The algebraic elements of $K$ extension of $F$ are a subfield of $K$












2












$begingroup$


the question I wanna try to answer is this:




The algebraic elements of $K$ extension of $F$ are a subfield of $K$




I know that:



An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.



My approach is that I have two algebraic elements $a,b in K$.
$a+b in K$ and also $ab in K$, and of course they belong to $S$ too.



Maybe I need to write a polynomial with the algebraic elements and show that is finite?



I've some difficult to glue together those pieces for a satisfying proof. Any hints?



Thanks






abstract-algebra







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    the question I wanna try to answer is this:




    The algebraic elements of $K$ extension of $F$ are a subfield of $K$




    I know that:



    An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.



    My approach is that I have two algebraic elements $a,b in K$.
    $a+b in K$ and also $ab in K$, and of course they belong to $S$ too.



    Maybe I need to write a polynomial with the algebraic elements and show that is finite?



    I've some difficult to glue together those pieces for a satisfying proof. Any hints?



    Thanks






    abstract-algebra







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      the question I wanna try to answer is this:




      The algebraic elements of $K$ extension of $F$ are a subfield of $K$




      I know that:



      An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.



      My approach is that I have two algebraic elements $a,b in K$.
      $a+b in K$ and also $ab in K$, and of course they belong to $S$ too.



      Maybe I need to write a polynomial with the algebraic elements and show that is finite?



      I've some difficult to glue together those pieces for a satisfying proof. Any hints?



      Thanks






      abstract-algebra







      share|cite|improve this question











      $endgroup$




      the question I wanna try to answer is this:




      The algebraic elements of $K$ extension of $F$ are a subfield of $K$




      I know that:



      An element $a in K$ is algebraic on $F$ if $f(a)=0$ for some polynomial $not=0$. Let $S$ be the subfield of algebraic elements of K. So as definition, $S in K$ because $K$ is an extension of S too. But if it's an algebraic extension, must be closed and of course finite.



      My approach is that I have two algebraic elements $a,b in K$.
      $a+b in K$ and also $ab in K$, and of course they belong to $S$ too.



      Maybe I need to write a polynomial with the algebraic elements and show that is finite?



      I've some difficult to glue together those pieces for a satisfying proof. Any hints?



      Thanks






      abstract-algebra




      abstract-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 10:09









      Bernard

      124k741116




      124k741116










      asked Dec 13 '18 at 9:56









      AlessarAlessar

      313115




      313115






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:22












          • $begingroup$
            I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:27












          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%2f3037823%2fthe-algebraic-elements-of-k-extension-of-f-are-a-subfield-of-k%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









          2












          $begingroup$

          Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:22












          • $begingroup$
            I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:27
















          2












          $begingroup$

          Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:22












          • $begingroup$
            I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:27














          2












          2








          2





          $begingroup$

          Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.






          share|cite|improve this answer









          $endgroup$



          Hint: Prove that $a in K$ is algebraic over $F$ iff $[F(a):F]$ is finite iff $a in L subseteq K$ with $[L:F]$ finite. Then prove that $a,b in K$ algebraic implies $[F(a,b):F]$ finite.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 10:03









          lhflhf

          167k11172404




          167k11172404












          • $begingroup$
            To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:22












          • $begingroup$
            I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:27


















          • $begingroup$
            To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:22












          • $begingroup$
            I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
            $endgroup$
            – Alessar
            Dec 13 '18 at 10:27
















          $begingroup$
          To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
          $endgroup$
          – Alessar
          Dec 13 '18 at 10:22






          $begingroup$
          To prove the first I can write an $f(x) = x^n + a_{n-1}x^{n-1}+...+a_1 x+a_0$ with $a_i$ algebraic elements, so I have a set $L={ a_0,a_1,...,a_{n-1}}$. This set is finite. The solutions of the polynomial lives in $K$ for some $s$ solution(s) of $f(x)$. Then $K(s)$ is a subfield of algebraic elements of $K$. The second assertion is that if we repeat for a $b$, we have anothe extention that is finite, so the proof is complete. Any error?
          $endgroup$
          – Alessar
          Dec 13 '18 at 10:22














          $begingroup$
          I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
          $endgroup$
          – Alessar
          Dec 13 '18 at 10:27




          $begingroup$
          I think that we could prove this also with a $f(x)$ composed of coefficient $a,b$ instead of the all $n-tuple$ set used previously. The idea is the same
          $endgroup$
          – Alessar
          Dec 13 '18 at 10:27


















          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%2f3037823%2fthe-algebraic-elements-of-k-extension-of-f-are-a-subfield-of-k%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

          How to change which sound is reproduced for terminal bell?

          Image
          3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
          Read more

          Can I use Tabulator js library in my java Spring + Thymeleaf project?

          Image
          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
          Read more

          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

          Image
          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
          Read more