Abstract Algebra | Polynomial Rings

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  • Опубліковано 17 січ 2025

КОМЕНТАРІ • 11

  • @Batman-vh9so
    @Batman-vh9so 4 роки тому +22

    Honestly best math channel on UA-cam! It'd be great if you could also do a series on algebraic and differential topology.

  • @michellejingdong
    @michellejingdong 2 роки тому +3

    Really nice and helpful channel! Thank you a lot!

  • @nailabenali7488
    @nailabenali7488 4 роки тому +4

    Can you do all the results about degre and roots of a polynomial? Also the fundamental theorem of algebra? And maybe show why a polynomial can't be considered as a funtion ( I mean not the same as his polynomial function) like in (Z/nZ)? Thank you for the great work!!

    • @MichaelPennMath
      @MichaelPennMath  4 роки тому +3

      I have several more polynomial videos waiting to be uploaded. I'll also think about some of those other topics as well.

    • @mrlimemil
      @mrlimemil 4 роки тому +1

      As part of these polynomial videos are you planning on doing a video the Heegner numbers?
      I think your videos on these topics have been great so far and look forward to any new videos you put out.

  • @ashimchakraborty2908
    @ashimchakraborty2908 4 роки тому +1

    @#Michael Penn, sir why did you take p(x)=x instead of p(x)=1? What is the point i mean p(x)q(x)=1 when p(x),q(x) are constant polynomial with p(x) and q(x) not equal to 0. Reply please sir

    • @MGSchmahl
      @MGSchmahl 3 роки тому +2

      The purpose was to prove that there exists a polynomial with no multiplicative inverse. p(x)=1 is trivially its own inverse, and any other 0-degree polynomial p(x)=c has an inverse p'(x)=1/c.
      Therefore, to provide a counterexample, we have to look to polynomials of degree at least 1, the simplest of which is p(x)=x.

    • @ashimchakraborty2908
      @ashimchakraborty2908 3 роки тому

      @@MGSchmahl Thanks

  • @mattiafontanelli3639
    @mattiafontanelli3639 11 днів тому

    Can someone explain me why if a polynomial g(x) is irriducible via omomorphism so the pre image h(x) of g(x) is irreducible

  • @el_ginga
    @el_ginga Рік тому

    Thanks Mr Zuckerberg