Exercise 15.5.1

Proof. If $[\alpha ]\in {\left(\mathbb{Z}[i]∕\mathit{\beta \mathbb{Z}}[i]\right)}^{\text{∗}}$, there is some $\gamma \in \mathbb{Z}[i]$ such that $[\alpha ][\gamma ]=[1]$. This shows that $1\in [\alpha ][\gamma ]=[\mathit{\alpha \gamma }]=\mathit{\alpha \gamma }+\mathit{\beta \mathbb{Z}}[i]$, so that

If $\xi \in \mathbb{Z}[i]$ divides $\alpha$ and $\beta$, then $\xi$ divides $1=\mathit{\alpha \gamma }-\mathit{\beta \delta }$, thus $\xi$ is a unit. This proves that $\alpha ,\beta$ are relatively prime.

Conversely, suppose that $\alpha ,\beta$ are relatively prime. Since $\mathbb{Z}[i]$ is a principal ideal domain, the ideal $\mathit{\alpha \mathbb{Z}}[i]+\mathit{\beta \mathbb{Z}}[i]$ is equal to $\mathit{\zeta \mathbb{Z}}[i]$ for some $\zeta \in \mathbb{Z}[i]$:

Then $\alpha =\alpha \cdot 1+\beta \cdot 0\in \mathit{\alpha \mathbb{Z}}[i]+\mathit{\beta \mathbb{Z}}[i]=\mathit{\zeta \mathbb{Z}}[i]$, thus $\alpha =\mathit{\zeta \lambda },\lambda \in \mathbb{Z}[i]$, and $\zeta \mid \alpha$. Similarly, $\beta =\alpha \cdot 0+\beta \cdot 1\in \mathit{\zeta \mathbb{Z}}[i]$, so that $\zeta \mid \beta$.

Since $\zeta \mid \alpha ,\zeta \mid \beta$, were $\alpha$ and $\beta$ are relatively prime, this implies that $\zeta$ is a unit, so that $\mathit{\zeta \mathbb{Z}}[i]=\mathbb{Z}[i]$, and

Then $1\in \mathbb{Z}[i]=\mathit{\alpha \mathbb{Z}}[i]+\mathit{\beta \mathbb{Z}}[i]$, thus there are some $\gamma ,\delta \in \mathbb{Z}[i]$ such that $1=\mathit{\alpha \gamma }+\mathit{\beta \delta }$. Since $[\beta ]=[0]$,

where $[\gamma ]\in \mathbb{Z}[i]∕\mathit{\beta \mathbb{Z}}[i]$. This proves that $[\alpha ]$ has an inverse $[\gamma ]\in \mathbb{Z}[i]∕\mathit{\beta \mathbb{Z}}[i]$:

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