Problems 2

Поділитися
Вставка
  • Опубліковано 19 лис 2024

КОМЕНТАРІ • 5

  • @mayur_krishna_devotee
    @mayur_krishna_devotee Рік тому +1

    An interesting way to think about the forward direction in Problem 5: If n is not a prime, then we can find 1 < a < n such that a divides n. If we consider the ideal generated by a, then it only contains multiples of a and as 1 < a < n, this ideal does NOT contain all of 1 to n. So this ideal is different from both the zero ideal and the unit ideal(the full ring) and hence one can use Problem 4 now to deduce that Z/nZ is not a field!

  • @maneeshgoutam9206
    @maneeshgoutam9206 4 роки тому +6

    good work

  • @karishmasharma15
    @karishmasharma15 Рік тому

    Thank you Sir 🙏

  • @Uzumaki_Naruto061
    @Uzumaki_Naruto061 2 роки тому +1

    sir I have a doubt, in problem 4 you said that Kernel cannot be R because 1 goes to 1, but what if all elements in R go to the zero element in R', it's still a ring homomorphism and it's not injective(assuming R' has some elements other than zero), why can this case not happen?

    • @mayur_krishna_devotee
      @mayur_krishna_devotee Рік тому

      He said in the beginning that we need to assume R' is not the zero ring, so that 1 and 0 are different in R'. So the zero map, which maps everything in R to R' is not a homomorphism because 1 in R has to be mapped to 1 in R' which is different from 0. So you cannot consider the zero map as a homomorphism as R' is assumed to be not the zero ring