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. 

The quotient ring you are after must be $\Bbb Z[i]/I$ where $I=(2+i)$, otherwise it would not tell you much about the status is the ideal $I$. You must know that multiplication by a complex number is a combination of rotation and scaling, and so multiplication by $2+i$ is the unique such operation that sends $1\in\Bbb C$ to $2+i$. Therefore the image of the grid (lattice) $\Bbb Z[i]\subset\Bbb C$ is the grid $I\subset\Bbb Z[i]$ with orthogonal spanning vectors $2+i$ and $(2+i)i=-1+2i$. The square spanned by those two vectors has area $|\begin{smallmatrix}2&-1\\1&\phantom+2\end{smallmatrix}|=5$, so the density of the points of $I$ should be smaller than that of $\Bbb Z[i]$ by a factor $5$. This is a not entirely rigorous argument showing that the quotient ring $\Bbb Z[i]/I$ should have $5$ elements. There aren't very many rings with $5$ elements, so you should be able to guess which $5$ elements you can choose as representatives of the classes in $\Bbb Z[i]/I$. Now go ahead and show that all elements of $\Bbb Z[i]$ can be transformed into one and only one of those $5$ elements by adding an element of $I$.
By the way, the quotient being a finite ring it will be a field if and only if it is an integral domain.: it is either both of them or none of them. In the case at hand it will be both (all rings with a prime number of elements are integral domains and also fields).
