The elements of the group ring $R = mathbbZ_4mathbbC_2$

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Let $mathbbC_2$ denote the group of order $2$ and let $mathbbZ_4$ denote the ring of integers modulo $4$. The group ring $R = mathbbZ_4mathbbC_2$ is commutative.



My problem is how to generate all the elements of this ring and the operations involved. I am a learner try to teach myself something to do with group rings.




group-theory ring-theory group-rings




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    Let $mathbbC_2$ denote the group of order $2$ and let $mathbbZ_4$ denote the ring of integers modulo $4$. The group ring $R = mathbbZ_4mathbbC_2$ is commutative.



    My problem is how to generate all the elements of this ring and the operations involved. I am a learner try to teach myself something to do with group rings.




    group-theory ring-theory group-rings




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $mathbbC_2$ denote the group of order $2$ and let $mathbbZ_4$ denote the ring of integers modulo $4$. The group ring $R = mathbbZ_4mathbbC_2$ is commutative.



      My problem is how to generate all the elements of this ring and the operations involved. I am a learner try to teach myself something to do with group rings.




      group-theory ring-theory group-rings




      share|cite|improve this question











      Let $mathbbC_2$ denote the group of order $2$ and let $mathbbZ_4$ denote the ring of integers modulo $4$. The group ring $R = mathbbZ_4mathbbC_2$ is commutative.



      My problem is how to generate all the elements of this ring and the operations involved. I am a learner try to teach myself something to do with group rings.





      group-theory ring-theory group-rings





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 26 at 21:22









      Sulayman

      18717




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          Let's denote by $0,1,2,3$ the elements of $mathbbZ_4$ and by $mathbf1$ (the identity) and $mathbfu$ the elements of $C_2$. An element of the group ring is of the form
          $$
          amathbf1+bmathbfu
          $$
          where $a,binmathbbZ_4$. So we have sixteen elements in all. The zero element is $0mathbf1+0mathbfu$, the identity is $1mathbf1+0mathbfu$.



          Addition is performed in the obvious way:
          $$
          (amathbf1+bmathbfu)+(cmathbf1+dmathbfu)=
          (a+c)mathbf1+(b+d)mathbfu
          $$
          Multiplication is performed with the usual rules:
          $$
          (amathbf1+bmathbfu)(cmathbf1+dmathbfu)=
          acmathbf1mathbf1+admathbf1mathbfu+bmathbfumathbf1+
          bcmathbfumathbfu=
          (ac+bd)mathbf1+(ad+bc)mathbfu
          $$
          because $mathbfumathbfu=mathbf1$ in $C_2$.



          Note that we haven't used any special property of $mathbbZ_4$, just that it is a ring. By definition, the elements of the group (in this case $C_2$) commute with the elements of the ring (here $mathbbZ_4$).






          share|cite|improve this answer





















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            Let's denote by $0,1,2,3$ the elements of $mathbbZ_4$ and by $mathbf1$ (the identity) and $mathbfu$ the elements of $C_2$. An element of the group ring is of the form
            $$
            amathbf1+bmathbfu
            $$
            where $a,binmathbbZ_4$. So we have sixteen elements in all. The zero element is $0mathbf1+0mathbfu$, the identity is $1mathbf1+0mathbfu$.



            Addition is performed in the obvious way:
            $$
            (amathbf1+bmathbfu)+(cmathbf1+dmathbfu)=
            (a+c)mathbf1+(b+d)mathbfu
            $$
            Multiplication is performed with the usual rules:
            $$
            (amathbf1+bmathbfu)(cmathbf1+dmathbfu)=
            acmathbf1mathbf1+admathbf1mathbfu+bmathbfumathbf1+
            bcmathbfumathbfu=
            (ac+bd)mathbf1+(ad+bc)mathbfu
            $$
            because $mathbfumathbfu=mathbf1$ in $C_2$.



            Note that we haven't used any special property of $mathbbZ_4$, just that it is a ring. By definition, the elements of the group (in this case $C_2$) commute with the elements of the ring (here $mathbbZ_4$).






            share|cite|improve this answer

























              up vote
              2
              down vote













              Let's denote by $0,1,2,3$ the elements of $mathbbZ_4$ and by $mathbf1$ (the identity) and $mathbfu$ the elements of $C_2$. An element of the group ring is of the form
              $$
              amathbf1+bmathbfu
              $$
              where $a,binmathbbZ_4$. So we have sixteen elements in all. The zero element is $0mathbf1+0mathbfu$, the identity is $1mathbf1+0mathbfu$.



              Addition is performed in the obvious way:
              $$
              (amathbf1+bmathbfu)+(cmathbf1+dmathbfu)=
              (a+c)mathbf1+(b+d)mathbfu
              $$
              Multiplication is performed with the usual rules:
              $$
              (amathbf1+bmathbfu)(cmathbf1+dmathbfu)=
              acmathbf1mathbf1+admathbf1mathbfu+bmathbfumathbf1+
              bcmathbfumathbfu=
              (ac+bd)mathbf1+(ad+bc)mathbfu
              $$
              because $mathbfumathbfu=mathbf1$ in $C_2$.



              Note that we haven't used any special property of $mathbbZ_4$, just that it is a ring. By definition, the elements of the group (in this case $C_2$) commute with the elements of the ring (here $mathbbZ_4$).






              share|cite|improve this answer























                up vote
                2
                down vote










                up vote
                2
                down vote









                Let's denote by $0,1,2,3$ the elements of $mathbbZ_4$ and by $mathbf1$ (the identity) and $mathbfu$ the elements of $C_2$. An element of the group ring is of the form
                $$
                amathbf1+bmathbfu
                $$
                where $a,binmathbbZ_4$. So we have sixteen elements in all. The zero element is $0mathbf1+0mathbfu$, the identity is $1mathbf1+0mathbfu$.



                Addition is performed in the obvious way:
                $$
                (amathbf1+bmathbfu)+(cmathbf1+dmathbfu)=
                (a+c)mathbf1+(b+d)mathbfu
                $$
                Multiplication is performed with the usual rules:
                $$
                (amathbf1+bmathbfu)(cmathbf1+dmathbfu)=
                acmathbf1mathbf1+admathbf1mathbfu+bmathbfumathbf1+
                bcmathbfumathbfu=
                (ac+bd)mathbf1+(ad+bc)mathbfu
                $$
                because $mathbfumathbfu=mathbf1$ in $C_2$.



                Note that we haven't used any special property of $mathbbZ_4$, just that it is a ring. By definition, the elements of the group (in this case $C_2$) commute with the elements of the ring (here $mathbbZ_4$).






                share|cite|improve this answer













                Let's denote by $0,1,2,3$ the elements of $mathbbZ_4$ and by $mathbf1$ (the identity) and $mathbfu$ the elements of $C_2$. An element of the group ring is of the form
                $$
                amathbf1+bmathbfu
                $$
                where $a,binmathbbZ_4$. So we have sixteen elements in all. The zero element is $0mathbf1+0mathbfu$, the identity is $1mathbf1+0mathbfu$.



                Addition is performed in the obvious way:
                $$
                (amathbf1+bmathbfu)+(cmathbf1+dmathbfu)=
                (a+c)mathbf1+(b+d)mathbfu
                $$
                Multiplication is performed with the usual rules:
                $$
                (amathbf1+bmathbfu)(cmathbf1+dmathbfu)=
                acmathbf1mathbf1+admathbf1mathbfu+bmathbfumathbf1+
                bcmathbfumathbfu=
                (ac+bd)mathbf1+(ad+bc)mathbfu
                $$
                because $mathbfumathbfu=mathbf1$ in $C_2$.



                Note that we haven't used any special property of $mathbbZ_4$, just that it is a ring. By definition, the elements of the group (in this case $C_2$) commute with the elements of the ring (here $mathbbZ_4$).







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 26 at 21:32









                egreg

                164k1180187




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