Extending an automorphism from $mathbb{Q}(sqrt{2})$ to an automorphism of $mathbb{Q}(sqrt[4]{2})$












1












$begingroup$


I am reading an introduction to abstract algebra by Keith Nicholson. There is a theorem that goes like this :




Let $sigma : F rightarrow{overline{F}}$ be an isomorphism of
fields, let $f in F[x]$ be a nonconstant polynomial, and let
$f^{sigma}$ be its image by $sigma$. If $E supset F$ is a splitting
field for $f$ and $bar{E} supset bar{F}$ is a splitting field for
$f^{sigma}$, there is an isomorphism $E rightarrow bar{E}$ that
extends $sigma$




But if I consider the extension $mathbb{Q}(sqrt[4]{2})/mathbb{Q}(sqrt{2})$, where we have that $mathbb{Q}(sqrt[4]{2})$ is the splitting field of $x^2-sqrt{2}$ over $mathbb{Q}(sqrt{2})$. Then consider the isomorphism $sigma : mathbb{Q}(sqrt{2}) rightarrow mathbb{Q}(sqrt{2})$ that sends $sqrt{2} mapsto -sqrt{2}$. So the theorem tells us that $sigma$ extends to an automorphism $hat{sigma}$ of $mathbb{Q}(sqrt[4]{2})$. But we we have to have $hat{sigma} : sqrt[4]{2} mapsto pm sqrt[4]{2}$ but then $sigma(sqrt{2})=sqrt{2}$ (so $hat{sigma}$ doesn't actually extend $sigma$, no?). What am I doing wrong here?






extension-field







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am reading an introduction to abstract algebra by Keith Nicholson. There is a theorem that goes like this :




    Let $sigma : F rightarrow{overline{F}}$ be an isomorphism of
    fields, let $f in F[x]$ be a nonconstant polynomial, and let
    $f^{sigma}$ be its image by $sigma$. If $E supset F$ is a splitting
    field for $f$ and $bar{E} supset bar{F}$ is a splitting field for
    $f^{sigma}$, there is an isomorphism $E rightarrow bar{E}$ that
    extends $sigma$




    But if I consider the extension $mathbb{Q}(sqrt[4]{2})/mathbb{Q}(sqrt{2})$, where we have that $mathbb{Q}(sqrt[4]{2})$ is the splitting field of $x^2-sqrt{2}$ over $mathbb{Q}(sqrt{2})$. Then consider the isomorphism $sigma : mathbb{Q}(sqrt{2}) rightarrow mathbb{Q}(sqrt{2})$ that sends $sqrt{2} mapsto -sqrt{2}$. So the theorem tells us that $sigma$ extends to an automorphism $hat{sigma}$ of $mathbb{Q}(sqrt[4]{2})$. But we we have to have $hat{sigma} : sqrt[4]{2} mapsto pm sqrt[4]{2}$ but then $sigma(sqrt{2})=sqrt{2}$ (so $hat{sigma}$ doesn't actually extend $sigma$, no?). What am I doing wrong here?






    extension-field







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am reading an introduction to abstract algebra by Keith Nicholson. There is a theorem that goes like this :




      Let $sigma : F rightarrow{overline{F}}$ be an isomorphism of
      fields, let $f in F[x]$ be a nonconstant polynomial, and let
      $f^{sigma}$ be its image by $sigma$. If $E supset F$ is a splitting
      field for $f$ and $bar{E} supset bar{F}$ is a splitting field for
      $f^{sigma}$, there is an isomorphism $E rightarrow bar{E}$ that
      extends $sigma$




      But if I consider the extension $mathbb{Q}(sqrt[4]{2})/mathbb{Q}(sqrt{2})$, where we have that $mathbb{Q}(sqrt[4]{2})$ is the splitting field of $x^2-sqrt{2}$ over $mathbb{Q}(sqrt{2})$. Then consider the isomorphism $sigma : mathbb{Q}(sqrt{2}) rightarrow mathbb{Q}(sqrt{2})$ that sends $sqrt{2} mapsto -sqrt{2}$. So the theorem tells us that $sigma$ extends to an automorphism $hat{sigma}$ of $mathbb{Q}(sqrt[4]{2})$. But we we have to have $hat{sigma} : sqrt[4]{2} mapsto pm sqrt[4]{2}$ but then $sigma(sqrt{2})=sqrt{2}$ (so $hat{sigma}$ doesn't actually extend $sigma$, no?). What am I doing wrong here?






      extension-field







      share|cite|improve this question









      $endgroup$




      I am reading an introduction to abstract algebra by Keith Nicholson. There is a theorem that goes like this :




      Let $sigma : F rightarrow{overline{F}}$ be an isomorphism of
      fields, let $f in F[x]$ be a nonconstant polynomial, and let
      $f^{sigma}$ be its image by $sigma$. If $E supset F$ is a splitting
      field for $f$ and $bar{E} supset bar{F}$ is a splitting field for
      $f^{sigma}$, there is an isomorphism $E rightarrow bar{E}$ that
      extends $sigma$




      But if I consider the extension $mathbb{Q}(sqrt[4]{2})/mathbb{Q}(sqrt{2})$, where we have that $mathbb{Q}(sqrt[4]{2})$ is the splitting field of $x^2-sqrt{2}$ over $mathbb{Q}(sqrt{2})$. Then consider the isomorphism $sigma : mathbb{Q}(sqrt{2}) rightarrow mathbb{Q}(sqrt{2})$ that sends $sqrt{2} mapsto -sqrt{2}$. So the theorem tells us that $sigma$ extends to an automorphism $hat{sigma}$ of $mathbb{Q}(sqrt[4]{2})$. But we we have to have $hat{sigma} : sqrt[4]{2} mapsto pm sqrt[4]{2}$ but then $sigma(sqrt{2})=sqrt{2}$ (so $hat{sigma}$ doesn't actually extend $sigma$, no?). What am I doing wrong here?






      extension-field




      extension-field






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 2 at 20:59









      roi_saumonroi_saumon

      45628




      45628






















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          Your mistake is expecting $sigma$ to extend to an automorphism of $mathbb{Q}(sqrt[4]{2})$. The theorem doesn't say that $sigma$ extends to an automorphism of $E$, it says that it extends to an isomorphism $Eto overline{E}$. And $E$ and $overline{E}$ are splitting fields of different polynomials, namely $f$ and $f^sigma$.



          We have $F = overline{F} = mathbb{Q}(sqrt{2})$, $sigma$ the automorphism of $F$ determined by $sqrt{2}to -sqrt{2}$, and $f$ the polynomial $x^2-sqrt{2}$. Then $f^sigma$ is $x^2 + sqrt{2}$. The splitting field $E$ of $f$ over $F$ is $mathbb{Q}(sqrt[4]{2})$, and the splitting field $overline{E}$ of $f^sigma$ over $overline{F}$ is $mathbb{Q}(sqrt[4]{2}i)$.



          And indeed, $E$ is isomorphic to $overline{E}$ by a map $widehat{sigma}$ determined by $sqrt[4]{2}mapsto sqrt[4]{2}i$ (note that $sqrt[4]{2}mapsto -sqrt[4]{2}i$ would also work). As expected, $widehat{sigma}$ extends $sigma$, since $$widehat{sigma}(sqrt{2}) = widehat{sigma}((sqrt[4]{2})^2) = (widehat{sigma}(sqrt[4]{2}))^2 = (sqrt[4]{2}i)^2 = -sqrt{2}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh, this is so subtle, thanks!
            $endgroup$
            – roi_saumon
            Jan 2 at 22:53











          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%2f3059970%2fextending-an-automorphism-from-mathbbq-sqrt2-to-an-automorphism-of-ma%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          Your mistake is expecting $sigma$ to extend to an automorphism of $mathbb{Q}(sqrt[4]{2})$. The theorem doesn't say that $sigma$ extends to an automorphism of $E$, it says that it extends to an isomorphism $Eto overline{E}$. And $E$ and $overline{E}$ are splitting fields of different polynomials, namely $f$ and $f^sigma$.



          We have $F = overline{F} = mathbb{Q}(sqrt{2})$, $sigma$ the automorphism of $F$ determined by $sqrt{2}to -sqrt{2}$, and $f$ the polynomial $x^2-sqrt{2}$. Then $f^sigma$ is $x^2 + sqrt{2}$. The splitting field $E$ of $f$ over $F$ is $mathbb{Q}(sqrt[4]{2})$, and the splitting field $overline{E}$ of $f^sigma$ over $overline{F}$ is $mathbb{Q}(sqrt[4]{2}i)$.



          And indeed, $E$ is isomorphic to $overline{E}$ by a map $widehat{sigma}$ determined by $sqrt[4]{2}mapsto sqrt[4]{2}i$ (note that $sqrt[4]{2}mapsto -sqrt[4]{2}i$ would also work). As expected, $widehat{sigma}$ extends $sigma$, since $$widehat{sigma}(sqrt{2}) = widehat{sigma}((sqrt[4]{2})^2) = (widehat{sigma}(sqrt[4]{2}))^2 = (sqrt[4]{2}i)^2 = -sqrt{2}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh, this is so subtle, thanks!
            $endgroup$
            – roi_saumon
            Jan 2 at 22:53
















          5












          $begingroup$

          Your mistake is expecting $sigma$ to extend to an automorphism of $mathbb{Q}(sqrt[4]{2})$. The theorem doesn't say that $sigma$ extends to an automorphism of $E$, it says that it extends to an isomorphism $Eto overline{E}$. And $E$ and $overline{E}$ are splitting fields of different polynomials, namely $f$ and $f^sigma$.



          We have $F = overline{F} = mathbb{Q}(sqrt{2})$, $sigma$ the automorphism of $F$ determined by $sqrt{2}to -sqrt{2}$, and $f$ the polynomial $x^2-sqrt{2}$. Then $f^sigma$ is $x^2 + sqrt{2}$. The splitting field $E$ of $f$ over $F$ is $mathbb{Q}(sqrt[4]{2})$, and the splitting field $overline{E}$ of $f^sigma$ over $overline{F}$ is $mathbb{Q}(sqrt[4]{2}i)$.



          And indeed, $E$ is isomorphic to $overline{E}$ by a map $widehat{sigma}$ determined by $sqrt[4]{2}mapsto sqrt[4]{2}i$ (note that $sqrt[4]{2}mapsto -sqrt[4]{2}i$ would also work). As expected, $widehat{sigma}$ extends $sigma$, since $$widehat{sigma}(sqrt{2}) = widehat{sigma}((sqrt[4]{2})^2) = (widehat{sigma}(sqrt[4]{2}))^2 = (sqrt[4]{2}i)^2 = -sqrt{2}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh, this is so subtle, thanks!
            $endgroup$
            – roi_saumon
            Jan 2 at 22:53














          5












          5








          5





          $begingroup$

          Your mistake is expecting $sigma$ to extend to an automorphism of $mathbb{Q}(sqrt[4]{2})$. The theorem doesn't say that $sigma$ extends to an automorphism of $E$, it says that it extends to an isomorphism $Eto overline{E}$. And $E$ and $overline{E}$ are splitting fields of different polynomials, namely $f$ and $f^sigma$.



          We have $F = overline{F} = mathbb{Q}(sqrt{2})$, $sigma$ the automorphism of $F$ determined by $sqrt{2}to -sqrt{2}$, and $f$ the polynomial $x^2-sqrt{2}$. Then $f^sigma$ is $x^2 + sqrt{2}$. The splitting field $E$ of $f$ over $F$ is $mathbb{Q}(sqrt[4]{2})$, and the splitting field $overline{E}$ of $f^sigma$ over $overline{F}$ is $mathbb{Q}(sqrt[4]{2}i)$.



          And indeed, $E$ is isomorphic to $overline{E}$ by a map $widehat{sigma}$ determined by $sqrt[4]{2}mapsto sqrt[4]{2}i$ (note that $sqrt[4]{2}mapsto -sqrt[4]{2}i$ would also work). As expected, $widehat{sigma}$ extends $sigma$, since $$widehat{sigma}(sqrt{2}) = widehat{sigma}((sqrt[4]{2})^2) = (widehat{sigma}(sqrt[4]{2}))^2 = (sqrt[4]{2}i)^2 = -sqrt{2}.$$






          share|cite|improve this answer











          $endgroup$



          Your mistake is expecting $sigma$ to extend to an automorphism of $mathbb{Q}(sqrt[4]{2})$. The theorem doesn't say that $sigma$ extends to an automorphism of $E$, it says that it extends to an isomorphism $Eto overline{E}$. And $E$ and $overline{E}$ are splitting fields of different polynomials, namely $f$ and $f^sigma$.



          We have $F = overline{F} = mathbb{Q}(sqrt{2})$, $sigma$ the automorphism of $F$ determined by $sqrt{2}to -sqrt{2}$, and $f$ the polynomial $x^2-sqrt{2}$. Then $f^sigma$ is $x^2 + sqrt{2}$. The splitting field $E$ of $f$ over $F$ is $mathbb{Q}(sqrt[4]{2})$, and the splitting field $overline{E}$ of $f^sigma$ over $overline{F}$ is $mathbb{Q}(sqrt[4]{2}i)$.



          And indeed, $E$ is isomorphic to $overline{E}$ by a map $widehat{sigma}$ determined by $sqrt[4]{2}mapsto sqrt[4]{2}i$ (note that $sqrt[4]{2}mapsto -sqrt[4]{2}i$ would also work). As expected, $widehat{sigma}$ extends $sigma$, since $$widehat{sigma}(sqrt{2}) = widehat{sigma}((sqrt[4]{2})^2) = (widehat{sigma}(sqrt[4]{2}))^2 = (sqrt[4]{2}i)^2 = -sqrt{2}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 21:20

























          answered Jan 2 at 21:14









          Alex KruckmanAlex Kruckman

          26.9k22556




          26.9k22556












          • $begingroup$
            Oh, this is so subtle, thanks!
            $endgroup$
            – roi_saumon
            Jan 2 at 22:53


















          • $begingroup$
            Oh, this is so subtle, thanks!
            $endgroup$
            – roi_saumon
            Jan 2 at 22:53
















          $begingroup$
          Oh, this is so subtle, thanks!
          $endgroup$
          – roi_saumon
          Jan 2 at 22:53




          $begingroup$
          Oh, this is so subtle, thanks!
          $endgroup$
          – roi_saumon
          Jan 2 at 22:53


















          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%2f3059970%2fextending-an-automorphism-from-mathbbq-sqrt2-to-an-automorphism-of-ma%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

          MongoDB - Not Authorized To Execute Command

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
          Read more

          How to fix TextFormField cause rebuild widget in Flutter

          Image
          .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
          Read more

          in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

          Image
          1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
          Read more