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Algebraic comments

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  • Kermit Rose
    Primes in algebraic number systems: Consider the extension to the integers, {A + B * sqrt(d) }, where A, B are variable integers, and d is a fixed integer. (a1
    Message 1 of 3 , Dec 4, 2009
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      Primes in algebraic number systems:

      Consider the extension to the integers,

      {A + B * sqrt(d) }, where A, B are variable integers, and d is a fixed
      integer.


      (a1 + b1 * sqrt(d) )*(a2 + b2 * sqrt(d))
      = (a1*a2 + b1 * b2 * d) + (a1*b2 + b1*a2) * sqrt(d)

      This multiplication rule is valid, even if d is a square integer.

      So, which numbers in the set

      {A + B sqrt(d)} are not composite?

      a + b sqrt(d) is composite if there exist a1,a2,b1,b2 such that

      a = a1*a2 + b1 * b2 * d,
      and
      b = a1*b2 + b1 * a2

      For example, when d = 10,


      (1 + sqrt(10)) * (2 + sqrt(10)) = 12 + 3 sqrt(10)

      However,

      12 + 3 sqrt(10) = 3 * (4 + sqrt(10)).

      This proves that {A + B sqrt(10)} does not have
      unique factorization. Prime theory is not as useful in this case.

      However, there is unique factorization for d = -1.

      There are a few other values of d that yield unique factorization.
      I do not know all of them.

      For the Gaussian integers, {A + B * sqrt(-1) },

      the primes are
      (1) the integer primes which are equal to 3 mod 4,
      and
      (2) numbers of the form a + bi, a-bi, where
      i = sqrt(-1), and
      a**2 + b**2 = p is an integer prime equal to 1 mod 4.

      This follows from the algebraic identity


      (a1 + b1 * sqrt(d) )*(a2 + b2 * sqrt(d))
      = (a1*a2 + b1 * b2 * d) + (a1*b2 + b1*a2) * sqrt(d)

      when d = -1,
      this identity becomes


      (a1 + b1 i )*(a2 + b2 i)
      = (a1*a2 - b1 * b2 ) + (a1*b2 + b1*a1) i

      In particular, if a1 = a2, and b1 = -b2,



      (a1 + b1 i )*(a1 - b1 i)
      = (a1**2 + b1**2)


      Because multiplication is commutative,

      [(a1 + b1 i )*(a1 - b1 i)]*[(a2 + b2 i )*(a2 - b2 i)]

      = [(a1 + b1 i )*(a2 + b2 i )]*[(a1 - b1 i)*(a2 - b2 i)]


      (a1**2 + b1**2)(a2**2+b2**2)
      = [(a1*a2 - b1 * b2 ) + (a1*b2 + b1*a2) i]
      *[(a1*a2 - b1 * b2 ) - (a1*b2 + b1*a2) i]

      = (a1*a2 - b1 * b2 )**2 + (a1*b2 + b1*a2)**2

      The sum of two integer squares is either prime or composite.

      If the sum of two integer squares is composite,
      it can be factored into prime integers, each of which
      is the sum of two squares.

      I do not recall the proof of this.

      Kermit
    • djbroadhurst
      ... False. If you allow zero to be a square, then 3^2 + 0^2 is a counterexample. If you do not allow zero to be a square then 3^2 + 6^2 is a counterexample.
      Message 2 of 3 , Dec 4, 2009
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        --- In primenumbers@yahoogroups.com,
        Kermit Rose <kermit@...> wrote:

        > If the sum of two integer squares is composite,
        > it can be factored into prime integers, each of which
        > is the sum of two squares.

        False.

        If you allow zero to be a square, then
        3^2 + 0^2 is a counterexample.
        If you do not allow zero to be a square then
        3^2 + 6^2 is a counterexample.
        Either way you claim is clearly wrong.

        David
      • mikeoakes2
        ... If d = 1 mod 4, then the integers of the field k(sqrt(d)) are indeed of your form but you must also allow A and B to be both half-integers. See for example
        Message 3 of 3 , Dec 4, 2009
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          --- In primenumbers@yahoogroups.com, Kermit Rose <kermit@...> wrote:
          >
          >
          > Primes in algebraic number systems:
          >
          > Consider the extension to the integers,
          >
          > {A + B * sqrt(d) }, where A, B are variable integers, and d is a
          > fixed integer.

          If d = 1 mod 4, then the integers of the field k(sqrt(d)) are indeed of your form but you must also allow A and B to be both half-integers.
          See for example Hardy & Wright, Theorem 238.

          Mike
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