Ring Homomorphisms and Ideals Part 1

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  • Опубликовано: 18 дек 2024

Комментарии • 8

  • @jozbornn
    @jozbornn 8 лет назад +9

    These are so fantastically prepared. There are about 27 hours of videos in this playlist. As soon as I'm finished I'll be going through your other series!

  • @LastvanLichtenGlorie
    @LastvanLichtenGlorie 6 лет назад +7

    Your presentations are absolutely beautiful. Thank you so dearly. Sincerely from Texas

  • @FullerHob
    @FullerHob 3 года назад +1

    Such a clear and intuitive way of teaching this. Great video.

  • @MsJundell
    @MsJundell 8 лет назад +4

    I am very happy that I found your channel. it helps me a lot for may courses.

  • @bonbonpony
    @bonbonpony 4 года назад

    07:40 What if I already have the two tables (orange and green), and what I want to find instead is the isomorphism φ that does the proper "relabelling"? Is there any algorithmic method of coming up with such φ? Or at least figuring out that there isn't any?

  • @AVINASHKUMAR-lc8vu
    @AVINASHKUMAR-lc8vu 7 лет назад

    Sir, i have a doubt. Is the following question even valid.
    "Consider two non zero rings R and R'. Give an example of a homomorphism f:R-----R' such that R has unity but R' does not have unity?. " Because if R' doesnot have unity then it is not even a ring.

    • @MuffinsAPlenty
      @MuffinsAPlenty 7 лет назад +5

      There are several possible definitions of a ring. Some people allow rings not to have unity. A more modern definition of a ring requires that it has unity. If your book/professor is asking for a ring without unity, they are using an older (but still commonly used) definition of a ring. If you want, you can think of this as an algebraic structure which satisfies all of the axioms of a ring except for the axiom requiring the multiplicative identity.
      A nice situation of a "ring without unity" is an ideal! For any ring R, if I is a nonzero proper ideal of R, then I is a "ring without unity." Can you think of an example of a homomorphism from a ring R to one of its nonzero proper ideals? (Notice that since your codomain has no multiplicative identity, you also drop the condition that a ring homomorphism must send the multiplicative identity to the multiplicative identity.)

  • @Akk-vr9lj
    @Akk-vr9lj 5 лет назад

    Hindi me kaise bolega