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Math Help - Prove Z[sqrt(d)] is not a UFD when d=1 mod 4

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    Prove Z[sqrt(d)] is not a UFD when d=1 mod 4

    Let d \equiv 1 \pmod{4} be a square-free integer with d \neq 1. By considering the factorisation of d-1, show that the ring \mathbb{Z}[\sqrt{d}] is never a Unique Factorisation Domain.
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    Quote Originally Posted by Boysilver View Post
    Let d \equiv 1 \pmod{4} be a square-free integer with d \neq 1. By considering the factorisation of d-1, show that the ring \mathbb{Z}[\sqrt{d}] is never a Unique Factorisation Domain.
    first show that 2 is irreducible: if 2=(m+n\sqrt{d})(r+s \sqrt{d}), then 4=(m^2-n^2d)(r^2-s^2d). but the equation x^2-y^2d= \pm 2 has no integer solution since y^2d \pm 2 \equiv y^2 \pm 2 \equiv 2 \ \text{or} \ 3 \mod 4.

    now, if \mathbb{Z}[\sqrt{d}] was a UFD, then, because 2 is irreducible, it'd have to be prime. therefore, since 2 \mid d-1 = (1 + \sqrt{d})(-1 + \sqrt{d}), we'd have either 2 \mid 1 + \sqrt{d} or 2 \mid -1 + \sqrt{d}, which is impossible.
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