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Let $\mathbb Z[i]$ denote the ring of Gaussian integers. For which of the following values of $n$ is the quotient ring $\mathbb Z[i]/n \mathbb Z[i]$ an integral domain?
$2,13,19,7$

How can I solve the problem?

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closed as off-topic by Jyrki Lahtonen, metamorphy, mrtaurho, Lee David Chung Lin, Hw Chu Jul 14 at 20:49

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    $\begingroup$ Why do you post two questions within 5 minutes with two different user names? $\endgroup$ – Phira Dec 5 '12 at 2:43
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    $\begingroup$ when n is a gaussian prime, so 19 and 17 (because they equal 3 mod 4). $\endgroup$ – user51427 Dec 5 '12 at 2:44
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$Z[i]/n Z[i]$ is an integral domain iff $n$ is a Gaussian prime.

No composite integer $n$ can be a Gaussian prime.

$2$ is not a Gaussian prime because $2=(1+i)(1-i)=(1+i)^2(-i)$.

Primes $p$ that are the sum of two squares are not Gaussian primes, because $p=a^2+b^2=(a+bi)(a-bi)$. So $p=2$ and $p\equiv 1\bmod 4$ are not Gaussian primes.

This leaves us with the primes that are not the sum of two squares, which are exactly those that are congruent to $3 \bmod 4$. These are indeed Gaussian primes.

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A small start: $(3+2i)(3-2i)$ is divisible by $13$.

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Hint $\rm\ \, R = \Bbb Z[{\it i}\,]\cong \Bbb Z[x]/(x^2\!+\!1)\:$ so $\rm\:R/2 \cong \Bbb Z_2[x]/(x^2\!+\!1) \cong \Bbb Z_2[x]/(x\!+\!1)^2\:$ is not a domain. Similarly, we easily compute that $\rm R/13 \cong \Bbb Z_{13}[x]/(x^2\!+\!1) \cong \Bbb Z_{13}[x]/(x\!-\!5)(x\!+\!5)\:$ is not a domain.

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