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

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

Let $\displaystyle d \equiv 1 \pmod{4}$ be a square-free integer with $\displaystyle d \neq 1$. By considering the factorisation of $\displaystyle d-1$, show that the ring $\displaystyle \mathbb{Z}[\sqrt{d}]$ is never a Unique Factorisation Domain.

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

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

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