Here's my new attempt at showing $Z[i]/(2+3i) \simeq Z/13Z$.
proof:
Let's define the natural homomorphism $$\phi: Z \to Z[i]/(2+3i)$$ where $$\phi(z)= z+ (2+3i)$$ It is easy to check that this is a ring homomorphism.
Claim 1: $Ker (\phi)= 13 Z$
proof: We will show both inclusions.
Let $z\in 13Z$ then $z=13k$ for some integer k. But $13k= (2+3i)(2k-3ik)$ and so $z \in (2+3i)$ and so $z+(2+3i)=0+(2+3i)$ as cosets and hence $\phi(z)= 0 +(2+3i)$ i.e. $z \in ker(\phi)$ and hence $13Z \subset ker(\phi)$
Now let's show the reverse inclusion.
Let $z\in ker(\phi)$, then $z \in (2+3i) \cap Z$. Then $z= (2+3i)(a+bi)$ for some integers a,b such that $z\in Z$, and so multiplying out we get $z= (2a-3b) + (2b+3a)i$ for some $a,b$ such that $z$ is an integer. Now for z to be an integer, the complex part must be $0$, hence $2b=-3a$. Now 3 is a prime dividing $2b$ which implies 3 divides b. Similarly we have that 2 divides a. Hence $b=3k$ and $a=2m$ for some integers k and m. Substituting into $2b=-3a$ we obtain $6k=-6m$ and hence $k= -m$ and hence $b=-3m$ and $a= 2m$ Finally substituting into $z=(2+3i)(a+bi)$ we get $z=(2+3i)(2m-3mi)= 13m$ and hence $z\in 13Z$ and we are done.
Claim 2: $\phi$ is surjective
I have also done this part but I would like to make sure my first claim is correct first.
Hence by the first isomorphism theorem, we have that $$Z/13Z \simeq Z[i]/(2+3i)$$ Thanks.
