Show that $(2+i)$ is a prime ideal

Consider the set Gaussian integer $\mathbb{Z}[i]$. Show that $(2+i)$ is a prime ideal.

I try to come out with a quotient ring such that the set Gaussian integers over the ideal $(2+i)$ is either a field or integral domain. But I failed to see what is the quotient ring.

Use that $\mathbb Z[i]\simeq \mathbb Z[X]/(X^2+1)$. Via this isomorphism $i$ corresponds to $\hat X$ (the residue class of $X$), and therefore the ideal $(2+i)$ corresponds to $(2+\hat X)=(2+X,X^2+1)/(X^2+1)$. Now $$\frac{\mathbb Z[X]/(X^2+1)}{(2+X,X^2+1)/(X^2+1)}\simeq \frac{\mathbb Z[X]}{(2+X,X^2+1)}\simeq \frac{\mathbb Z[X]/(2+X)}{(2+X,X^2+1)/(2+X)}.$$ But $\mathbb Z[X]/(2+X)\simeq \mathbb Z$, by sending $X$ to $-2$, so $$\frac{\mathbb Z[X]/(2+X)}{(2+X,X^2+1)/(2+X)}\simeq \mathbb Z/((-2)^2+1)=\mathbb Z/(5).$$

Remark. This answer has the "advantage" that doesn't use the arithmetical properties of $\mathbb Z[i]$ and can be used in many other situations, like this one.

$\mathbb{Z}[i]$ is a unique factorization domain, so it is equivalent to show that $2+i$ is irreductible in $\mathbb{Z}[i]$. If $2+i=ab$ with $a,b \in \mathbb{Z}[i]$, then $5=N(2+i)=N(a)N(b)$ (where $N : z \mapsto z \overline{z}$). Therefore, $N(a)=1$ or $N(b)=1$. But $z \in \mathbb{Z}[i]$ is invertible iff $N(z)=1$.

If you really want to show that $\mathbb{Z}[i]/(2+i)$ is an integral domain, consider the ring homomorphism $\phi : \left\{ \begin{array}{ccc} \mathbb{Z}[i] & \to & \mathbb{Z}_5 \\ a+ib & \mapsto & \overline{a-2b} \end{array} \right.$; its kernel is $(2+i)$ so $\mathbb{Z}[i]/(2+i)$ is the field $\mathbb{F}_5$.

Remark: The choice of $\phi$ is motivated by the following property: in $\mathbb{Z}[i]/(2+i)$, $i=-2$ and $5=1+(-2)^2=1+i^2=0$.

$\mathbb{Z}[i]$ is an euclidian ring. So, it is principal. In a principal ring, the factorization is unique.

If $(2+i)=(a+ib)(c+id)$,

$\lvert 2+i \rvert^2=\lvert a+ib \rvert^2 \lvert c+id \rvert^2$ $\implies 4+1=(a^2+b^2)(c^2+d^2) \implies a^2+b^2=1$ or $c^2+d^2=1 \implies a+ib$ or $c+id$ invertible.

So, $2+ib$ is prime.

If $zz'\in (2+ib)$, $2+ib$ divides $z$ or $z'$ because the factorization is unique.