Results 1 to 4 of 4

Thread: Algebraic integers and UFDs

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    130

    Algebraic integers and UFDs

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

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

    I know that 2 is not a prime in $\displaystyle Z[\sqrt d]$. 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 $\displaystyle d \leq -3$ I don't have a handy characterization (sum of 2 squares or something like that).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    If $\displaystyle d \equiv 1 mod4$ we have that there exists $\displaystyle p\in \mathbb{Z}$ such that $\displaystyle d=4p+1$ and so $\displaystyle (1+\sqrt{d} )(1- \sqrt{d} )= 1-(4p+1)=-4p$ and this last one has a factorization in integers since it's even.

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

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Jose27 View Post
    If $\displaystyle d \equiv 1 mod4$ we have that there exists $\displaystyle p\in \mathbb{Z}$ such that $\displaystyle d=4p+1$ and so $\displaystyle (1+\sqrt{d} )(1- \sqrt{d} )= 1-(4p+1)=-4p$ and this last one has a factorization in integers since it's even.

    If $\displaystyle d<-2$ and $\displaystyle d$ odd then $\displaystyle (1+\sqrt{d} )(1- \sqrt{d} ) = -d+1$ and this one is even and so it has non-trivial factorization in integers. If $\displaystyle d$ is even take $\displaystyle (2+\sqrt{d} )(2-\sqrt{d} )$
    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 $\displaystyle a=\frac{1+\sqrt{d}}{2} \notin \mathbb{Z}[\sqrt{d}]$ and $\displaystyle a^2 - a + \frac{1 - d}{4} = 0.$ thus if $\displaystyle d \equiv 1 \mod 4,$ then $\displaystyle \mathbb{Z}[\sqrt{d}]$ won't be integrally closed and hence it cannot be a UFD.

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

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


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

  4. #4
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Another example of an integral domain that is not U.F.D is

    $\displaystyle K = F[x^2, y^2, xy]$, 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 $\displaystyle x^2$ is irreducible in K, but it is not a prime element of K.

    If it were a prime element, $\displaystyle x^2 | {(xy)}^2 $ implies $\displaystyle x^2 | xy$, which is impossible.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Gauss' Lemma and algebraic integers
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: May 25th 2010, 08:23 PM
  2. Are algebraic integers dense in the reals?
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Jun 24th 2009, 12:19 PM
  3. Algebraic Integers trouble...
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: May 8th 2009, 10:02 AM
  4. Algebraic integers
    Posted in the Math Challenge Problems Forum
    Replies: 0
    Last Post: Apr 11th 2009, 02:35 PM
  5. Algebraic integers ring
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Apr 1st 2008, 11:31 AM

Search Tags


/mathhelpforum @mathhelpforum