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Re: [PrimeNumbers] primes of polynomials

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  • Phil Carmody
    ... First you need to specify the ring in which you re working. You do not do so. One cannot assume Z[x] given that many of the examples we ve seen have been
    Message 1 of 14 , Apr 2 11:44 AM
      --- gordon_as_number <gordon_as_number@...> wrote:
      > Only a comment.
      >
      > If you want to prove a polynomials P(x) is prime,

      First you need to specify the ring in which you're working.
      You do not do so.
      One cannot assume Z[x] given that many of the examples we've seen have
      been instead in Q[x].

      > isnt enough to determine the assoiated roots.
      >
      > E.g. P(x)=x^2+2
      >
      > The number N is P(x) which can base expanded in
      > some base b as
      >
      > N=\sum_{i=0}^m c_i b^i .

      Bases ought to be irrelevant. If they're not, you're doing something wrong.

      > Determining the roots and the coefficients I claim
      > is order \ln^3(N), which is a rough estimate here
      > because I didnt let N vary

      Order(funtion(N)) only makes sense if you let N vary.
      You've reached that level of illucidity again...

      > and I didnt include what
      > the value of m is. So you have to polynomials evaluated
      > in the different bases b

      Yup - illucid. Polynomials can be evaluated, but their value is
      independent of any base.

      > with the same coefficients c_i.

      The c_i, as you introduced them, are not coefficients.

      > The number of real c.c. pairs (x-\alpha_i)*(x-\alpha_j)^\star
      > = integer or (x-\alpha_i) an integer span the prime factors.

      With no firm foundation to build on, you've completely lost me now.
      No point in continuing.

      Phil

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