiv_p b^3$, then $a^6equiv_5 b^6 equiv a^5b implies pmid a^5(a-b) implies pmid a implies pmid b implies a=b$ a contradiction.






                  share|cite|improve this answer









                  $endgroup$



                  Let $p= 5k+r$ where $rin{0,2,3,4}$. Let us prove that $${0^5-2,1^5-2,...,(p-1)^5-2}=_{pmod p} {0,1,2,...,p-1}$$



                  Say there exist $ane b in mathbb{F}_p$ such that $$a^5-2equiv _p b^5-2 implies a^5equiv _pb^5$$
                  Since by Fermat theorem we have $a^{5k+r-1}equiv _p1$ we deduce:$$a^{5k}equiv _pb^{5k}implies a^{r-1}equiv_pb^{r-1} $$



                  Case 1: $r=0$ (so $p=5$) then $a^5equiv_5 a$ and $b^5equiv_5 b$ so $aequiv _5b$ a contradiciton since $ane b$.



                  Case 2: $r=2$ then $aequiv_p b$ a contradiciton since $ane b$.



                  Case 3: $r=3$ then $a^2equiv_p b^2$, then since $ane b$ we have $aequiv_p -b$ but then $a^5equiv_5 -b^5 implies pmid 2a^5 implies pmid a implies pmid b implies a=b$ a contradiction.



                  Case 4: $r=4$ then $a^3equiv_p b^3$, then $a^6equiv_5 b^6 equiv a^5b implies pmid a^5(a-b) implies pmid a implies pmid b implies a=b$ a contradiction.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 21:00









                  greedoidgreedoid

                  46.1k1160117




                  46.1k1160117






























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