The division property in integer-like sets

In 'Elements of Number Theory' John Stillwell considers the division property in the integer-like sets of numbers Z[i] (the Gaussian integers, p107) and Z[√-2] (an example of a set of quadratic integers, p119). In the case of the Gaussian integers it is relatively easy to visualize 'multiples' of integers as forming a square grid as in Figure 6.1. However, for the quadratic integers Z[√-2] this doesn't appear to be so easy and you wonder how the grid in Figure 7.1 arises and how Pythagoras' theorem in this case can be applied in the proof.

I will attempt to go through both cases and explain how I see it. Firstly, the Gaussian integers. Suppose μ and β are the Gaussian integers
$$ \mu=m+ni...(1)$$
$$ \beta=r+si...(2)$$

where m, n, r and s are elements of Z and i=√-1. Then
$$\mu\beta=m\beta+n\beta i...(3)$$
Now we have that
$$\beta i=ri+si^{2}=-s+ri...(4)$$
If we think of β and βi as vectors on the complex plane, then β is the vector (r,s) and βi is the vector (-s,r). So, as Stillwell says, i rotates the vector β anticlockwise through 90 degrees. We can see that using the standard dot product of these vectors
$$(r,s)\cdot(-s,r)=r(-s)+sr=0...(5)$$
as expected for mutually orthogonal vectors. We can see therefore that equation (3) is telling us that μβ is the sum of real multiples of β and βi. We can therefore think of this as a square grid with β and βi as base vectors, as anticipated.

Now what about the quadratic integers Z[√-2]? Suppose that this time μ and β are the quadratic integers
$$ \mu=a+b\sqrt{-2}...(6)$$
$$ \beta=c+d\sqrt{-2}...(7)$$
where a, b, c, and d are elements of Z. Then
$$\mu\beta=a\beta+b\beta\sqrt{-2}...(8)$$
What is β√-2? We have
$$\beta\sqrt{-2}=c\sqrt{-2}+d(\sqrt{-2})^{2}=-2d+c\sqrt{-2}...(9)$$
If we again think about μ and β as being vectors, then β is the vector (c,d) and  β√-2 is the vector (-2d,c). In the normal sense it does not look like these two vectors are orthogonal, however, if we define the dot product of two vectors (w,x) and (y,z) to be
$$(w,x)\cdot(y,z)=wy+2xz...(10)$$
then
$$(c,d)\cdot(-2d,c)=-2cd+2dc=0...(11)$$
and so we retrieve the orthogonality of β and β√-2. The definition in equation (10) also fits in with the definition of the norm since
$$norm\beta=(c,d)\cdot(c,d)=c^{2}+2d^{2}...(12)$$
which is as defined by Stillwell (p120). Hence we can again think that equation (8) is telling us that μβ is the sum of real multiples of β and β√-2; i.e. we can think of a grid with β and β√-2 being base vectors, as described in the book (Figure 7.1). Note, however, that this is not a square grid as β and β√-2 are not the same length.

I want to go one step further and show that the proof on page 119 doesn't have to rely on Pythagoras' Theorem. We can see from Figure 7.1 that |ρ| is less than or equal to the length of the vector from the origin to the centre of the figure. That is the vector
$$\frac{\beta}{2}+\frac{\beta\sqrt{-2}}{2}...(13)$$
which is
$$\frac{1}{2}(c,d)+\frac{1}{2}(-2d,c)=\frac{1}{2}(c-2d,d+c)...(14)$$
It follows that
$$\left|\rho\right|^{2}\le\frac{1}{2}(c-2d,d+c)\cdot\frac{1}{2}(c-2d,d+c)...(15)$$
Expanding the RHS we get
$$ \frac{1}{2}(c-2d,d+c)\cdot\frac{1}{2}(c-2d,d+c)=\frac{1}{4}((c-2d)^{2}+2(d+c)^{2})=\frac{1}{4}(c^{2}-4cd+4d^{2}+2d^{2}+4cd+2c^{2})...(16)$$
and so
$$ \frac{1}{2}(c-2d,d+c)\cdot\frac{1}{2}(c-2d,d+c)=\frac{1}{4}(3c^{2}+6d^{2})=\frac{3}{4}(c^{2}+2d^{2})=\frac{3}{4}\left|\beta\right|^{2}...(17)$$
Hence we have
$$\left|\rho\right|^{2}\le\frac{3}{4}\left|\beta\right|^{2}...(18)$$
as obtained in the book.