TITLE: Reason why in Gaussian integers, norm divisibility may not lead to divisibility.
QUESTION [5 upvotes]: It is taken as true (with a very easy proof) for Gaussian integers, that for $\alpha, \beta \in \mathbb {Z}[i]$, if $\beta \mid \alpha$ then $N_{\beta} \mid N_\alpha$ in $\mathbb {Z}$. It would be an aid for easier check of $\beta \mid \alpha$ in $\mathbb {Z}$, provided the reverse is true, but it is usually not true, as the following example shows:
$$\alpha = 14+3i,\qquad\beta= 4 +5i.$$
Here, $N_{\beta} = 41$, $N_{\alpha}=205$, but the ratio $\frac{\alpha}{\beta} = \frac{14+3i}{4+5i}$, after rationalization $ = \frac{71}{41} -\frac{58}{41}i \notin \mathbb {Z}$.
There should be some reason behind this, I hope, that can be expressed geometrically and/or algebraically .

REPLY [4 votes]: From your choice of example and notation I wonder if this is coming from pages 2 and 3 of http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf or the similarity is a coincidence. 
In any case, the basic reason you should not expect norm divisibility to imply actual divisibility in $\mathbf Z[i]$ is that norms can lose essential information.  There are Gaussian integers with the same norm that do not have the same factors, such as $1+2i$ and $1-2i$, which both have norm $5$ but neither divides the other (in fact they are relatively prime in $\mathbf Z[i]$). In general you can not reconstruct $\alpha$ up to unit multiple in $\mathbf Z[i]$ from knowledge of ${\rm N}(\alpha)$, so why should you expect divisibility of norms to imply divisibility of elements? 
(In $\mathbf Z[i]$ there are some situations where information about the norm is equivalent to information about the original element, such as being invertible. In certain cases there is equivalence for divisibility, such as $(1+i) \mid \alpha \Leftrightarrow 2 \mid {\rm N}(\alpha)$, but you should not expect such things to be true in general.)
Consider an analogue with polynomials and degrees: instead of ${\rm N}(\alpha\beta) = {\rm N}(\alpha){\rm N}(\beta)$ in $\mathbf Z[i]$ we have $\deg(fg) = \deg f + \deg g$ in $\mathbf R[x]$, so a polynomial analogue of $\alpha \mid \beta \Rightarrow {\rm N}(\alpha) \mid {\rm N}(\beta)$ is $f \mid g \Rightarrow\deg f \leq \deg g$. If I told you two polynomials $f$ and $g$ in $\mathbf R[x]$ satisfy $\deg f \leq \deg g$, would you expect $f$ should divide $g$?