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Finding all units in Z[sqrt(2)]

  1. Feb 28, 2010 #1
    How would I show that every unit in Z[sqrt(2)] is of the form +/- (1 +/- sqrt(2) )^n ?

    I can show these are all units, but I can't show every unit is one of these. From some research, I'm aware this is a special case of Dirichlet's Unit Theorem, but that is far above the level I'm working at.

    Any help would be appreciated :)

    Thanks
     
  2. jcsd
  3. Mar 2, 2010 #2
    Every element of [tex]Z[\sqrt{2}][/tex] is of the form [tex]a+b\sqrt{2}[/tex], a,b integers.

    Suppose [tex]a+b\sqrt{2}[/tex] is a unit. Then: [tex](a+b\sqrt{2})(c+d\sqrt{2})=1[/tex].

    So:[tex]ac+(bc+ad)\sqrt{2}+2bd=1[/tex]. So already you know that [tex]bc=-ad[/tex].

    Now just try manipulating the expression you are left with: [tex](ac+2bd=1)[/tex] using this fact. You should be able to show that [tex]a,b,c,d[/tex] must all be either 1 or -1.
     
  4. Aug 13, 2010 #3
    The problem is they don't have to. For example, [tex]3+2\sqrt{2[/tex] is unit in [tex]\mathbb{Z}[\sqrt{2} ][/tex]. I'm also trying to solve this problem right now.

    Edit: I think I got it. Let [tex]\nu(a+b\sqrt{2})=a^2-2b^2[/tex]. It's multiplicative, so [tex]z[/tex] is unit iff [tex]\nu(z)=\pm 1[/tex]. It's easy to see that signs of [tex]a[/tex] and [tex]b[/tex] doesn't matter when it comes to invertibility, so we can assume [tex]z>0[/tex].

    Lemma Smallest element of [tex]\mathbb{Z}[\sqrt{2}][/tex] with [tex]z>1[/tex] and [tex]\nu(z)=\pm 1[/tex] is [tex]1+\sqrt{2}[/tex]
    Proof
    Let [tex]a,b\in \mathbb{N}[/tex]. If such element is of type [tex]a-b\sqrt{2}>1[/tex], then [tex]a+b\sqrt{2}>1[/tex] and so [tex]\nu({a-b\sqrt{2}})>1[/tex]. Similarly if it's [tex]-a+b\sqrt{2}[/tex]. It has to be [tex]a+b\sqrt{2}[/tex]. Case with one of a, b equal zero is easily checked by hand. This proves the lemma.

    Now suppose [tex]\nu(z)=1[/tex], [tex]z > 0[/tex]. Let [tex] z_0 = 1+\sqrt{2}[/tex]. There exists integer k (positive or negative) with [tex]z_0^k \leq z < z_0^{k+1}[/tex]. Easily to check, [tex]1\leq z z_0^{-k} < z_0[/tex]. If [tex]z z_0^{-k}\neq 1[/tex], it contradicts minimality of [tex]z_0[/tex], and so [tex]z=z_0^k[/tex].

    Hope I haven't messed up too badly.
     
    Last edited: Aug 14, 2010
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