Abstract Algebra 13.4: A Polynomial Factor Ring

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  • Опубликовано: 7 фев 2025
  • In this video, we consider a more complicated example of a factor ring, and show how it is effectively the same as the complex numbers.

Комментарии •

  • @n.trushaev5132
    @n.trushaev5132 4 года назад +21

    Great video. I have been getting legitimately frustrated and angry because I'm doing online classes now due to the ongoing corona lockdown, and my prof. did not thoroughly cover the details regarding the structure of polynomial factor rings or provide any lecture material, but basically told us to use our textbook to work out the computational details. It's very frustrating because textbooks usually don't go very deep into the details of their computations or examples, and there are very few video resources available for abstract algebra and other mid-to-high level mathematics.

    • @patrickjones1510
      @patrickjones1510  4 года назад +7

      I'm glad you found it helpful. This video series was created when I was forced to teach the class online, and I figured I'd put them up for everyone exactly for people in your situation.

    • @isabel5154
      @isabel5154 4 года назад

      @@patrickjones1510 I'm in the same situation, now I have to learn everything online. Your videos are really helping. Thanks!!

  • @benjaminmartin7349
    @benjaminmartin7349 6 месяцев назад

    Thank you great video explaining how quotients in polynomial rings actually works.

  • @darpi_
    @darpi_ 3 года назад

    I had been stuck here for a while. Thank you so much for clearing my doubt! I have my algebra exam tomorrow wish me luck !

  • @algebraentodaspartes
    @algebraentodaspartes 3 года назад +2

    Thanks a lot, I'm starting to study rings on my own and this helped me a lot with the examples in my book.

  • @OpelsiSq
    @OpelsiSq 2 года назад

    I was blown away when i realized about it being isomorphic to complex numbers. Somehow i didn't notice x^2=-1 (which means x=i) until you metioned it)

    • @patrickjones1510
      @patrickjones1510  2 года назад +1

      That is one of the things that I find fascinating about math in general, and abstract algebra in particular. You can be studying something from an entirely different area of math, and find that it is intimately tied together with a topic that at first seems completely unrelated.

  • @focusmaestro4013
    @focusmaestro4013 Год назад

    Amazing professor..... Brilliant work!! Thanks a lot

  • @sfratini
    @sfratini Год назад

    Excellent explanation!

  • @profesorjan7614
    @profesorjan7614 Год назад

    What a great video! Thanks

  • @hussainibnali1530
    @hussainibnali1530 Год назад

    Thank you soo much for cleaning my dought's 😊 I need this

  • @Mesohornet11
    @Mesohornet11 Год назад

    you are the best!

  • @rem.31
    @rem.31 2 года назад

    ty🖤

  • @37695543xs
    @37695543xs 4 года назад

    good, now i understand why it can be treat as complex numbers

  • @beetsreliable
    @beetsreliable Год назад

    Amazing! Thank you

  • @athair937
    @athair937 Год назад

    Hi; i take this video as my main reference on the subject, so well explained that i keep coming back from time to time; i would like to ask if it could be right to say there is a hint on the absortion property of ideals in it? i mean an absortion property such that operations on the quotient turn back to the ideal as the zero-class(?), in the sense of an additive (group) zero class (?) thanks a lot once again!!

  • @silesiaball9505
    @silesiaball9505 2 года назад +2

    I have a question, because I'm not sure if I understand this.
    Why can we say that x^2+1 is effectively 0?

    • @patrickjones1510
      @patrickjones1510  2 года назад

      The 0 for a ring is the additive identity. It is confusing because the elements of our factor ring are huge sets of polynomials, but in this case the additive identity is the ring of all multiples of x^2+1. The true additive identity is , but to understand the simplification trick, it is easier to just consider the "defining polynomial" x^2+1 as being 0.

  • @sdfdsf4162
    @sdfdsf4162 2 года назад

    awesome tnx

  • @steliostoulis1875
    @steliostoulis1875 Год назад

    Thank you

  • @rodilrun9195
    @rodilrun9195 4 года назад

    Very useful sir.

  • @chriswinchell1570
    @chriswinchell1570 4 года назад

    Well done. Thanks.

  • @ibrahimislam8759
    @ibrahimislam8759 3 года назад +1

    Dear Patrick ,....I have a question from this topic could you please help me,....my question is: what is ring ℤₚ[x]) / ⟨x²⟩,,...I mean to what ring it looks like which is easy to see,....is it isomorphic to ℤₚ² ?I have strong intuttion that it is isomorphic to ℤₚ² but can't prove it,.....if it is not isomorphic to ℤₚ²,then to which ring it is isomorphic?

    • @ibrahimislam8759
      @ibrahimislam8759 3 года назад

      Since the factor x² does not break into distinct factors i cannot use Chinese remainder theorem here

  • @athair937
    @athair937 2 года назад

    A rather dumb question probably: what are some applications of a polinomyal quotient ring (beside the homomorphism' theorem.that states that a (given) ring homomorphism is isomorphic to the quotient ring defined by the ideal? Thanks

    • @patrickjones1510
      @patrickjones1510  2 года назад +1

      I'm honestly not the best person to answer that question; I typically teach lower level courses than this one, so I'm not up on all the details (especially after a couple of years since I went through this). I can't really point to a specific use of this particular topic. However, a lot of this is building its way up to Galois theory, which has many clever applications.

    • @athair937
      @athair937 2 года назад

      @@patrickjones1510 thank you! Great video; greetings

    • @athair937
      @athair937 2 года назад

      @@patrickjones1510 Hi; your video prompted me to get into the first stages of study of quotient rings, which had always been quite alien to me and quite hard to pin down, so i really thank you; perhaps you could tell some reference book on the subject? Greetings

  • @anonymous-random
    @anonymous-random 4 года назад +1

    omg Thank u so much!!

  • @girlsinacoma
    @girlsinacoma 6 месяцев назад

    Now prove there’s an inverse 😢