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. 

$\mathbb{Z}[i]$ is a Abelian group of rank $2$: that is, $\{ 1, i \}$ is a basis and every element can be written uniquely as $a + bi$ with $a,b \in \mathbb{Z}$.
$\langle 2 + i \rangle$ is also a Abelian group of rank $2$. Can we find a basis for it?
Well, we know every element in it can be written uniquely $(2+i)x$ for $x \in \mathbb{Z}[i]$, which in turn means it can be written uniquely as $(2+i)(a+bi)$.
So a basis is $\{ 2+i, -1+2i \} $.
Let's stuff it in a matrix so that its (integer) rowspace is $\langle 2+i \rangle$, and then do (integer) row reduction to simplify (in general, think "Euclidean algorithm" to reduce a column):
$$ \left( \begin{matrix}2 & 1 \\ -1 & 2 \end{matrix} \right)
\to \left( \begin{matrix}2 & 1 \\ -5 & 0 \end{matrix} \right)
\to \left( \begin{matrix}2 & 1 \\ 5 & 0 \end{matrix} \right)
$$
(note that multiplying a row by $-1$ is fine, because that's invertible! Dividing a row by $5$ would be bad, though, because it's not an integer row operation. Multiplying a row by $2$ would also be bad, because that's not invertible)
I did row reduction from right to  left, because I wanted the bottom row to have its entries on the left, not on the right.
So a simplified basis for our ideal is $\{ 5, i + 2 \}$. Suddenly, the identity of the ring $\mathbb{Z}[i] / \langle 2+i\rangle = \mathbb{Z}[i] / \langle 5, i+2\rangle$ becomes clear!
