Reason why in Gaussian integers, norm divisibility may not lead to divisibility.

Question

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 .

Answer 1

In $\Bbb Z$, all of the multiples of $n$ are easy to visualize: they form a regular sequence of points along the number line, spaced $n$ units apart. That's the subring $n\Bbb Z$. It's an expanded copy of $\Bbb Z$, contained within $\Bbb Z$.

In $\Bbb Z[i]$, something similar happens. Within the lattice of Gaussian integers, all of the multiples of $\alpha$ form a grid that looks like an enlargement of the entire grid; it's formed by taking $\Bbb Z$-linear combinations of $\alpha$ and $i\alpha$. You should draw a couple of these to get a feel for it. The subring $(1+i)\Bbb Z[i]$ is nice to look at, and you should try something like $(3+2i)\Bbb Z[i]$ as well.

Now, multiples of $\alpha=(3+2i)$ (norm 13) are just points in that subring/sublattice. Look at it, and then look at the sublattice generated by $\overline{\alpha}=(3-2i)$ as well. Note that they do not contain the same points. Any point in the lattice generated by $\alpha$ is a multiple of $\alpha$; any point in the lattice generated by $\overline{\alpha}$, but not in the one generated by $\alpha$, is not a multiple of alpha, even though its norm is a multiple of $13$. (The points that these two lattices have in common are precisely the Gaussian integer multiple of $13$.)

Does that help?

Answer 2

It will help to think in terms of the factorizations of $\alpha$ and $\beta$ into irreducible elements. Now, the irreducibles of $\mathbb{Z}[i]$ consist of a unit of $\mathbb{Z}[i]$ times one of:

• $1+i$
• $a \pm bi$ where $0 < b < a$ and $a^2 + b^2 = p$ is a prime integer (i.e. irreducible as an element of $\mathbb{Z}$) with $p \equiv 1\pmod{4}$. (And for each such prime $p$, there exists a unique such pair $a, b$.)
• $p$ where $p \equiv 3\pmod{4}$ is a prime integer.

Now, $\beta \mid \alpha$ if and only if the exponent of each irreducible in the factorization of $\alpha$ is greater than or equal to the corresponding exponent in $\beta$. However, looking at the norms of irreducibles, you see that for irreducibles of the second type above, $a + bi$ and $a - bi$ have the same norm. Therefore, taking the norm loses the distinction between these two irreducibles, but the divisibility condition does make the distinction (since $a+bi$ is not a unit of $\mathbb{Z}[i]$ times $a-bi$).

In the original example, $\alpha = 14 + 3i$ has irreducible factorization $\alpha = (2-i) (5+4i)$ while $\beta = 4+5i$ is irreducible itself and is equivalent to $\beta = i (5-4i)$. Therefore, the failure of $\beta$ to divide $\alpha$ comes from the distinction between $5-4i$ and $5+4i$ which the norm collapses.

Answer 3

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 $g$?