(I think I understand how to show $\mathbb{Z}[i]/\langle5\rangle \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$ using the Chinese remainder theorem.)
I don't know how to "identify the structure of the additive group" of $\mathbb{Z}[i]/\langle 3 + 4i \rangle$. The answer is apparently $\mathbb{Z}/25\mathbb{Z}$, and the solutions say to show that $$ \phi : \mathbb{Z}[i]/\langle 3 + 4i \rangle \to \mathbb{Z}/25\mathbb{Z} \qquad \text{given by} \qquad a + ib \mapsto a - 7b $$ is an isomorphism. There are a few things I note:
- $7^2 = -1 \mod{25}$
- $\phi(3 + 4i) = 0$
- $25 = 3^2 + 4^2$
But, I can't put the pieces together. What I want to know is:
- Why is it 25 in $25 \mathbb{Z}$?
- How can we guess to send $a + ib$ to $a - 7b$?
