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?
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?
$Z[i]/n Z[i]$ 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.
A small start: $(3+2i)(3-2i)$ is divisible by $13$.
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.