Fundamental Theorem on Homomorphisms - Application












1












$begingroup$


I want to show



1.) For $n,min mathbb Z$ with $m|n$ there exists a ring homomorphism between $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$.



I know that there is the canonical homomorphism $f: mathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. The Fundamental theorem on homomorphisms for rings now allows me, because $nmathbb{Z}$ is an ideal in $mathbb Z$, to conclude the existence of a ring homomorphism $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. But here is the problem: $mathbb{Z}/nmathbb{Z}$ has to be a subset of $ker(f)$ for that conclusion. Is that the case and if so: why?



2.) A restriction of the above Homomorphism on the unity groups $(mathbb Z/nmathbb Z)^* rightarrow (mathbb Z/mmathbb Z)^*$ is a group homomorphism.



I believe that one can solve this easily if 1.) is solved.






abstract-algebra ring-theory modules ideals







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I want to show



    1.) For $n,min mathbb Z$ with $m|n$ there exists a ring homomorphism between $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$.



    I know that there is the canonical homomorphism $f: mathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. The Fundamental theorem on homomorphisms for rings now allows me, because $nmathbb{Z}$ is an ideal in $mathbb Z$, to conclude the existence of a ring homomorphism $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. But here is the problem: $mathbb{Z}/nmathbb{Z}$ has to be a subset of $ker(f)$ for that conclusion. Is that the case and if so: why?



    2.) A restriction of the above Homomorphism on the unity groups $(mathbb Z/nmathbb Z)^* rightarrow (mathbb Z/mmathbb Z)^*$ is a group homomorphism.



    I believe that one can solve this easily if 1.) is solved.






    abstract-algebra ring-theory modules ideals







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I want to show



      1.) For $n,min mathbb Z$ with $m|n$ there exists a ring homomorphism between $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$.



      I know that there is the canonical homomorphism $f: mathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. The Fundamental theorem on homomorphisms for rings now allows me, because $nmathbb{Z}$ is an ideal in $mathbb Z$, to conclude the existence of a ring homomorphism $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. But here is the problem: $mathbb{Z}/nmathbb{Z}$ has to be a subset of $ker(f)$ for that conclusion. Is that the case and if so: why?



      2.) A restriction of the above Homomorphism on the unity groups $(mathbb Z/nmathbb Z)^* rightarrow (mathbb Z/mmathbb Z)^*$ is a group homomorphism.



      I believe that one can solve this easily if 1.) is solved.






      abstract-algebra ring-theory modules ideals







      share|cite|improve this question











      $endgroup$




      I want to show



      1.) For $n,min mathbb Z$ with $m|n$ there exists a ring homomorphism between $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$.



      I know that there is the canonical homomorphism $f: mathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. The Fundamental theorem on homomorphisms for rings now allows me, because $nmathbb{Z}$ is an ideal in $mathbb Z$, to conclude the existence of a ring homomorphism $mathbb{Z}/nmathbb{Z} rightarrow mathbb{Z}/mmathbb{Z}$. But here is the problem: $mathbb{Z}/nmathbb{Z}$ has to be a subset of $ker(f)$ for that conclusion. Is that the case and if so: why?



      2.) A restriction of the above Homomorphism on the unity groups $(mathbb Z/nmathbb Z)^* rightarrow (mathbb Z/mmathbb Z)^*$ is a group homomorphism.



      I believe that one can solve this easily if 1.) is solved.






      abstract-algebra ring-theory modules ideals




      abstract-algebra ring-theory modules ideals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 at 15:39









      Antonios-Alexandros Robotis

      9,75241640




      9,75241640










      asked Jan 3 at 15:25









      KingDingelingKingDingeling

      1366




      1366






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Why not study the canonical map $mathbb{Z}xrightarrow{varphi} mathbb{Z}/mmathbb{Z}$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbb{Z}/kervarphicong mathbb{Z}/mmathbb{Z}$. Because $m|n$, we get that for $kin nmathbb{Z}$, i.e. $k=ell n=ell rm$ for some $kin mathbb{Z}$, we have
          $$varphi(k) =[k]_m=[ell r m]_m=0 pmod{m}.$$
          So, $nmathbb{Z}subseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetilde{varphi}:nmathbb{Z}to mmathbb{Z}$. This is defined by $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer and for taking the time :)
            $endgroup$
            – KingDingeling
            Jan 3 at 16:02











          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%2f3060672%2ffundamental-theorem-on-homomorphisms-application%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












          $begingroup$

          Why not study the canonical map $mathbb{Z}xrightarrow{varphi} mathbb{Z}/mmathbb{Z}$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbb{Z}/kervarphicong mathbb{Z}/mmathbb{Z}$. Because $m|n$, we get that for $kin nmathbb{Z}$, i.e. $k=ell n=ell rm$ for some $kin mathbb{Z}$, we have
          $$varphi(k) =[k]_m=[ell r m]_m=0 pmod{m}.$$
          So, $nmathbb{Z}subseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetilde{varphi}:nmathbb{Z}to mmathbb{Z}$. This is defined by $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer and for taking the time :)
            $endgroup$
            – KingDingeling
            Jan 3 at 16:02
















          2












          $begingroup$

          Why not study the canonical map $mathbb{Z}xrightarrow{varphi} mathbb{Z}/mmathbb{Z}$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbb{Z}/kervarphicong mathbb{Z}/mmathbb{Z}$. Because $m|n$, we get that for $kin nmathbb{Z}$, i.e. $k=ell n=ell rm$ for some $kin mathbb{Z}$, we have
          $$varphi(k) =[k]_m=[ell r m]_m=0 pmod{m}.$$
          So, $nmathbb{Z}subseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetilde{varphi}:nmathbb{Z}to mmathbb{Z}$. This is defined by $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your answer and for taking the time :)
            $endgroup$
            – KingDingeling
            Jan 3 at 16:02














          2












          2








          2





          $begingroup$

          Why not study the canonical map $mathbb{Z}xrightarrow{varphi} mathbb{Z}/mmathbb{Z}$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbb{Z}/kervarphicong mathbb{Z}/mmathbb{Z}$. Because $m|n$, we get that for $kin nmathbb{Z}$, i.e. $k=ell n=ell rm$ for some $kin mathbb{Z}$, we have
          $$varphi(k) =[k]_m=[ell r m]_m=0 pmod{m}.$$
          So, $nmathbb{Z}subseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetilde{varphi}:nmathbb{Z}to mmathbb{Z}$. This is defined by $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$






          share|cite|improve this answer











          $endgroup$



          Why not study the canonical map $mathbb{Z}xrightarrow{varphi} mathbb{Z}/mmathbb{Z}$ given by the projection $nmapsto [n]_m$? The first isomorphism theorem tells us that $mathbb{Z}/kervarphicong mathbb{Z}/mmathbb{Z}$. Because $m|n$, we get that for $kin nmathbb{Z}$, i.e. $k=ell n=ell rm$ for some $kin mathbb{Z}$, we have
          $$varphi(k) =[k]_m=[ell r m]_m=0 pmod{m}.$$
          So, $nmathbb{Z}subseteqkervarphi$ and the factorization theorem tells us that we get an induced map $widetilde{varphi}:nmathbb{Z}to mmathbb{Z}$. This is defined by $widetilde{varphi}(a+nmathbb{Z})=varphi(a)=[a]_m.$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 3 at 15:40

























          answered Jan 3 at 15:34









          Antonios-Alexandros RobotisAntonios-Alexandros Robotis

          9,75241640




          9,75241640












          • $begingroup$
            Thank you for your answer and for taking the time :)
            $endgroup$
            – KingDingeling
            Jan 3 at 16:02


















          • $begingroup$
            Thank you for your answer and for taking the time :)
            $endgroup$
            – KingDingeling
            Jan 3 at 16:02
















          $begingroup$
          Thank you for your answer and for taking the time :)
          $endgroup$
          – KingDingeling
          Jan 3 at 16:02




          $begingroup$
          Thank you for your answer and for taking the time :)
          $endgroup$
          – KingDingeling
          Jan 3 at 16:02


















          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%2f3060672%2ffundamental-theorem-on-homomorphisms-application%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

          Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

          Image
          28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
          Read more

          Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

          Image
          up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
          Read more

          A Topological Invariant for $pi_3(U(n))$

          Image
          3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
          Read more