Surjective ring homomorphism from $\mathbb{Z}[i] \to \mathbb{Z}_q$

abstract-algebraeuclidean-domaingaussian-integersring-homomorphismring-theory

So I've got this question that is asking me to show that the map $\phi: \mathbb{Z}[i] \to \mathbb{Z}_q$ such that $\phi(r + is) = [r] +[s][n]$ is a surjective ring homomorphism where $0 \neq n \in \mathbb{Z}$ and let $q = n^2+ 1$ denote the equivalence class of $r \in \mathbb{Z} \, \text{mod} \, q$ by $[r]$.
So far I think I was able to show the homomorphism (I would appreciate if I could know whether I have done it correctly or not) but I couldn't really figure out how to show the map is a surjection. I would appreciate any guidance.

my work for the homomorphism
for $z = a+bi$ and $w=c+di$ where $a,b,c,d \in \mathbb{Z}$ then

firstly:
$ \phi(z+w) = \phi((a+c)+i(b+d)) = [a+c] + [b+d][n]$
$ = [a] +[c] +[b][n]+[d][n] $
$ = \phi(z) +\phi(w)$

secondly :
$ \phi(zw) = \phi((ac-bd)+i(ad+bc)) = [ac-bd] + [ad+bc][n]$
$ = [ac] +[-bd] +[ad][n]+[bc][n] $
$ = \phi(z) \phi(w)$
given that the congruence class for $bd$ is equal to $ -bd$.

I intuitively know that the homomorphism is mapping every element of $\mathbb{Z}[i]$ to $\mathbb{Z}_q$ in addition there are elements that when mapped they lie in the same congruence class so it would make the map a surjection but I'm not sure whether this intuition is correct and additionally how can I go and prove it.

I was further thinking how would I go about finding the kernel of $\phi$.

thanks in advance for your guidance

Best Answer

Your argument that $\phi$ is homomorphic is correct, except for your claim that

the congruence class for $bd$ is equal to $−bd$

That's not true in general. However, if you just compute $\phi(z)\phi(w)=([a]+[bn])([c]+[dn])$, you'll see that it's equal to $[ac-bd] + [ad+bc][n]=\phi(zw)$.

As for surjectivity: doesn't $[r]+[0][n]=[r]$ cover everything?

$\\$

EDIT for elaboration:

Another way to prove that $\phi$ is a surjective homomorphism is to notice that $\phi$ is essentially just

$\dfrac{\mathbb{Z}[x]}{\langle x^2+1\rangle}\to\dfrac{\mathbb{Z}[x]}{\langle x^2+1,\;q\rangle}\to\dfrac{\mathbb{Z}_q[x]}{\langle x^2+1\rangle}=\dfrac{\mathbb{Z}_q[x]}{\langle\, (x+n)(x-n)\,\rangle}\to\dfrac{\mathbb{Z}_q[x]}{\langle x-n\rangle}\to\mathbb{Z}_q[n]=\mathbb{Z}_q $

where each homomorphism is defined in the obvious way, equating that which is represented by the variable in each square bracket. By inspection these homomorphisms are surjective.

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