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Math Help - Eisenstein integers

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    Last edited by aaa0909; June 5th 2008 at 10:34 PM.
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    Quote Originally Posted by aaa0909 View Post
    a) the rational prime 2 is an Eisenstein prime
    The proof for (b) will do this one.

    b) a rational prime of the form 3k+2, where k is a positive integer, is an Eisenstein prime
    Say that p = \alpha \beta where \alpha , \beta are non-units. It follows that N(\alpha),N(\beta) > 1 and so p^2 = N(\alpha)N(\beta) thus  p = N(\alpha). Let \alpha = a+b\omega then p = a^2 - ab + b^2. Thus, 4p = (2a-b)^2 + 3b^2 which means p\equiv (2a-b)^2 (\bmod 3). If 3\not | p then p\equiv 1(\bmod 3) but 1 is the only square mod 3. Thus, rational primes which are of form 3k+2 are Eisenstein primes.

    c) a rational prime of the form 3k+1, where k is a positive integers, factors into the product of two primes that are not associates of one another
    Say that p has form 3k+1. By using quadradic reciprocity it follows that (-3/p) = 1. Thus, there is an integer n so that n^2 \equiv -3(\bmod p) thus p|(n^2+3). Thus, p|(n+i\sqrt{3})(n-i\sqrt{3}) = (n+1+2\omega)(n-1-2\omega). If were prime then it would divide both those factors but that is impossible since then p|[(n+1+2\omega)+(n-1-2\omega)] \implies p|2n, since p\not = 2 it means p|n. This is impossible and so p is reducible. Thus, p = \alpha \beta. We have p^2 = N(\alpha)N(\beta) and so N(\alpha) = N(\beta) = p. Since the norm is a rational prime it means \alpha,\beta are prime. Note, p = N(\alpha)  = \alpha \bar \alpha. And so p factors into two prime \alpha , \bar \alpha which are not associate.
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