Results 1 to 5 of 5

Math Help - Show x is a unit iff N(x)=1 in Z[w]

  1. #1
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2

    Show x is a unit iff N(x)=1 in Z[w]

    Let <br />
w = \frac {-1}{2} + \frac {3^{1/2}}{2}i<br />
, let <br />
N:Z[w] \rightarrow Z \ by\  N(a+bw) = a^2 - ab + b^2<br />
and suppose that x is in Z[w]. Show that x is a unit iff N(x)=1.

    My proof so far:

    Assume xy = 1 for y in Z[w].

    1 = N(1) = N(xy) = N(x)N(y), implies that N(x) = N(y) = 1.

    Conversely, let x = a + bw with N(x) = 1.

    Then N(a+bw) = 1 = a^2 - ab + b^2

    Then that should equals to (a+bw)(???)

    Any hints? Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by tttcomrader View Post
    Let <br />
w = \frac {-1}{2} + \frac {3^{1/2}}{2}i<br />
, let <br />
N:Z[w] \rightarrow Z \ by\  N(a+bw) = a^2 - ab + b^2<br />
and suppose that x is in Z[w]. Show that x is a unit iff N(x)=1.

    My proof so far:

    Assume xy = 1 for y in Z[w].

    1 = N(1) = N(xy) = N(x)N(y), implies that N(x) = N(y) = 1.

    Conversely, let x = a + bw with N(x) = 1.

    Then N(a+bw) = 1 = a^2 - ab + b^2

    Then that should equals to (a+bw)(???)

    Any hints? Thanks.
    You're looking for an element (???) such that (a+bw)(???) = 1, where \textstyle w=-\frac12+\frac{\sqrt3}2i. The standard way of doing this is to find the inverse of a+bw by rationalising the denominator:

    \frac1{a+bw} = \frac{a+b\bar{w}}{(a+bw)(a+b\bar{w})}, where \bar{w} is the complex conjugate of w. When you multiply out the expression in the denominator, you'll discover the reason for defining N(a+bw) in that way.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Assuming that N is a euclidean norm, then here is my proof..

    If x is a unit, then xx^{-1} = 1. So,
    1 = N(1) = N(xx^{-1}) = N(x)N(x^{-1}). Since, for any x, N(x) \geq 0, we have N(x) = 1.

    Coversely, suppose N(x) = 1. Let x = a + bw, where a,b \in \mathbb{Z}. Now, N(x) = N(a + bw) = a^2 - ab + b^2 = 1
    This means that (a,b) forms a solution set
    \{ (\pm 1,0), (0, \pm 1), (-1,-1) \}.

    For (a,b) = (\pm 1,0) \implies x = \pm 1 which are units.
    For (a,b) = (0, \pm 1) \implies x = \pm w = \frac{\mp 1}{2} + \frac{ \pm \sqrt 3}{2}i which are also units.
    For (a,b) = (\pm 1,\pm 1) \implies x = \pm 1 \pm  1w = -1 + \frac{1}{2} - \frac{\sqrt 3}{2}i = \frac{-1}{2} - \frac{\sqrt 3}{2}i which is also a unit.

    In all cases, x is a unit. QED.
    Last edited by kalagota; January 28th 2008 at 07:11 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2
    I was working with Opalg's proof, and now I got to this point:

    1=a^2-ab+b^2=(a+bw)( \frac {a+b \bar {w}}{a^2-ab+b^2} )

    But ain't I looking for (a+bw)(u+vw) for some integers u and v? And how do I turn that into w? Thank you!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by tttcomrader View Post
    I was working with Opalg's proof, and now I got to this point:

    1=a^2-ab+b^2=(a+bw)( \frac {a+b \bar {w}}{a^2-ab+b^2} )

    But ain't I looking for (a+bw)(u+vw) for some integers u and v? And how do I turn that into w? Thank you!
    Don't forget that you're assuming that N(a+bw)=1. So (a+bw)(a+b \bar {w}) = 1. You also know that \textstyle\bar{w} = -\frac12-\frac{\sqrt3}2i = -1-w. Therefore (a+bw)(a-b-bw) = 1, and you can take u=ab and v=b.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. unit length unit rise ratio
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: October 16th 2011, 09:45 PM
  2. How to show work (Unit circle)
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 14th 2009, 06:09 PM
  3. Replies: 3
    Last Post: September 17th 2009, 08:35 PM
  4. show unit normal vector..
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 12th 2009, 07:22 PM
  5. Replies: 1
    Last Post: February 9th 2009, 08:13 PM

Search Tags


/mathhelpforum @mathhelpforum