Finding a single generator for $\ker(\varphi)$; $\varphi : \mathbb{Z}[i] \to \mathbb{F}_{13}$ is defined by $\varphi (a+bi) = a + 5b (\mod{13}).$

Question

Given that $\varphi : \mathbb{Z}[i] \to \mathbb{F}_{13}$ is defined by $\varphi (a+bi) = a + 5b (\mod{13})$, I am trying to find the single generator for $ \ker (\varphi)$. I have already shown that $\ker (\varphi)$ is the ideal generated by $13$ and $i-5.$

I then tried

$$ \frac{13}{-5+i} \cdot \frac{-5-i}{-5-i} = \frac{-65-13i}{26} = -2\frac{1}{2} - \frac{1}{2}i. $$

I am not sure how to proceed from here, so any help is greatly appreciated.

Answer 1

You have $13=(3+2i)(3-2i)$ and $5-i=(3+2i)(1-i)$ hence $$13\Bbb Z[i]+(5-i)\Bbb Z[i]\subseteq(3+2i)\Bbb Z[i]$$ On the other hand $3+2i=13-2(5-i)$, hence $$(3+2i)\Bbb Z[i]\subseteq 13\Bbb Z[i]+(5-i)\Bbb Z[i]$$ Consequently, $$(3+2i)\Bbb Z[i]=13\Bbb Z[i]+(5-i)\Bbb Z[i]$$ that's $3+2i$ generates the kernel of your homomorphism.