Prove that if in a ring R, x^3=x for all x in R, then ring R is commutative.

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  • Опубликовано: 7 янв 2025

Комментарии • 13

  • @MathsICU
    @MathsICU  Год назад +2

    If x^2=0 for some x in a ring R and x^3=x for every x in R, then x=x^3=x.x^2=x.0=0.

  • @AmanDhiman_0043
    @AmanDhiman_0043 2 месяца назад

    Very helpful 🎉🎉🎉

  • @Himachali__priya
    @Himachali__priya Месяц назад +1

    Thank you

  • @MathsICU
    @MathsICU  Год назад

    at 9:00, we have used the property a + (-a) =0, as R is a group with respect to +. (Not the cancellation laws.)

  • @DescryGames
    @DescryGames Год назад +2

    Can we prove it using one-one onto, polynomial ring?

  • @vansh12766
    @vansh12766 8 месяцев назад

    wonderful sir smj aa gya ❤❤ apka awaz acha hai 😊

  • @lost8886
    @lost8886 11 месяцев назад

    At 19:00 why cant we replace x^2 by x-x^2 directly as x^2 can be anything.

  • @MrHolmes7
    @MrHolmes7 Год назад +1

    Where does (x^2y - x^2yx^2)^2 come from?
    Can I just make up anything that will help match up x^3 = x, or is that expression something I should already know?

    • @MathsICU
      @MathsICU  Год назад

      It is just an expression which has been used to solve x^3=x for the commutativeness of the ring and nothing else.

  • @nawazsarif3600
    @nawazsarif3600 Год назад

    What will happen if (x²y-x²yx²) become divisor of zero ?
    Can we then write (x²y-x²yx²)²=0 => (x²y-x²yx²)=0 ???

    • @nawazsarif3600
      @nawazsarif3600 Год назад

      For example, in (Z8, +,.)
      4²=0 but 4 nonzero

    • @MathsICU
      @MathsICU  Год назад

      It is only possible when ring satisfies the property x^3=x for all x in a ring.
      But in Z8 , 2^3=8=0 not equal to 2.Hence Z8 does not satisfy the property.

    • @MathsICU
      @MathsICU  Год назад

      If x^2=0 for some x in a ring R and x^3=x for every x in R, then x=x^3=x.x^2=x.0=0.