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

votes


















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









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056258%2f7-13-independent-units-in-the-local-field-mathbbq-3%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056258%2f7-13-independent-units-in-the-local-field-mathbbq-3%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Wiesbaden

                      Image
                      Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
                      Read more

                      Marschland

                      Image
                      Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
                      Read more

                      Dieringhausen

                      Image
                      Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                      Read more