Show that $\psi$ is surjective

Question

I have read a lot answers about this (for example here, here, and here), but I want to prove this directly consider the canonical homomorphism, defined by, \begin{align*} \psi : \mathbb{Z} &\hookrightarrow \mathbb{Z}[i]/ I \\ x &\mapsto \psi(x):= x+I \end{align*} where $I=\langle a+bi\rangle$ and $(a,b)=1$, but I need to show $\psi$ is surjective, 'cause I'm able to show that if $\psi$ is surjective then $\ker \psi=\langle a^2+b^2\rangle$, and for the first isomorphism theorem it follows that $\mathbb{Z}[i]/\langle a+bi\rangle\cong \mathbb{Z}_{a^2+b^2}$. In one of this post Matt E give and hint but I don't understand how finite works here. Any hint to show $\psi$ is surjective? Thank you guys. I think it's enough to show that there exists a $m\in \mathbb{Z}$ such that $\psi(m)=m+I=i+I$, 'cause then given $c+di+I\in\mathbb{Z}[i]/I$ then with $x:=c+dm\in\mathbb{Z}$ it's such that $\psi(x)=x+I=c+dm+I=c+di+I$. But how to show that $i\in \text{Im }\psi$?

Answer 1

To show that $m+I = i+I$ for some $m\in\mathbb Z$ it suffices to show that $m-i\in I$ or, taking $m$ for $-m$, that $m+i\in I$ for some $m\in\mathbb{Z}$. Choose $r, s\in\mathbb Z$ such that $ar+bs=1$; then $$(a+bi)s + (a+bi)ri = (as-br) + (ar+bs)i = (as-br)+i\in I.$$ Take $m = as-br$.