1. ## Ideals

Suppose $p$ is irreducible in $\mathbb{Z}[i]$.

1. Let $A_n = (p^n)$ (the ideal generated by $p^n$). Prove $\mathbb{Z}[i]/(p) \cong A_n/A_{n+1}$ as additive abelian groups.

2. Show $|\mathbb{Z}[i]/(p^n)| = |\mathbb{Z}[i]/(p)|^n$.

2. So far I've got

1. $A_n/A_{n+1} = \{a+(p^{n+1})|a\in (p^n)\}$
and $a=q\cdot p^{n+1} + r$ where $r=0$ or $N(r) < N(p^{n+1})$.

We also know $a = b\cdot p^n$ for some $b\in \mathbb{Z}[i]$.

I'm lost after this...

3. $R = \mathbb{Z}[i]/(p) = \{a+(p)|a\in\mathbb{Z}[i]\}$.

$S = (p^n)/(p^{n+1}) = \{a+(p^{n+1})|a\in (p^n)\} = \{a\cdot p^n+(p^{n+1})|a\in\mathbb{Z}[i]\}$.

Define $\varphi: R \rightarrow S$ such that $a \mapsto p^n\cdot a$.

$\varphi(0)=0$ and $\varphi(a+b) = p^n(a+b) = p^n\cdot a + p^n\cdot b = \varphi(a)+\varphi(b)$, hence $\varphi$ is a homomorphism.

Suppose $\varphi(a) = \varphi(b)$, then $p^n\cdot a = p^n\cdot b$ which forces $a = b$ since we're working in an integral domain. Thus $\varphi$ is injective.

given $b\in S \Longrightarrow b = p^n\cdot a$ for some $a\in \mathbb{Z}[i]$. So choose said $a$ to get $\varphi(a) = b$. Hence $\varphi$ is surjective.

Therefore $\varphi$ is an isomorphism with for these additive groups.

Finally we get $\mathbb{Z}[i]/(p) \cong (p^n)/(p^{n+1})$ as additive abelian groups.

**For some reason I'm not too confident in this solution, so someone else might want to look it over.**

4. Hint for #2:

Use induction and use #1 in your yielding case.