how to determine End$(mathbb{Q},+)$












4












$begingroup$


I think every homomorphism is determined by the map of $ operatorname{id} $. So End$(mathbb{Q},+)$ is isomorphic to $mathbb{Q}$ but I am not sure it's right.






abstract-algebra







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Both are correct, so you just need to prove the claims.
    $endgroup$
    – Tobias Kildetoft
    Dec 27 '14 at 15:38
















4












$begingroup$


I think every homomorphism is determined by the map of $ operatorname{id} $. So End$(mathbb{Q},+)$ is isomorphic to $mathbb{Q}$ but I am not sure it's right.






abstract-algebra







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Both are correct, so you just need to prove the claims.
    $endgroup$
    – Tobias Kildetoft
    Dec 27 '14 at 15:38














4












4








4


1



$begingroup$


I think every homomorphism is determined by the map of $ operatorname{id} $. So End$(mathbb{Q},+)$ is isomorphic to $mathbb{Q}$ but I am not sure it's right.






abstract-algebra







share|cite|improve this question











$endgroup$




I think every homomorphism is determined by the map of $ operatorname{id} $. So End$(mathbb{Q},+)$ is isomorphic to $mathbb{Q}$ but I am not sure it's right.






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 Jan 18 at 3:57









user549397

1,5101418




1,5101418










asked Dec 27 '14 at 15:26









litter boylitter boy

261




261








  • 2




    $begingroup$
    Both are correct, so you just need to prove the claims.
    $endgroup$
    – Tobias Kildetoft
    Dec 27 '14 at 15:38














  • 2




    $begingroup$
    Both are correct, so you just need to prove the claims.
    $endgroup$
    – Tobias Kildetoft
    Dec 27 '14 at 15:38








2




2




$begingroup$
Both are correct, so you just need to prove the claims.
$endgroup$
– Tobias Kildetoft
Dec 27 '14 at 15:38




$begingroup$
Both are correct, so you just need to prove the claims.
$endgroup$
– Tobias Kildetoft
Dec 27 '14 at 15:38










2 Answers
2






active

oldest

votes


















2












$begingroup$

Your claims are true; if you want to prove that ever homomorphism is determined by $1$, notice that if $f:mathbb Qrightarrow mathbb Q$ is a homomorphism, then we clearly have $f(-n)=-f(n)$ and $f(n+1)=f(n)+f(1)$. Using these identities we determine the value of $f$ over $mathbb Z$ given the value of $f(1)$. Moreover, we can prove that
$$fleft(underbrace{frac{p}q+ldots+frac{p}q}_{qtext{ times}}right)=underbrace{fleft(frac{p}qright)+ldots+fleft(frac{p}qright)}_{qtext{ times}}$$
where the left hand side equals $f(p)$, which is an integer and thus already determined and, knowing that there is only one rational solution to
$$c=underbrace{r+ldots+r}_{ntext{ times}}$$
suffices to determine $fleft(frac{p}qright)$. From here, since $f(1)$ could be any rational and $(f+g)(1)$ would be the sum of the associated rational values of $f$ and $g$, it's clear that End($mathbb Q,+$) is isomorphic to $mathbb Q$, since the map $fmapsto f(1)$ is an isomorphism between the two.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Define $phi: mathbb{Q} rightarrow mathbb{Q}, 1 mapsto r.$ Then for each $n in mathbb{N}, phi(n) = nr.$ Also $phi(-n) = -phi(n) = -nr.$ Thus, $phi(n) = nr, forall n in mathbb{Z}.$ Take $frac{m}{n} in mathbb{Q}, m > 0$ We can write $frac{m}{n} = frac{1}{n} + frac{1}{n} + cdots + frac{1}{n}$ ($m$ summands). So $phi(frac{m}{n}) = m phi(frac{1}{n}).$ So in particular, for $m = n,$ we get that $phi(frac{1}{n}) = frac{r}{n}.$ Thus $phi(frac{r}{n}) = frac{mr}{n}.$ Similarly for $m < 0.$ Hence $phi(frac{m}{n}) = frac{mr}{n}, forall frac{m}{n} in mathbb{Q}.$



    On the other hand, for any $r in mathbb{Q}, phi: mathbb{Q} rightarrow mathbb{Q}$ defined by $1 mapsto r$ is a group homomorphism.






    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%2f1082495%2fhow-to-determine-end-mathbbq%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












      $begingroup$

      Your claims are true; if you want to prove that ever homomorphism is determined by $1$, notice that if $f:mathbb Qrightarrow mathbb Q$ is a homomorphism, then we clearly have $f(-n)=-f(n)$ and $f(n+1)=f(n)+f(1)$. Using these identities we determine the value of $f$ over $mathbb Z$ given the value of $f(1)$. Moreover, we can prove that
      $$fleft(underbrace{frac{p}q+ldots+frac{p}q}_{qtext{ times}}right)=underbrace{fleft(frac{p}qright)+ldots+fleft(frac{p}qright)}_{qtext{ times}}$$
      where the left hand side equals $f(p)$, which is an integer and thus already determined and, knowing that there is only one rational solution to
      $$c=underbrace{r+ldots+r}_{ntext{ times}}$$
      suffices to determine $fleft(frac{p}qright)$. From here, since $f(1)$ could be any rational and $(f+g)(1)$ would be the sum of the associated rational values of $f$ and $g$, it's clear that End($mathbb Q,+$) is isomorphic to $mathbb Q$, since the map $fmapsto f(1)$ is an isomorphism between the two.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Your claims are true; if you want to prove that ever homomorphism is determined by $1$, notice that if $f:mathbb Qrightarrow mathbb Q$ is a homomorphism, then we clearly have $f(-n)=-f(n)$ and $f(n+1)=f(n)+f(1)$. Using these identities we determine the value of $f$ over $mathbb Z$ given the value of $f(1)$. Moreover, we can prove that
        $$fleft(underbrace{frac{p}q+ldots+frac{p}q}_{qtext{ times}}right)=underbrace{fleft(frac{p}qright)+ldots+fleft(frac{p}qright)}_{qtext{ times}}$$
        where the left hand side equals $f(p)$, which is an integer and thus already determined and, knowing that there is only one rational solution to
        $$c=underbrace{r+ldots+r}_{ntext{ times}}$$
        suffices to determine $fleft(frac{p}qright)$. From here, since $f(1)$ could be any rational and $(f+g)(1)$ would be the sum of the associated rational values of $f$ and $g$, it's clear that End($mathbb Q,+$) is isomorphic to $mathbb Q$, since the map $fmapsto f(1)$ is an isomorphism between the two.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Your claims are true; if you want to prove that ever homomorphism is determined by $1$, notice that if $f:mathbb Qrightarrow mathbb Q$ is a homomorphism, then we clearly have $f(-n)=-f(n)$ and $f(n+1)=f(n)+f(1)$. Using these identities we determine the value of $f$ over $mathbb Z$ given the value of $f(1)$. Moreover, we can prove that
          $$fleft(underbrace{frac{p}q+ldots+frac{p}q}_{qtext{ times}}right)=underbrace{fleft(frac{p}qright)+ldots+fleft(frac{p}qright)}_{qtext{ times}}$$
          where the left hand side equals $f(p)$, which is an integer and thus already determined and, knowing that there is only one rational solution to
          $$c=underbrace{r+ldots+r}_{ntext{ times}}$$
          suffices to determine $fleft(frac{p}qright)$. From here, since $f(1)$ could be any rational and $(f+g)(1)$ would be the sum of the associated rational values of $f$ and $g$, it's clear that End($mathbb Q,+$) is isomorphic to $mathbb Q$, since the map $fmapsto f(1)$ is an isomorphism between the two.






          share|cite|improve this answer









          $endgroup$



          Your claims are true; if you want to prove that ever homomorphism is determined by $1$, notice that if $f:mathbb Qrightarrow mathbb Q$ is a homomorphism, then we clearly have $f(-n)=-f(n)$ and $f(n+1)=f(n)+f(1)$. Using these identities we determine the value of $f$ over $mathbb Z$ given the value of $f(1)$. Moreover, we can prove that
          $$fleft(underbrace{frac{p}q+ldots+frac{p}q}_{qtext{ times}}right)=underbrace{fleft(frac{p}qright)+ldots+fleft(frac{p}qright)}_{qtext{ times}}$$
          where the left hand side equals $f(p)$, which is an integer and thus already determined and, knowing that there is only one rational solution to
          $$c=underbrace{r+ldots+r}_{ntext{ times}}$$
          suffices to determine $fleft(frac{p}qright)$. From here, since $f(1)$ could be any rational and $(f+g)(1)$ would be the sum of the associated rational values of $f$ and $g$, it's clear that End($mathbb Q,+$) is isomorphic to $mathbb Q$, since the map $fmapsto f(1)$ is an isomorphism between the two.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '14 at 16:23









          Milo BrandtMilo Brandt

          39.8k476140




          39.8k476140























              1












              $begingroup$

              Define $phi: mathbb{Q} rightarrow mathbb{Q}, 1 mapsto r.$ Then for each $n in mathbb{N}, phi(n) = nr.$ Also $phi(-n) = -phi(n) = -nr.$ Thus, $phi(n) = nr, forall n in mathbb{Z}.$ Take $frac{m}{n} in mathbb{Q}, m > 0$ We can write $frac{m}{n} = frac{1}{n} + frac{1}{n} + cdots + frac{1}{n}$ ($m$ summands). So $phi(frac{m}{n}) = m phi(frac{1}{n}).$ So in particular, for $m = n,$ we get that $phi(frac{1}{n}) = frac{r}{n}.$ Thus $phi(frac{r}{n}) = frac{mr}{n}.$ Similarly for $m < 0.$ Hence $phi(frac{m}{n}) = frac{mr}{n}, forall frac{m}{n} in mathbb{Q}.$



              On the other hand, for any $r in mathbb{Q}, phi: mathbb{Q} rightarrow mathbb{Q}$ defined by $1 mapsto r$ is a group homomorphism.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Define $phi: mathbb{Q} rightarrow mathbb{Q}, 1 mapsto r.$ Then for each $n in mathbb{N}, phi(n) = nr.$ Also $phi(-n) = -phi(n) = -nr.$ Thus, $phi(n) = nr, forall n in mathbb{Z}.$ Take $frac{m}{n} in mathbb{Q}, m > 0$ We can write $frac{m}{n} = frac{1}{n} + frac{1}{n} + cdots + frac{1}{n}$ ($m$ summands). So $phi(frac{m}{n}) = m phi(frac{1}{n}).$ So in particular, for $m = n,$ we get that $phi(frac{1}{n}) = frac{r}{n}.$ Thus $phi(frac{r}{n}) = frac{mr}{n}.$ Similarly for $m < 0.$ Hence $phi(frac{m}{n}) = frac{mr}{n}, forall frac{m}{n} in mathbb{Q}.$



                On the other hand, for any $r in mathbb{Q}, phi: mathbb{Q} rightarrow mathbb{Q}$ defined by $1 mapsto r$ is a group homomorphism.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Define $phi: mathbb{Q} rightarrow mathbb{Q}, 1 mapsto r.$ Then for each $n in mathbb{N}, phi(n) = nr.$ Also $phi(-n) = -phi(n) = -nr.$ Thus, $phi(n) = nr, forall n in mathbb{Z}.$ Take $frac{m}{n} in mathbb{Q}, m > 0$ We can write $frac{m}{n} = frac{1}{n} + frac{1}{n} + cdots + frac{1}{n}$ ($m$ summands). So $phi(frac{m}{n}) = m phi(frac{1}{n}).$ So in particular, for $m = n,$ we get that $phi(frac{1}{n}) = frac{r}{n}.$ Thus $phi(frac{r}{n}) = frac{mr}{n}.$ Similarly for $m < 0.$ Hence $phi(frac{m}{n}) = frac{mr}{n}, forall frac{m}{n} in mathbb{Q}.$



                  On the other hand, for any $r in mathbb{Q}, phi: mathbb{Q} rightarrow mathbb{Q}$ defined by $1 mapsto r$ is a group homomorphism.






                  share|cite|improve this answer









                  $endgroup$



                  Define $phi: mathbb{Q} rightarrow mathbb{Q}, 1 mapsto r.$ Then for each $n in mathbb{N}, phi(n) = nr.$ Also $phi(-n) = -phi(n) = -nr.$ Thus, $phi(n) = nr, forall n in mathbb{Z}.$ Take $frac{m}{n} in mathbb{Q}, m > 0$ We can write $frac{m}{n} = frac{1}{n} + frac{1}{n} + cdots + frac{1}{n}$ ($m$ summands). So $phi(frac{m}{n}) = m phi(frac{1}{n}).$ So in particular, for $m = n,$ we get that $phi(frac{1}{n}) = frac{r}{n}.$ Thus $phi(frac{r}{n}) = frac{mr}{n}.$ Similarly for $m < 0.$ Hence $phi(frac{m}{n}) = frac{mr}{n}, forall frac{m}{n} in mathbb{Q}.$



                  On the other hand, for any $r in mathbb{Q}, phi: mathbb{Q} rightarrow mathbb{Q}$ defined by $1 mapsto r$ is a group homomorphism.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 27 '14 at 16:25









                  KrishKrish

                  6,32411020




                  6,32411020






























                      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%2f1082495%2fhow-to-determine-end-mathbbq%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

                      Mario Kart Wii

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
                      Read more

                      What does “Dominus providebit” mean?

                      Image
                      5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
                      Read more

                      File:Tiny Toon Adventures Wacky Sports JP Title.png

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! File:Tiny Toon Adventures Wacky Sports JP Title.png From The Cutting Room Floor Jump to: navigation, search File File history File usage No higher resolution available. Tiny_Toon_Adventures_Wacky_Sports_JP_Title.png ‎ (160 × 144 pixels, file size: 3 KB, MIME type: image/png ) This image is a screenshot of a(n) Game Boy title screen. File history Click on a date/time to view the file as it appeared at that time. Date/Time Thumbnail Dimensions User Comment current 21:31, 20 August 2016 160 × 144 (3 KB) TheBluegeek (Talk | contribs) 17:25, 29 August 2013 160 × 144 (3 KB) Luigi-San (Talk | contribs) {{titlescreen|Game Boy}} You cannot overwrite this file. File usage The following page links to this file: Tiny Toon Advent
                      Read more