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.
$\rm \Bbb Z\stackrel{h}{\to}\, \Bbb Z[{\it i}\,]/(2\!+\!{\it i})\:$ is $\rm\,\color{#0b0}{\bf onto,\:}$ by $\rm\:mod\,\ 2\!+\!{\it i}:\ {\it i}\,\equiv -2\phantom{\dfrac{|}{|}}\!\Rightarrow\:a\!+\!b{\it i}\,\equiv a\!+\!b(-2)\in \Bbb Z\ $
$\rm n\in ker\ h\iff 2\!+\!{\it i}\,\mid n\iff\phantom{\dfrac{|}{|_|}}\!\!\!\!\!\!\! \dfrac{n}{2\!+\!{\it i}}\, =\, \dfrac{(2\!-\!{\it i})n}5\,\in\, \Bbb Z[{\it i}\,] \iff \color{#c00}5\mid n\ $
So $\rm\,\ \Bbb Z[{\it i}\,]/({\it i}+\!2)\, \color{#0b0}{\bf =\ Im\:h}\,\cong\, \Bbb Z/ker\:h \,=\, \Bbb Z/\color{#c00}5\ $ is a domain, so $\rm\: ({\it i}+\!2)\:$ is prime. $\ \ $ QED
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).
$\mathbb{Z}[i]$ 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!
