Number elements in $\Bbb{Z}[i]/\langle3+i\rangle$

Question

I need to find out the number of elements in the quotient ring $\Bbb{Z}[i]/\langle3+i\rangle$. For $a,b\in\Bbb{Z}$, I can write any elements in the quotient ring as $$a+bi+\langle3+i\rangle=(a-3b)+\langle3+i\rangle$$ As $10=(3+i)(3-i)$, I can say that order of any element in the quotient ring is divisible by $10$. But I can't proceed further. Please help.

Answer 1

Continuing your argument...

Since $a+bi+\langle3+i\rangle=(a-3b)+\langle3+i\rangle$, we can try $\phi: \mathbb Z[i] \to \mathbb Z / 10 \mathbb Z$ given by $a+bi \mapsto a-3b \bmod 10$.

It is easy to prove that $\phi$ is a surjective ring homomorphism. You just need to prove that $\ker \phi = \langle3+i\rangle$, which is immediate.

Answer 2

Hint: $\Bbb{Z}[i]$ is a Euclidean domain with respect to the norm $N(a+bi)=a^2+b^2$. Therefore, every element of $\Bbb{Z}[i]$ is equivalent mod $3+i$ to an element with norm less than $10$.