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Gauss Composition? and a naive composition law

  1. Dec 13, 2011 #1
    What exactly is gauss composition? I've heard of Manjul Bhargava's work, which apparently generalized gauss composition, but what is gauss composition? I would like to add that I've been thinking about quadratics polynomials with rational coefficients, and I discovered this composition law that turns the set of quadratic polynomials into an abelian group. Let f(x)=ax^2+bx+c and g(x)=zx^2+dx+r be two quadratic polynomials with rational coefficients. Denote the set of quadratic polynomials with rational coefficients by T{x}. Then the composition law %:T{x} X T{x}--->T{x} defined by f(x)%g(x)=azx^2+bdx+cr turns T{x} into an abelian group. This has probably already been figured out before, but an interesting note!


    Sincerely,
    mathguy

    EDIT:(simple explanations please, thank you.)
     
  2. jcsd
  3. Dec 16, 2011 #2

    Stephen Tashi

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    Getting a good answer to a question is somewhat a matter of luck. There might be an expert on Gauss composition on the forum who is chomping at the bit to answer such a very general question. If no such expert turns up, I suggest you ask a more specific question. This PDF looks interesting: http://www.google.com/url?sa=t&rct=...sg=AFQjCNFCUMwAwetrjbw_3lkt373P3ppmJQ&cad=rja

    It tells what Gauss thought that Gauss composition was. If you have a specific question about something in it, you might lure me or some other non-Gauss-composition student into reading it and trying to answer. (I haven't read it yet.)

    According to that PDF, Gauss composition is a ternary operation, not a binary operation. As to the Abelian group idea, how are you going to define inverses?
     
  4. Dec 16, 2011 #3
    "simple explanations please"

    No simple explanation that I can find.

    A few papers that might explain it to you can be found such as
    The shaping of Arithmetic after C.F. Gauss's Disquisitiones Arithmeticae

    A copy of Disquisitiones Arithmeticae converted to English can also be had, though a bit pricy.
     
  5. Dec 19, 2011 #4
    ok, so its a ternary operation rather than a binary. I will look into that pdf you have. With regards to the abelian group idea, let f(x)=ax^2+bx+c be a quadratic polynomial with rational coefficients. Let g(x)=(1/a)x^2+(1/b)x+1/c. Then, f(x)%g(x)=a(1/a)x^2+b(1/b)x+c(1/c)=x^2+x+1. x^2+x+1 is the identity, because if f(x)=ax^2+bx+c and t(x)=x^2+x+1, then f(x)%t(x)=a(1)x^2+b(1)x+c(1)=ax^2+bx+c=f(x), and t(x)%f(x)=1ax^2+1bx+1c=ax^2+bx+c=f(x).

    EDIT: I see now. 0 can't be one of the coefficients. So if f(x)=ax^2+bx+c AND if neither b nor c equals 0, then the set along with the naive composition forms an abelian group.

    EDIT(again): In http://www.icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_13.pdf, Bhargava says that Gauss laid down a remarkable law of composition on integral binary quadratic forms. Did he find several?
     
    Last edited: Dec 19, 2011
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