Question 5

Given \(R = \mathbb{Z}[i]\) and \(N(a+bi) = a^2 + b^2 = \left\lvert a+bi\right\rvert^2\) where \(\left\lvert\cdot\right\rvert\) denotes complex modulus, considering the natural inclusion \(\mathbb{Z}[i]\subset \mathbb{C}\).

To show that \(R\) is an Euclidean domain, let \(z, w\in R\) be arbitrary and we want to know if there exists solution for \[ z = qw + r \] where \(q,r\in R\) with \(r = 0\) or \(\left\lvert r\right\rvert < \left\lvert w\right\rvert\).

Solving in \(\mathbb{C}\), a priori we know that \[ \frac{z}{w} = q + \frac{r}{w} \] with \(\left\lvert\frac{r}{w}\right\rvert < 1\). Now treat \(\frac{z}{w}\) as just a complex number and we know that we can choose \(q\in \mathbb{Z}[i]\) satisfying \(\left\lvert\frac{z}{w} - q\right\rvert < 1\) (in fact it can be \(\frac{\sqrt{2}}{2}\)). Now we choose \(r = z - qw\) and as \(\left\lvert\frac{r}{w}\right\rvert < 1\), the condition is satisfied.

Question 6

For both finding gcd and solving for Bézout’s inequality, we do Division algorithm,

\[\begin{align*} 11 + 7i &= (1 - i)(3 + 7i) + (1 + 3i) \\ 3 + 7i &= 2(1 + 3i) + (1 + i) \\ 1 + 3i &= (2 + i)(1 + i) \end{align*}\]

1. \(\gcd(11 + 7i, 3 + 7i) = 1 + i\)

2. \[\begin{align*} 1 + i &= (3 + 7i) - 2(1 + 3i) \\ &= (3 + 7i) + 2(1 - i)(3 + 7i) - 2(11 + 7i) \\ &= (3 - 2i)(3 + 7i) - 2(11 + 7i) \end{align*}\]