Algebraic integers and UFDs

robeuler · #1 $Old$ November 2nd, 2009, 05:32 AM

Let $d$ be an integer (square free, not equal to one)

Prove that if $d=1(mod4)$ or $d \leq -3$ then $Z[\sqrt d]$ is not a UFD.

I know that 2 is not a prime in $Z[\sqrt d]$ . I also know that if d is one mod three we can rewrite it as the sum of 2 squares. But I don't know how to come up with explicit d. For $d \leq -3$ I don't have a handy characterization (sum of 2 squares or something like that).

Jose27 · #2 $Old$ November 2nd, 2009, 07:15 PM

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 $(2+\sqrt{d} )(2-\sqrt{d} )$

NonCommAlg · #3 $Old$ November 3rd, 2009, 04:23 AM

Quote:

Originally Posted by Jose27 $View Post$

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 $2 \mid n+\sqrt{d}.$

aliceinwonderland · #4 $Old$ November 3rd, 2009, 10:06 PM

Another example of an integral domain that is not U.F.D is

$K = F[x^2, y^2, xy]$ , where F is a field. The proof of showing that K is not U.F.D is similar to the above proof.

We see that $x^2$ is irreducible in K, but it is not a prime element of K.

If it were a prime element, $x^2 | {(xy)}^2$ implies $x^2 | xy$ , which is impossible.