Why must an automorphism of an extension of $mathbb{Q}$ send 1 to a rational number?












2












$begingroup$


This is from a video I was watching that claimed this:




If $phi$ is an automorphism of an extension field $F$ of $mathbb{Q}$, then $phi(q)=q$ for all $qinmathbb{Q}$.




The proof started by assuming that $phi(1)=q$ for some $qinmathbb{Q}$. I understand the stuff that followed. But the explanation never said why we don't have to consider the case in which $phi$ sends $1$ to something in $Fsetminusmathbb{Q}$.



How come we know an automorphism of an extension of $mathbb{Q}$ must send $1$ to something in $mathbb{Q}$ to begin with?



(Thanks in advance.)






field-theory extension-field







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
    $endgroup$
    – J. W. Tanner
    Jan 10 at 23:54










  • $begingroup$
    $phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
    $endgroup$
    – reuns
    Jan 10 at 23:58
















2












$begingroup$


This is from a video I was watching that claimed this:




If $phi$ is an automorphism of an extension field $F$ of $mathbb{Q}$, then $phi(q)=q$ for all $qinmathbb{Q}$.




The proof started by assuming that $phi(1)=q$ for some $qinmathbb{Q}$. I understand the stuff that followed. But the explanation never said why we don't have to consider the case in which $phi$ sends $1$ to something in $Fsetminusmathbb{Q}$.



How come we know an automorphism of an extension of $mathbb{Q}$ must send $1$ to something in $mathbb{Q}$ to begin with?



(Thanks in advance.)






field-theory extension-field







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
    $endgroup$
    – J. W. Tanner
    Jan 10 at 23:54










  • $begingroup$
    $phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
    $endgroup$
    – reuns
    Jan 10 at 23:58














2












2








2


2



$begingroup$


This is from a video I was watching that claimed this:




If $phi$ is an automorphism of an extension field $F$ of $mathbb{Q}$, then $phi(q)=q$ for all $qinmathbb{Q}$.




The proof started by assuming that $phi(1)=q$ for some $qinmathbb{Q}$. I understand the stuff that followed. But the explanation never said why we don't have to consider the case in which $phi$ sends $1$ to something in $Fsetminusmathbb{Q}$.



How come we know an automorphism of an extension of $mathbb{Q}$ must send $1$ to something in $mathbb{Q}$ to begin with?



(Thanks in advance.)






field-theory extension-field







share|cite|improve this question









$endgroup$




This is from a video I was watching that claimed this:




If $phi$ is an automorphism of an extension field $F$ of $mathbb{Q}$, then $phi(q)=q$ for all $qinmathbb{Q}$.




The proof started by assuming that $phi(1)=q$ for some $qinmathbb{Q}$. I understand the stuff that followed. But the explanation never said why we don't have to consider the case in which $phi$ sends $1$ to something in $Fsetminusmathbb{Q}$.



How come we know an automorphism of an extension of $mathbb{Q}$ must send $1$ to something in $mathbb{Q}$ to begin with?



(Thanks in advance.)






field-theory extension-field




field-theory extension-field






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 10 at 23:46









anonanon444anonanon444

1756




1756








  • 3




    $begingroup$
    Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
    $endgroup$
    – J. W. Tanner
    Jan 10 at 23:54










  • $begingroup$
    $phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
    $endgroup$
    – reuns
    Jan 10 at 23:58














  • 3




    $begingroup$
    Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
    $endgroup$
    – J. W. Tanner
    Jan 10 at 23:54










  • $begingroup$
    $phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
    $endgroup$
    – reuns
    Jan 10 at 23:58








3




3




$begingroup$
Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
$endgroup$
– J. W. Tanner
Jan 10 at 23:54




$begingroup$
Doesn't a field automorphism have to map the multiplicative identity to the multiplicative identity?
$endgroup$
– J. W. Tanner
Jan 10 at 23:54












$begingroup$
$phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
$endgroup$
– reuns
Jan 10 at 23:58




$begingroup$
$phi(1) = phi(1/1) = phi(1)/ phi(1)=1$
$endgroup$
– reuns
Jan 10 at 23:58










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let



$phi:F to F tag 1$



be any field automorphism of the field



$F supset Bbb Q; tag 2$



then



$(phi(1))^2 = phi(1) phi(1) = phi(1^2) = phi(1), tag 3$



which implies



$phi(1) in {0, 1 }; tag 4$



now



$phi(1) = 0 Longrightarrow phi(c) = phi(c cdot 1) = phi(c)phi(1) = phi(c) cdot 0 = 0, tag 5$



which we rule out since $phi$ is a field automorphism; thus we see just why



$phi(1) = 1; tag 6$



it follows that



$n in Bbb N Longrightarrow phi(n) = phi(underbrace{1 + 1 + ldots + 1}_{text{ n times}}) = underbrace{phi(1) + phi(1) + ldots + phi(1)}_{text{n times}} = n; tag 7$



and also for $n in N$,



$n + (-n) = 0 Longrightarrow phi(n) + phi(-n) = 0 Longrightarrow phi(-n) = -phi(n), tag 8$



since in addition $phi(0) = 0$ (always!), it follows from (7) and (8) that



$z in Bbb Z Longrightarrow phi(z) = z; tag 9$



thus, for



$dfrac{r}{s} in Bbb Q, ; r, s in Bbb Z, tag{10}$



$s phi left (dfrac{r}{s} right ) = phi(s)phi left (dfrac{r}{s} right ) = phi left ( sdfrac{r}{s} right ) = phi(r) = r, tag{11}$



whence



$phi left (dfrac{r}{s} right ) = dfrac{r}{s}, tag{12}$



and we have shown that



$forall q in Bbb Q, ; phi(q) = q, tag{13}$



as per request. $OEDelta$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    I have no idea why specifically $qin Bbb Q$. However, we know any homomorphism sends $1$ to an idempotent element. In a field there are only two idempotent elements. So $phi(1)=1$ is quite easy to prove. And from there $phi(q)=q$ for any $qin Bbb Q$ follows.






    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%2f3069325%2fwhy-must-an-automorphism-of-an-extension-of-mathbbq-send-1-to-a-rational-nu%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









      2












      $begingroup$

      Let



      $phi:F to F tag 1$



      be any field automorphism of the field



      $F supset Bbb Q; tag 2$



      then



      $(phi(1))^2 = phi(1) phi(1) = phi(1^2) = phi(1), tag 3$



      which implies



      $phi(1) in {0, 1 }; tag 4$



      now



      $phi(1) = 0 Longrightarrow phi(c) = phi(c cdot 1) = phi(c)phi(1) = phi(c) cdot 0 = 0, tag 5$



      which we rule out since $phi$ is a field automorphism; thus we see just why



      $phi(1) = 1; tag 6$



      it follows that



      $n in Bbb N Longrightarrow phi(n) = phi(underbrace{1 + 1 + ldots + 1}_{text{ n times}}) = underbrace{phi(1) + phi(1) + ldots + phi(1)}_{text{n times}} = n; tag 7$



      and also for $n in N$,



      $n + (-n) = 0 Longrightarrow phi(n) + phi(-n) = 0 Longrightarrow phi(-n) = -phi(n), tag 8$



      since in addition $phi(0) = 0$ (always!), it follows from (7) and (8) that



      $z in Bbb Z Longrightarrow phi(z) = z; tag 9$



      thus, for



      $dfrac{r}{s} in Bbb Q, ; r, s in Bbb Z, tag{10}$



      $s phi left (dfrac{r}{s} right ) = phi(s)phi left (dfrac{r}{s} right ) = phi left ( sdfrac{r}{s} right ) = phi(r) = r, tag{11}$



      whence



      $phi left (dfrac{r}{s} right ) = dfrac{r}{s}, tag{12}$



      and we have shown that



      $forall q in Bbb Q, ; phi(q) = q, tag{13}$



      as per request. $OEDelta$.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Let



        $phi:F to F tag 1$



        be any field automorphism of the field



        $F supset Bbb Q; tag 2$



        then



        $(phi(1))^2 = phi(1) phi(1) = phi(1^2) = phi(1), tag 3$



        which implies



        $phi(1) in {0, 1 }; tag 4$



        now



        $phi(1) = 0 Longrightarrow phi(c) = phi(c cdot 1) = phi(c)phi(1) = phi(c) cdot 0 = 0, tag 5$



        which we rule out since $phi$ is a field automorphism; thus we see just why



        $phi(1) = 1; tag 6$



        it follows that



        $n in Bbb N Longrightarrow phi(n) = phi(underbrace{1 + 1 + ldots + 1}_{text{ n times}}) = underbrace{phi(1) + phi(1) + ldots + phi(1)}_{text{n times}} = n; tag 7$



        and also for $n in N$,



        $n + (-n) = 0 Longrightarrow phi(n) + phi(-n) = 0 Longrightarrow phi(-n) = -phi(n), tag 8$



        since in addition $phi(0) = 0$ (always!), it follows from (7) and (8) that



        $z in Bbb Z Longrightarrow phi(z) = z; tag 9$



        thus, for



        $dfrac{r}{s} in Bbb Q, ; r, s in Bbb Z, tag{10}$



        $s phi left (dfrac{r}{s} right ) = phi(s)phi left (dfrac{r}{s} right ) = phi left ( sdfrac{r}{s} right ) = phi(r) = r, tag{11}$



        whence



        $phi left (dfrac{r}{s} right ) = dfrac{r}{s}, tag{12}$



        and we have shown that



        $forall q in Bbb Q, ; phi(q) = q, tag{13}$



        as per request. $OEDelta$.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Let



          $phi:F to F tag 1$



          be any field automorphism of the field



          $F supset Bbb Q; tag 2$



          then



          $(phi(1))^2 = phi(1) phi(1) = phi(1^2) = phi(1), tag 3$



          which implies



          $phi(1) in {0, 1 }; tag 4$



          now



          $phi(1) = 0 Longrightarrow phi(c) = phi(c cdot 1) = phi(c)phi(1) = phi(c) cdot 0 = 0, tag 5$



          which we rule out since $phi$ is a field automorphism; thus we see just why



          $phi(1) = 1; tag 6$



          it follows that



          $n in Bbb N Longrightarrow phi(n) = phi(underbrace{1 + 1 + ldots + 1}_{text{ n times}}) = underbrace{phi(1) + phi(1) + ldots + phi(1)}_{text{n times}} = n; tag 7$



          and also for $n in N$,



          $n + (-n) = 0 Longrightarrow phi(n) + phi(-n) = 0 Longrightarrow phi(-n) = -phi(n), tag 8$



          since in addition $phi(0) = 0$ (always!), it follows from (7) and (8) that



          $z in Bbb Z Longrightarrow phi(z) = z; tag 9$



          thus, for



          $dfrac{r}{s} in Bbb Q, ; r, s in Bbb Z, tag{10}$



          $s phi left (dfrac{r}{s} right ) = phi(s)phi left (dfrac{r}{s} right ) = phi left ( sdfrac{r}{s} right ) = phi(r) = r, tag{11}$



          whence



          $phi left (dfrac{r}{s} right ) = dfrac{r}{s}, tag{12}$



          and we have shown that



          $forall q in Bbb Q, ; phi(q) = q, tag{13}$



          as per request. $OEDelta$.






          share|cite|improve this answer











          $endgroup$



          Let



          $phi:F to F tag 1$



          be any field automorphism of the field



          $F supset Bbb Q; tag 2$



          then



          $(phi(1))^2 = phi(1) phi(1) = phi(1^2) = phi(1), tag 3$



          which implies



          $phi(1) in {0, 1 }; tag 4$



          now



          $phi(1) = 0 Longrightarrow phi(c) = phi(c cdot 1) = phi(c)phi(1) = phi(c) cdot 0 = 0, tag 5$



          which we rule out since $phi$ is a field automorphism; thus we see just why



          $phi(1) = 1; tag 6$



          it follows that



          $n in Bbb N Longrightarrow phi(n) = phi(underbrace{1 + 1 + ldots + 1}_{text{ n times}}) = underbrace{phi(1) + phi(1) + ldots + phi(1)}_{text{n times}} = n; tag 7$



          and also for $n in N$,



          $n + (-n) = 0 Longrightarrow phi(n) + phi(-n) = 0 Longrightarrow phi(-n) = -phi(n), tag 8$



          since in addition $phi(0) = 0$ (always!), it follows from (7) and (8) that



          $z in Bbb Z Longrightarrow phi(z) = z; tag 9$



          thus, for



          $dfrac{r}{s} in Bbb Q, ; r, s in Bbb Z, tag{10}$



          $s phi left (dfrac{r}{s} right ) = phi(s)phi left (dfrac{r}{s} right ) = phi left ( sdfrac{r}{s} right ) = phi(r) = r, tag{11}$



          whence



          $phi left (dfrac{r}{s} right ) = dfrac{r}{s}, tag{12}$



          and we have shown that



          $forall q in Bbb Q, ; phi(q) = q, tag{13}$



          as per request. $OEDelta$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 11 at 3:46

























          answered Jan 11 at 0:35









          Robert LewisRobert Lewis

          47.7k23067




          47.7k23067























              0












              $begingroup$

              I have no idea why specifically $qin Bbb Q$. However, we know any homomorphism sends $1$ to an idempotent element. In a field there are only two idempotent elements. So $phi(1)=1$ is quite easy to prove. And from there $phi(q)=q$ for any $qin Bbb Q$ follows.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                I have no idea why specifically $qin Bbb Q$. However, we know any homomorphism sends $1$ to an idempotent element. In a field there are only two idempotent elements. So $phi(1)=1$ is quite easy to prove. And from there $phi(q)=q$ for any $qin Bbb Q$ follows.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I have no idea why specifically $qin Bbb Q$. However, we know any homomorphism sends $1$ to an idempotent element. In a field there are only two idempotent elements. So $phi(1)=1$ is quite easy to prove. And from there $phi(q)=q$ for any $qin Bbb Q$ follows.






                  share|cite|improve this answer









                  $endgroup$



                  I have no idea why specifically $qin Bbb Q$. However, we know any homomorphism sends $1$ to an idempotent element. In a field there are only two idempotent elements. So $phi(1)=1$ is quite easy to prove. And from there $phi(q)=q$ for any $qin Bbb Q$ follows.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 10 at 23:49









                  ArthurArthur

                  117k7116200




                  117k7116200






























                      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%2f3069325%2fwhy-must-an-automorphism-of-an-extension-of-mathbbq-send-1-to-a-rational-nu%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

                      Human spaceflight

                      Image
                      Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
                      Read more

                      Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                      Image
                      0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
                      Read more

                      張江高科駅

                      Image
                      張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
                      Read more