1. ## Ideals

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

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

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

2. So far I've got

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

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

I'm lost after this...

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

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

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

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

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

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

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

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