Example of reduced ring which is not integral domain

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So, a reduced ring has no nonzero nilpotent elements, while integral domain has no nonzero zero divisors. Of course, the latter is way stronger then the first condition, implying that every integral domain needs to be reduced ring. But, having $(exists n geq 2) x^n = 0 Longleftrightarrow x = 0$ does not up to my knowledge imply $xy=0 Longleftrightarrow (x = 0 vee y = 0)$.



That is why I need example of reduced ring which is not integral domain. Thanks!




commutative-algebra




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    $Bbb Z/6Bbb Z$?
    – Lord Shark the Unknown
    Jul 22 at 17:10











  • Try $mathbb R[x]/(x^2-x)$.
    – Mike Earnest
    Jul 22 at 17:14














up vote
0
down vote

favorite












So, a reduced ring has no nonzero nilpotent elements, while integral domain has no nonzero zero divisors. Of course, the latter is way stronger then the first condition, implying that every integral domain needs to be reduced ring. But, having $(exists n geq 2) x^n = 0 Longleftrightarrow x = 0$ does not up to my knowledge imply $xy=0 Longleftrightarrow (x = 0 vee y = 0)$.



That is why I need example of reduced ring which is not integral domain. Thanks!




commutative-algebra




share|cite|improve this question















  • 2




    $Bbb Z/6Bbb Z$?
    – Lord Shark the Unknown
    Jul 22 at 17:10











  • Try $mathbb R[x]/(x^2-x)$.
    – Mike Earnest
    Jul 22 at 17:14












up vote
0
down vote

favorite









up vote
0
down vote

favorite











So, a reduced ring has no nonzero nilpotent elements, while integral domain has no nonzero zero divisors. Of course, the latter is way stronger then the first condition, implying that every integral domain needs to be reduced ring. But, having $(exists n geq 2) x^n = 0 Longleftrightarrow x = 0$ does not up to my knowledge imply $xy=0 Longleftrightarrow (x = 0 vee y = 0)$.



That is why I need example of reduced ring which is not integral domain. Thanks!




commutative-algebra




share|cite|improve this question











So, a reduced ring has no nonzero nilpotent elements, while integral domain has no nonzero zero divisors. Of course, the latter is way stronger then the first condition, implying that every integral domain needs to be reduced ring. But, having $(exists n geq 2) x^n = 0 Longleftrightarrow x = 0$ does not up to my knowledge imply $xy=0 Longleftrightarrow (x = 0 vee y = 0)$.



That is why I need example of reduced ring which is not integral domain. Thanks!





commutative-algebra





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 22 at 17:07









nikola

557214




557214







  • 2




    $Bbb Z/6Bbb Z$?
    – Lord Shark the Unknown
    Jul 22 at 17:10











  • Try $mathbb R[x]/(x^2-x)$.
    – Mike Earnest
    Jul 22 at 17:14












  • 2




    $Bbb Z/6Bbb Z$?
    – Lord Shark the Unknown
    Jul 22 at 17:10











  • Try $mathbb R[x]/(x^2-x)$.
    – Mike Earnest
    Jul 22 at 17:14







2




2




$Bbb Z/6Bbb Z$?
– Lord Shark the Unknown
Jul 22 at 17:10





$Bbb Z/6Bbb Z$?
– Lord Shark the Unknown
Jul 22 at 17:10













Try $mathbb R[x]/(x^2-x)$.
– Mike Earnest
Jul 22 at 17:14




Try $mathbb R[x]/(x^2-x)$.
– Mike Earnest
Jul 22 at 17:14










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$Bbb Z[X,Y]/(XY)$ has no non-zero nilpotents, but $XY = 0$.






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    up vote
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    accepted










    $Bbb Z[X,Y]/(XY)$ has no non-zero nilpotents, but $XY = 0$.






    share|cite|improve this answer

























      up vote
      1
      down vote



      accepted










      $Bbb Z[X,Y]/(XY)$ has no non-zero nilpotents, but $XY = 0$.






      share|cite|improve this answer























        up vote
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        accepted







        up vote
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        down vote



        accepted






        $Bbb Z[X,Y]/(XY)$ has no non-zero nilpotents, but $XY = 0$.






        share|cite|improve this answer













        $Bbb Z[X,Y]/(XY)$ has no non-zero nilpotents, but $XY = 0$.







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        answered Jul 22 at 17:10









        Kenny Lau

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