Isomorphism from $Z[i]/(p)$ to $(p^n)/(p^{n+1})$ where $p$ is a irreducible element .

Question

Let $p\in\Bbb{Z}[i]$ be an irreducible element. Prove that for any positive integer $n\ge0$ the ideal $(p^{n+1})$ is an ideal in $(p^{n})$, and prove that multiplication by $p^n$ induces a isomorphism (defined above) between $\Bbb{Z}[i]/(p)$ and $(p^n)/(p^{n+1})$ as additive abelian groups

What I thought , was that $\Bbb{Z}[i]/(p)$ will be isomorphic to a field with $p^2$ elements.So if I define an isomorphism $$\phi:\ \Bbb{Z}[i]\ \longrightarrow\ \Bbb{Z}[i]/(p),$$ where the kernel is the ideal $(p)$ and the other elements are going to be $(c+id )+(p)$ where $0\le c,d \le (p-1)$ .

Now $(p^n)= p^n\Bbb{Z}[i]$ and $(p^{n+1}) = p^{n+1}\Bbb{Z}[i]$. Now $(p^{n+1}) $ is going to be a subring of $(p^n)$ and if we define an isomorphism $\sigma:(p^n)-> (p^n)/(p^{n+1})$.Here the kernel will be $(p^n)$ and the other elements will be $(a+ib)p^n$ where $0\le a,b \le (p-1)$.I will probably need to show that this mapping is isomorphic to a field of size $p^2$

So this is what I have tried to do precisely.This is a question from Dummit and Foote (pg-293 Q-7(a)) .What I did not understand is what does multiplication by $p^n$ mean . Instead of writing a complete answer I would recommend some hints .

Answer 1

There are a few problems with what you have written:

1. If $p$ is an integer then indeed $\Bbb{Z}[i]/(p)$ is a field of order $p^2$. However, if $p$ is not an integer then $\Bbb{Z}[i]/(p)$ is a field of order $N(p)$.
2. There does not exist an isomorphism between $(p^n)$ and $(p^n)/(p^{n+1})$. In fact by your earlier remarks, the former is infinite and the latter is finite.
3. Why are you interested in such an isomorphism? You are asked to show that $\Bbb{Z}[i]/(p)$ is isomorphic to $(p^n)/(p^{n+1})$ as abelian groups.

I will rephrase the original question from the book to be more explicit, in hopes of clarifying what it asks:

Let $p\in\Bbb{Z}[i]$ be an irreducible element and $n$ a positive integer. Show that the map $$\Bbb{Z}[i]/(p)\ \longrightarrow\ (p^n)/(p^{n+1}):\ x\ \longmapsto\ p^nx,$$ is an isomorphism of additive abelian groups.

Answer 2

The map defined by

$\sigma:x-> p^nx$ is an isomorphic map from $Z[i]/(p)->(p^n)/(p^{n+1})$. We first try to see whether the map is homomorphic or not

$\sigma((a+ib + c+id)) = p^{n}(a+c) + p^{n}(b+d)i = \sigma(a+ib) + \sigma(c+id)$.

1)injectivity: $\sigma(x)=\sigma(y).p^n(x)= p^n(y) .x=y$

2) surjectivity:

In $ (p^n)/(p^{n+1})$ the elements will be of the form $ p^n +(x+iy)$ where $0\le x,y\le N(p)-1$, which will be surjective as the cosets in $Z[i]/(p)$ will be of the form $ (a+ib) $ where $0 \le a,b \le N(p)-1$.