I was given to prove that $\frac{Z[i]}{<a+ib>}$ $\cong$ $\frac{Z}{<a^2+b^2>}$ , when gcd(a,b) = 1.
I am using the mapping f: Z $\to$ $\frac{Z[i]}{<a+ib>}$ given by f(x)=x+< a+ib >. I have proved that the homomorphism is onto. But I am not being able to show that ker f = < $a^2+b^2$ >. If x $\in$ ker f , then a+ib |x . hence a-ib | x. thus $ a^2+b^2$ | $x^2$. From here how can I conclude that $a^2+b^2$ |x ? Please give some hint. Thank you.
