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Math Help - Algebraic integers and UFDs

  1. #1
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    Algebraic integers and UFDs

    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).
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  2. #2
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    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} )
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  3. #3
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    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}.
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  4. #4
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    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.
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