Is it true that the order of any quotient ring $\mathbb Z[i]/\langle a+ib \rangle $ is $a^2+b^2$ ? (where not both $a,b$ are zero)

Question

Is it true that the order of any quotient ring $\mathbb Z[i]/\langle a+ib \rangle $ is $a^2+b^2$ ? ( I know it is atmost finite ) Please help . Thanks in advance .

Answer 1

Elementary answer. The ring $\mathbb Z[i]$ is the lattice generated by the vectors $1$ and $i$. The ideal $\mathfrak a = (a+bi)$ is the subgroup generated by the vectors $a+bi$ and $ia -b$, or by the basis $A = \tiny \begin{pmatrix} a&-b\\b&a\end{pmatrix}$. Therefore, (for instance by the elementary divisor theorem), the order of the group $\mathbb Z[i]/\mathfrak a$ is $\det(A)$, which is conveniently $a^2+b^2$.

Answer 2

Hint: Leaving off the multiplicative structures, $\mathbb Z[i]$ becomes a free $\mathbb Z$ module with basis $\{1,i\}$, and the ideal $\langle a+ib \rangle$ becomes a submodule of it generated by $(a,b)$ and $(-b,a)$.