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.
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You can say that additive order of any element in the quotient ring divides $10$. – lhf Apr 18 '18 at 11:44
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1The title has a typo. – Dietrich Burde Apr 18 '18 at 11:48
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Possible duplicate of https://math.stackexchange.com/questions/23358/quotient-ring-of-gaussian-integers – lhf Apr 18 '18 at 13:10
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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 $10$.
lhf
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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.
lhf
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