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To check whether $\langle 3+i\rangle$ is a maximal ideal or not in the ring of Gaussian integers $\mathbb{Z}[i]$.

Attempt: $\mathbb{Z}[i]/\langle 3+i\rangle$ is isomorphic to $10 \mathbb{Z}$ which is not a prime ideal $\implies 10 \mathbb{Z}$ is not a maximal ideal. Hence $\langle 3+i\rangle$ is not a maximal ideal in the ring of Gaussian integers $\mathbb{Z}[i]$. Is this correct?

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$\mathbb{Z}[i]/(3+i) = \mathbb{Z}[x]/(x^2+1,3+x)=\mathbb{Z}/(3^2+1)=\mathbb{Z}/10$ is no integral domain, hence $(3+i)$ is not prime.

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    $\begingroup$ I got three downvotes in a short period of time, and this post here got one of them. Why? $\endgroup$ – Martin Brandenburg Jun 19 '15 at 22:43
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Since the norm of $3+i$ is neither prime nor the square of a prime, we know that $3+i$ is not irreducible.

Indeed $3+i=(1+i)(2-i)$

In particular, $(1+i)+\langle 3+i\rangle$ is a zero divisor in the quotient ring.

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