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    Hi,

    I have to find the smallest positive prime p in Z such that p = 1mod4 and p is not irreducible in R = Z[sqrt(-3)].

    So we need to find non units a,b such that for example 13 = ab, but how can this be done if 13 is prime?

    Taking the norm function gives 169 = N(a)N(b) so we need to find N(a) = N(b) = 13.

    Solving N(a) = 13 eg u^2 + 3v^2 = 13 gives u = +- 1 and v = +- 2 but I am not sure what this means.

    If 1 doesn't count because it's a unit then 37 gives solutions u = +-5 and v = +-2.

    Any help is appreciated.
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    Take a solution to u^2+3v^2 = 13 and form z = u + v \sqrt{-3}. What is z \bar z?
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    (Original post by DFranklin)
    Take a solution to u^2+3v^2 = 13 and form z = u + v \sqrt{-3}. What is z \bar z?
    7/3, not sure how you formed that or what it means.

    I thought prime elements are irreducible in an integral domain.
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    (Original post by Determinism)
    7/3, not sure how you formed that or what it means.
    I think maybe you were thrown by me using TeX without tags (was on mobile). With the tags, I said to form z = u + v \sqrt{-3}. And really this is *why* you look for solutions to u^2+3v^2 = 13 in the first place.

    I thought prime elements are irreducible in an integral domain.
    Just because p is prime in Z doesn't mean it is also prime in Z[\sqrt{-3}].

    It's a long while since I did this stuff, and obviously different courses may teach in a different order. But the relationship between primes in Z and solutions to N(z) = p is normally discussed as pretty much the first thing in lectures after introducing the idea of a norm. Your current posts leave me feeling that you are trying to answer questions without having covered/understood the related material.
 
 
 
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