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NonCommAlg · #3 November 3rd, 2009, 03:23 AM

Quote:

Originally Posted by Jose27

If $d \equiv 1 mod4$ we have that there exists $p\in \mathbb{Z}$ such that $d=4p+1$ and so $(1+\sqrt{d} )(1- \sqrt{d} )= 1-(4p+1)=-4p$ and this last one has a factorization in integers since it's even.

If $d<-2$ and $d$ odd then $(1+\sqrt{d} )(1- \sqrt{d} ) = -d+1$ and this one is even and so it has non-trivial factorization in integers. If $d$ is even take $(2+\sqrt{d} )(2-\sqrt{d} )$

you forgot that a domain is not a UFD if some element can be written as product of irreducible elements in more than one way.

1) we have $a=\frac{1+\sqrt{d}}{2} \notin \mathbb{Z}[\sqrt{d}]$ and $a^2 - a + \frac{1 - d}{4} = 0.$ thus if $d \equiv 1 \mod 4,$ then $\mathbb{Z}[\sqrt{d}]$ won't be integrally closed and hence it cannot be a UFD.

2) if $d < -2,$ then $2$ is an irreducible element of $\mathbb{Z}[\sqrt{d}].$ (very easy to see!) so if $\mathbb{Z}[\sqrt{d}]$ was a UFD, then $2$ would have to be prime. now choose $n \in \mathbb{Z}$ such that $n^2-d$ is an even number.

so $2 \mid (n-\sqrt{d})(n+\sqrt{d})$ and thus either $2 \mid n + \sqrt{d}$ or $2 \mid n - \sqrt{d},$ which is obviously impossible. Q.E.D.

Remark: clearly $2 \mid n + \sqrt{d}$ if and only if $2 \mid n - \sqrt{d}.$ so that "either ... or" statement in 2) is not necessary and we could have just said $2 \mid n+\sqrt{d}.$

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