Fundamental Theorem on Homomorphisms - Application












1














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







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    1














    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



























      1












      1








      1







      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















      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 2 days ago









      Antonios-Alexandros Robotis

      9,40741640




      9,40741640










      asked 2 days ago









      KingDingeling

      505




      505






















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          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























          • Thank you for your answer and for taking the time :)
            – KingDingeling
            yesterday











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          1 Answer
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          active

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          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























          • Thank you for your answer and for taking the time :)
            – KingDingeling
            yesterday
















          2














          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























          • Thank you for your answer and for taking the time :)
            – KingDingeling
            yesterday














          2












          2








          2






          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














          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 2 days ago

























          answered 2 days ago









          Antonios-Alexandros Robotis

          9,40741640




          9,40741640












          • Thank you for your answer and for taking the time :)
            – KingDingeling
            yesterday


















          • Thank you for your answer and for taking the time :)
            – KingDingeling
            yesterday
















          Thank you for your answer and for taking the time :)
          – KingDingeling
          yesterday




          Thank you for your answer and for taking the time :)
          – KingDingeling
          yesterday


















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