How to show that the kernel of this ring homomorphism $\phi: \mathbb{Z}[i]\to\mathbb{Z}/5\mathbb{Z}$ is the principal ideal $(2+i)$?

Question

In order to prove that $Z[i]/(2+i)\cong \mathbb{Z}/5\mathbb{Z}$, I defined the homomorphism $\phi(a+bi)=(a-2b)\mod 5$. It is easy to show that $2+i\in\ker(\phi)$, but I'm unsure how to prove the reverse inclusion.

I have the following so far: Suppose $a+bi\mapsto 0$. Then $a-2b\equiv 0\mod 5$, so $5$ divides $a-2b$.

I'm wondering if I can use the fact that $2+i$ is irreducible in $\mathbb{Z}[i]$.

Answer 1

Hint: Going from where you left off, suppose that $a-2b = 5n$; then $a+bi = 2b+5n+bi$. Now, why is this a multiple of $2+i$?

Answer 2

Now show that the quotient has five elements (for example, by considering $\mathbb Z[i]$ as a free abelian group of rank two and the ideal $(2+i)$ as the subgroup generated by $(2,1)$ and $(-1,2)$, and doing row and column reduction i.e. Smith normal form). Any surjective or injective function between sets of the same cardinality is bijective.