7, 13 independent units in the local field $mathbb{Q}_3$












2












$begingroup$


Why do $7, 13$ generate a rank $2$ subgroup of the group of units of $mathbb{Q}_3$?
I.e. if $7^a= 13^b$ in the local field $mathbb{Q}_3$ and $a, b$ are integers, then $(a, b)=(0,0)$.
(This claim is made on p.75 of Washington’s Introduction to Cyclotomic Fields)






algebraic-number-theory







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    If "$a,b$ are integers", what does it have to do with $mathbb{Q}_3$?
    $endgroup$
    – metamorphy
    Dec 29 '18 at 21:05










  • $begingroup$
    @metamorphy I’m not sure I follow what you are asking, but I the question is trivial by unique factorization if we are in $mathbb{Q}$, and maybe it’s also trivial in $mathbb{Q}_3$ but I don’t see why
    $endgroup$
    – usr0192
    Dec 29 '18 at 21:10






  • 1




    $begingroup$
    This is an equation in $mathbb{Q}$ ($=mathbb{Q}capmathbb{Q}_3$). As you know the solution in $mathbb{Q}$... you're done.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 21:29












  • $begingroup$
    Yeah, i was being stupid
    $endgroup$
    – usr0192
    Dec 29 '18 at 21:29
















2












$begingroup$


Why do $7, 13$ generate a rank $2$ subgroup of the group of units of $mathbb{Q}_3$?
I.e. if $7^a= 13^b$ in the local field $mathbb{Q}_3$ and $a, b$ are integers, then $(a, b)=(0,0)$.
(This claim is made on p.75 of Washington’s Introduction to Cyclotomic Fields)






algebraic-number-theory







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    If "$a,b$ are integers", what does it have to do with $mathbb{Q}_3$?
    $endgroup$
    – metamorphy
    Dec 29 '18 at 21:05










  • $begingroup$
    @metamorphy I’m not sure I follow what you are asking, but I the question is trivial by unique factorization if we are in $mathbb{Q}$, and maybe it’s also trivial in $mathbb{Q}_3$ but I don’t see why
    $endgroup$
    – usr0192
    Dec 29 '18 at 21:10






  • 1




    $begingroup$
    This is an equation in $mathbb{Q}$ ($=mathbb{Q}capmathbb{Q}_3$). As you know the solution in $mathbb{Q}$... you're done.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 21:29












  • $begingroup$
    Yeah, i was being stupid
    $endgroup$
    – usr0192
    Dec 29 '18 at 21:29














2












2








2


1



$begingroup$


Why do $7, 13$ generate a rank $2$ subgroup of the group of units of $mathbb{Q}_3$?
I.e. if $7^a= 13^b$ in the local field $mathbb{Q}_3$ and $a, b$ are integers, then $(a, b)=(0,0)$.
(This claim is made on p.75 of Washington’s Introduction to Cyclotomic Fields)






algebraic-number-theory







share|cite|improve this question









$endgroup$




Why do $7, 13$ generate a rank $2$ subgroup of the group of units of $mathbb{Q}_3$?
I.e. if $7^a= 13^b$ in the local field $mathbb{Q}_3$ and $a, b$ are integers, then $(a, b)=(0,0)$.
(This claim is made on p.75 of Washington’s Introduction to Cyclotomic Fields)






algebraic-number-theory




algebraic-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 21:01









usr0192usr0192

1,209412




1,209412








  • 3




    $begingroup$
    If "$a,b$ are integers", what does it have to do with $mathbb{Q}_3$?
    $endgroup$
    – metamorphy
    Dec 29 '18 at 21:05










  • $begingroup$
    @metamorphy I’m not sure I follow what you are asking, but I the question is trivial by unique factorization if we are in $mathbb{Q}$, and maybe it’s also trivial in $mathbb{Q}_3$ but I don’t see why
    $endgroup$
    – usr0192
    Dec 29 '18 at 21:10






  • 1




    $begingroup$
    This is an equation in $mathbb{Q}$ ($=mathbb{Q}capmathbb{Q}_3$). As you know the solution in $mathbb{Q}$... you're done.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 21:29












  • $begingroup$
    Yeah, i was being stupid
    $endgroup$
    – usr0192
    Dec 29 '18 at 21:29














  • 3




    $begingroup$
    If "$a,b$ are integers", what does it have to do with $mathbb{Q}_3$?
    $endgroup$
    – metamorphy
    Dec 29 '18 at 21:05










  • $begingroup$
    @metamorphy I’m not sure I follow what you are asking, but I the question is trivial by unique factorization if we are in $mathbb{Q}$, and maybe it’s also trivial in $mathbb{Q}_3$ but I don’t see why
    $endgroup$
    – usr0192
    Dec 29 '18 at 21:10






  • 1




    $begingroup$
    This is an equation in $mathbb{Q}$ ($=mathbb{Q}capmathbb{Q}_3$). As you know the solution in $mathbb{Q}$... you're done.
    $endgroup$
    – metamorphy
    Dec 29 '18 at 21:29












  • $begingroup$
    Yeah, i was being stupid
    $endgroup$
    – usr0192
    Dec 29 '18 at 21:29








3




3




$begingroup$
If "$a,b$ are integers", what does it have to do with $mathbb{Q}_3$?
$endgroup$
– metamorphy
Dec 29 '18 at 21:05




$begingroup$
If "$a,b$ are integers", what does it have to do with $mathbb{Q}_3$?
$endgroup$
– metamorphy
Dec 29 '18 at 21:05












$begingroup$
@metamorphy I’m not sure I follow what you are asking, but I the question is trivial by unique factorization if we are in $mathbb{Q}$, and maybe it’s also trivial in $mathbb{Q}_3$ but I don’t see why
$endgroup$
– usr0192
Dec 29 '18 at 21:10




$begingroup$
@metamorphy I’m not sure I follow what you are asking, but I the question is trivial by unique factorization if we are in $mathbb{Q}$, and maybe it’s also trivial in $mathbb{Q}_3$ but I don’t see why
$endgroup$
– usr0192
Dec 29 '18 at 21:10




1




1




$begingroup$
This is an equation in $mathbb{Q}$ ($=mathbb{Q}capmathbb{Q}_3$). As you know the solution in $mathbb{Q}$... you're done.
$endgroup$
– metamorphy
Dec 29 '18 at 21:29






$begingroup$
This is an equation in $mathbb{Q}$ ($=mathbb{Q}capmathbb{Q}_3$). As you know the solution in $mathbb{Q}$... you're done.
$endgroup$
– metamorphy
Dec 29 '18 at 21:29














$begingroup$
Yeah, i was being stupid
$endgroup$
– usr0192
Dec 29 '18 at 21:29




$begingroup$
Yeah, i was being stupid
$endgroup$
– usr0192
Dec 29 '18 at 21:29










2 Answers
2






active

oldest

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1












$begingroup$

If $7^a=13^b$ in $mathbb{Q}_3$, then $7^{-a}=13^{-b}$ so we may assume $a$ nonnegative. If $b>0$ then $7^a=13^b$ in $mathbb{Z}$, which is impossible; if $b<0$ then $7^a 13^{-b}=1$, again in $mathbb{Z}$, which is again impossible. Thus $b=0$, and therefore $a=0$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Alternatively, since the claim is true in $Bbb Q$, and since $Bbb Q^times toBbb Q^times_p$ is an injection, it’s true in $Bbb Q_p$ as well.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      active

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      1












      $begingroup$

      If $7^a=13^b$ in $mathbb{Q}_3$, then $7^{-a}=13^{-b}$ so we may assume $a$ nonnegative. If $b>0$ then $7^a=13^b$ in $mathbb{Z}$, which is impossible; if $b<0$ then $7^a 13^{-b}=1$, again in $mathbb{Z}$, which is again impossible. Thus $b=0$, and therefore $a=0$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        If $7^a=13^b$ in $mathbb{Q}_3$, then $7^{-a}=13^{-b}$ so we may assume $a$ nonnegative. If $b>0$ then $7^a=13^b$ in $mathbb{Z}$, which is impossible; if $b<0$ then $7^a 13^{-b}=1$, again in $mathbb{Z}$, which is again impossible. Thus $b=0$, and therefore $a=0$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          If $7^a=13^b$ in $mathbb{Q}_3$, then $7^{-a}=13^{-b}$ so we may assume $a$ nonnegative. If $b>0$ then $7^a=13^b$ in $mathbb{Z}$, which is impossible; if $b<0$ then $7^a 13^{-b}=1$, again in $mathbb{Z}$, which is again impossible. Thus $b=0$, and therefore $a=0$.






          share|cite|improve this answer









          $endgroup$



          If $7^a=13^b$ in $mathbb{Q}_3$, then $7^{-a}=13^{-b}$ so we may assume $a$ nonnegative. If $b>0$ then $7^a=13^b$ in $mathbb{Z}$, which is impossible; if $b<0$ then $7^a 13^{-b}=1$, again in $mathbb{Z}$, which is again impossible. Thus $b=0$, and therefore $a=0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 29 '18 at 21:23









          metamorphymetamorphy

          3,7021721




          3,7021721























              1












              $begingroup$

              Alternatively, since the claim is true in $Bbb Q$, and since $Bbb Q^times toBbb Q^times_p$ is an injection, it’s true in $Bbb Q_p$ as well.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Alternatively, since the claim is true in $Bbb Q$, and since $Bbb Q^times toBbb Q^times_p$ is an injection, it’s true in $Bbb Q_p$ as well.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Alternatively, since the claim is true in $Bbb Q$, and since $Bbb Q^times toBbb Q^times_p$ is an injection, it’s true in $Bbb Q_p$ as well.






                  share|cite|improve this answer









                  $endgroup$



                  Alternatively, since the claim is true in $Bbb Q$, and since $Bbb Q^times toBbb Q^times_p$ is an injection, it’s true in $Bbb Q_p$ as well.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 29 '18 at 21:41









                  LubinLubin

                  45.3k44688




                  45.3k44688






























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