Define $\phi: \mathbb{Z}[i] \to \mathbb{F}_{13}$ by $\phi(a + bi) = a + 5b \mod 13$ with $\mathbb{Z}[i] := \{ a + bi \mid a,b \in \mathbb{Z}\}$. Prove that $\text{Ker}(\phi)$ is generated by $13$ and $i-5$ and that $\text{Ker}(\phi)$ is a principal ideal.
My attempt:
So I wrote out the definitions and got $$ \text{Ker}(\phi) = \{x \in \mathbb{Z}[i] \mid \phi(x) = 0 \mod 13\} = \{a + bi \in \mathbb{Z}[i] \mid \phi(a + bi) = 0 \mod 13, a,b \in \mathbb{Z}\}. $$ and $$ (13, i-5) = \{13r_1 + (i-5)r_2 \mid r_1,r_2 \in \mathbb{Z}[i]\}. $$ So we want that $$ \{a + bi \in \mathbb{Z}[i] \mid \phi(a + bi) = 0 \mod 13, a,b \in \mathbb{Z}\} = \{13r_1 + (i-5)r_2 \mid r_1,r_2 \in \mathbb{Z}[i]\}. $$ $"\supseteq"$ Let $r_1 = c + di$ and $r_2 = e + fi$, with arbitrary $c,d,e,f \in \mathbb{Z}$, then every element in $(13, i-5)$ is written like \begin{align*} 13r_1 + (i-5)r_2 &= 13(c+di) + (i-5)(e + fi) \\ &= 13c + 13di + ei - f - 5e - 5fi \\ &= 13c - f - 5e + (13d + e - 5f)i. \end{align*} So for every arbitrary element from $(13, i-5)$, we can say \begin{align*} \phi(13c - f - 5e + (13d + e - 5f)i) &= 13c - f - 5e + 5(13d + e - 5f) \mod 13 \\ &= \underbrace{13c}_{= 0 \mod 13} - f - 5e + \underbrace{5 \cdot 13d}_{=0 \mod 13} + 5e -25f \mod 13 \\ &= \underbrace{- 26f}_{= 0 \mod 13} \, \, \underbrace{- 5e + 5e}_{= 0 \mod 13} \\ &= 0 \mod 13 \end{align*} So we conclude that every arbitrary element that is generated by 13 and $i-5$ is also in the kernel of $\phi$.
$"\subseteq"$ So here is where I can't come any further. I tried to write the kernel out like I did with the previous set, so I try to find $a,b \in \mathbb{Z}$ such that $\overline{a + 5b} = \overline{0} \implies \overline{a = -5b}$. But I can't find an expression for $a$ and $b$. I must admit, working with mods was never my strongest suit.
As for the principal kernel, I don't know how to take the first step. I can't just think of an element that generates the whole kernel? Maybe this will be easier once I find an expression for $a$ and $b$.
