ring contains a subring isomorphic to $Z$ implies it is integral domain?

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If a ring contains a subring isomorphic to $Z$,does it necessarily imply that the ring must be an integral domain?



I know that it must have characteristic 0.However I cannot proceed further.




ring-theory




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  • This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
    – Bernard
    Aug 28 at 14:36














up vote
0
down vote

favorite












If a ring contains a subring isomorphic to $Z$,does it necessarily imply that the ring must be an integral domain?



I know that it must have characteristic 0.However I cannot proceed further.




ring-theory




share|cite|improve this question




















  • This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
    – Bernard
    Aug 28 at 14:36












up vote
0
down vote

favorite









up vote
0
down vote

favorite











If a ring contains a subring isomorphic to $Z$,does it necessarily imply that the ring must be an integral domain?



I know that it must have characteristic 0.However I cannot proceed further.




ring-theory




share|cite|improve this question












If a ring contains a subring isomorphic to $Z$,does it necessarily imply that the ring must be an integral domain?



I know that it must have characteristic 0.However I cannot proceed further.





ring-theory





share|cite|improve this question











share|cite|improve this question




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asked Aug 28 at 14:34









Anwi

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  • This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
    – Bernard
    Aug 28 at 14:36
















  • This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
    – Bernard
    Aug 28 at 14:36















This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
– Bernard
Aug 28 at 14:36




This means every algebra over a field is an integral domain. Counter-example: $K[X]/(X^2)$.
– Bernard
Aug 28 at 14:36










2 Answers
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In general, no. Take $mathbbZtimes R$, where $R$ is a unit ring. It is not an integral domain, since $(1,0).(0,1)=(0,0)=0$. But it contains a subring isomorphic to the integers.






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    No; consider the ring $mathbb Z[epsilon]/(epsilon^2)$. That is, the ring consisting of elements of the form
    $$
    a + bepsilonqquad a,binmathbb Z,
    $$
    with obvious addition and multiplication given by
    $$
    (a + bepsilon)(c + depsilon) = ac + (ad + bc)epsilon.
    $$
    Then $epsilon^2 = 0$, even though $epsilon neq 0$.






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      2 Answers
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      2 Answers
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      active

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      up vote
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      accepted










      In general, no. Take $mathbbZtimes R$, where $R$ is a unit ring. It is not an integral domain, since $(1,0).(0,1)=(0,0)=0$. But it contains a subring isomorphic to the integers.






      share|cite|improve this answer
























        up vote
        0
        down vote



        accepted










        In general, no. Take $mathbbZtimes R$, where $R$ is a unit ring. It is not an integral domain, since $(1,0).(0,1)=(0,0)=0$. But it contains a subring isomorphic to the integers.






        share|cite|improve this answer






















          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          In general, no. Take $mathbbZtimes R$, where $R$ is a unit ring. It is not an integral domain, since $(1,0).(0,1)=(0,0)=0$. But it contains a subring isomorphic to the integers.






          share|cite|improve this answer












          In general, no. Take $mathbbZtimes R$, where $R$ is a unit ring. It is not an integral domain, since $(1,0).(0,1)=(0,0)=0$. But it contains a subring isomorphic to the integers.







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          answered Aug 28 at 14:37









          José Carlos Santos

          120k16101182




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              up vote
              1
              down vote













              No; consider the ring $mathbb Z[epsilon]/(epsilon^2)$. That is, the ring consisting of elements of the form
              $$
              a + bepsilonqquad a,binmathbb Z,
              $$
              with obvious addition and multiplication given by
              $$
              (a + bepsilon)(c + depsilon) = ac + (ad + bc)epsilon.
              $$
              Then $epsilon^2 = 0$, even though $epsilon neq 0$.






              share|cite|improve this answer
























                up vote
                1
                down vote













                No; consider the ring $mathbb Z[epsilon]/(epsilon^2)$. That is, the ring consisting of elements of the form
                $$
                a + bepsilonqquad a,binmathbb Z,
                $$
                with obvious addition and multiplication given by
                $$
                (a + bepsilon)(c + depsilon) = ac + (ad + bc)epsilon.
                $$
                Then $epsilon^2 = 0$, even though $epsilon neq 0$.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  No; consider the ring $mathbb Z[epsilon]/(epsilon^2)$. That is, the ring consisting of elements of the form
                  $$
                  a + bepsilonqquad a,binmathbb Z,
                  $$
                  with obvious addition and multiplication given by
                  $$
                  (a + bepsilon)(c + depsilon) = ac + (ad + bc)epsilon.
                  $$
                  Then $epsilon^2 = 0$, even though $epsilon neq 0$.






                  share|cite|improve this answer












                  No; consider the ring $mathbb Z[epsilon]/(epsilon^2)$. That is, the ring consisting of elements of the form
                  $$
                  a + bepsilonqquad a,binmathbb Z,
                  $$
                  with obvious addition and multiplication given by
                  $$
                  (a + bepsilon)(c + depsilon) = ac + (ad + bc)epsilon.
                  $$
                  Then $epsilon^2 = 0$, even though $epsilon neq 0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 28 at 14:36









                  Mees de Vries

                  14.1k12348




                  14.1k12348



























                       

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