To prove isomorphism and thereby find Quotient Ring of Gaussian Integers

Question

I am required to prove the following:

$\Bbb Z[i]/\langle a+bi\rangle\,\cong\Bbb Z/(a^2+b^2)\Bbb Z=:\Bbb Z_{a^2+b^2}$, where $\gcd(a,b)=1$.

While I have looked into the various solutions given here Quotient rings of Gaussian integers I wish to find a function $\phi:\Bbb Z[i] \to\Bbb Z_{a^2+b^2}$ such that $\phi$ is an onto ring homomorphism. According to the second answer in the aforementioned link I take such $\phi (x + yi) = x-(ab)^{-1}y$ but I am unable to prove the multiplicativity of the ring homomorphism $\phi$ and its surjectivity.

PS. I want to prove the result using first isomorphism theorem.

Answer 1

The required homomorphism $\varphi:\Bbb Z[i]\to\Bbb Z/\langle a^2+b^2\rangle$ is defined by: $$\varphi(x+iy)=x-avy+\langle a^2+b^2\rangle$$ where $v\in\Bbb Z$ is an inverse of $b$ modulo $a^2+b^2$ (note that $b$ is invertible because $\gcd\{a,b\}=1$).

To prove this, recall that $\Bbb Z[i]\cong\Bbb Z[x]/\langle x^2+1\rangle$ where $x$ is an indeterminate. By universal property of polynomial ring $\Bbb Z[x]$, there exists one and only one ring homomorphism \begin{align}\tag1 \Bbb Z[x]&\to\Bbb Z/\langle a^2+b^2\rangle& x&\mapsto -av+\langle a^2+b^2\rangle \end{align} Since $(-av)^2+1=(a^2+b^2)v^2$, we have $(-av)^2\equiv-1\pmod{a^2+b^2}$, hence (1) gives rise to one and only one ring homomorphism $\varphi:\Bbb Z[i]\to\Bbb Z/\langle a^2+b^2\rangle$ such that $\varphi(i)=-av+\langle a^2+b^2\rangle$.

Clearly, for every $n\in\Bbb Z$ we have $\varphi(n)=n\varphi(1)=n+\langle a^2+b^2\rangle$, hence $\varphi$ is surjective. On the other hand, if $x+iy\in\ker\varphi$, then $x\equiv avy\pmod{a^2+b^2}$, hence $$x+iy=(x-avy)+(a+ib)vy$$ is a mutliple of $a+ib$ in $\Bbb Z[i]$.