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I have the following assignment:

consider the map

$$|\cdot|:\mathbb{Z}[i]\longrightarrow \mathbb{N},\qquad |a+ib|:=a^2+b^2$$

1) Prove that $|\alpha|<|\beta|$ iff $|\alpha|\leq |\beta|-1$ and $|\alpha|<1$ iff $\alpha=0$

2) Let $\alpha,\beta\in\mathbb{Z}[i],\beta\neq 0$. Prove that the map $f:\mathbb{Z}[i]\longrightarrow\mathbb{Z}[i], f(\gamma):=\alpha-\gamma\beta$ is the composition of a dilatation by the factor $\sqrt{|\beta|}$, a rotation (angle?) and a translation.

3) Deduce that there exists $\gamma\in\mathbb{Z}[i]$ such that $|f(\gamma)|$ is strictly smaller than $|\beta|$.

$\textbf{Hint:}$ compare the size of a cell of the lattice $f(\mathbb{Z}[i])$ and the size of the set of points whose distance to $0$ is $\leq\sqrt{|\beta|}$.

What i did: point 1) is a trivial consequence of the fact that the norm takes integer non negative values. For point 2), I use complex multiplication of numbers which is: multiply absolute values and add angles. For point 3), i'm actually waiting for a miracle... I suppose i should prove that there exists a cell in $f(\mathbb{Z}[i])$ intersecting the open ball centered at the origin with radius $\sqrt{|\beta|}$, but i have no idea how to write down this. Only thing i noticed is that $f$ acts with a rotation, which does not affect distance from the origin, so that the only changes in $|\gamma|$ come from dilatation and by adding $\alpha$.

Could someone put me on the right direction?

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Hints (or not...)

1) This follows at once from the fac that $\;\forall\alpha\in\Bbb Z[i]\;,\;\;|\alpha|\in\Bbb N\,$

2) We have that

$$f=f_\alpha\circ f_\gamma\;,\;\;\text{with}\;\;f_\gamma(x):=\gamma x\;,\;\;f_\alpha(x)=\alpha-x$$

3) The closed unit circle centerd at the origin is compact and thus there're only a finite number of elements of any lattice (say, $\,\Bbb Z[i]\,$ ...?) intersecting it

Now you can dilate (shrink, in scientific talk...:) ) elements as shown in (2) so that their inside that circle...

The basic assumption here is that $\,\beta\neq0\,$ , I presume...

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  • $\begingroup$ thanks for answer, even if i can't still understand what to do :))...my problem is that i can't dilate and rotate as i prefer, because $\alpha$ and $\beta$ are fixed in the definiton of $f$, so how have i to choose $\gamma$ such that $f(\gamma)$ goes inside unit circle? $\endgroup$ – bateman May 23 '13 at 16:29
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    $\begingroup$ Oh, I see...perhaps you should make that clear in your post. Lemme see if I can come up with something in the next few minutes $\endgroup$ – DonAntonio May 23 '13 at 16:32
  • $\begingroup$ @bateman I cannot understand your point. If $\alpha$ and $\beta$ are fixed, then we can dilate and translate as we prefer, to find $\gamma$, right? What have I missed? $\endgroup$ – awllower May 23 '13 at 17:06
  • $\begingroup$ @awllower no, because translation is translation by $\alpha$ and rotation is rotation by the argument of $\beta$. I'm looking for a $\gamma$ such that $\alpha-\beta\gamma$ has modulus less than modulus of $\beta$, so i think the expression of such $\gamma$ depends on $\alpha$ and $\beta$ $\endgroup$ – bateman May 23 '13 at 17:40
  • $\begingroup$ @bateman So what is the problem? The expression of $\gamma$ cannot depend upon $\alpha$ and $\beta$? Thanks for the clarification still. $\endgroup$ – awllower May 24 '13 at 7:10

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