Describe the structure of factor ring $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]/(2+\sqrt{5})$. I'm really confused with this question...
I know, that

$ \mathbb{Z}[\frac{1+\sqrt{5}}{2}]=\left \{ \frac{a + b\sqrt{5}}{2} \enspace  | \enspace a,b \in \mathbb{Z},  \enspace a\equiv b\pmod 2 \right \} $
$ \mathbb{Z}[\frac{1+\sqrt{5}}{2}]=A[\sqrt{5}] $

But I don't understand how to describe the ring.
In general, how should I solve problems like this?
 A: Note that
$$
(2+\sqrt5)(2-\sqrt5)=4-5=-1.
$$
It follows that $2+\sqrt5$ is a unit in the ring $\mathbf Z[\frac{1+\sqrt5}2]$. The quotient ring in question is the zero ring.
But that does not say how you solve such a problem in general.
In general you can do the following. First note that
$$
\mathbf Z[\tfrac{1+\sqrt5}{2}]/(2+\sqrt5)\cong\mathbf Z[x]/(x^2-x-1)
$$
by sending $x$ to $(1+\sqrt 5)/2$, and using the fact that the polynomial $x^2-x-1$ vanishes at $(1+\sqrt5)/2$. It follows that
$$
\mathbf Z[\tfrac{1+\sqrt5}{2}]/(2+\sqrt5)\cong(\mathbf Z[x]/(x^2-x-1))/(2x+1)
$$
since under the preceding isomorphism it is $2x+1$ that is sent to $(1+\sqrt5)/2$.
Then use that a double quotient can be reduced to a single quotient:
$$
(\mathbf Z[x]/(x^2-x-1))/(2x+1)\cong\mathbf Z[x]/(x^2-x-1,2x+1).
$$
Then make a double quotient the other way around
$$
\mathbf Z[x]/(x^2-x-1,2x+1)\cong
(\mathbf Z[x]/(2x+1))/(x^2-x-1).
$$
Note that
$$
\mathbf Z[x]/(2x+1)\cong\mathbf Z[\tfrac12],
$$
the ring you obtain from $\mathbf Z$ by adjoining $\tfrac12$. In this isomorphism, $x$ is sent to $-\tfrac12$. Hence, the ideal $(x^2-x-1)$ is sent to
the ideal $(-\tfrac14)$ in $\mathbf Z[\tfrac12]$. Now, $-\tfrac14$ is invertible in the latter ring. Therefore
$$
(\mathbf Z[x]/(2x+1))/(x^2-x-1)\cong\mathbf Z[\tfrac12]/(\tfrac14)=\{0\}.
$$
A: You can do linear algebra.
Viewing the ring as a $\mathbb{Z}$-module (a.k.a. abelian group), it is not hard to see that it is a free module with basis $\{ 1, \frac{1+\sqrt{5}}{2} \}$.
Similarly, by multiplying by the basis elements not hard to see that the ideal $(2 + \sqrt{5})$ is a free $\mathbb{Z}$-module with basis $\{ 2 + \sqrt{5},\frac{7 + 3 \sqrt{5}}{2} \}$.
If we use coordinates relative to the chosen basis for the ring, the ideal is the rowspace of the matrix
$$ \left( \begin{matrix}
1 & 2 \\ 2 & 3
\end{matrix} \right)$$
e.g.  we have $2 + \sqrt{5} = 1 \cdot 1 + 2 \cdot\frac{1 + \sqrt{5}}{2}$.
By performing (integer) linear operations, we can row reduce the basis for the ideal to get
$$ \left( \begin{matrix}
1 & 0 \\ 0 & 1
\end{matrix} \right)$$
That is, the ideal $(2 + \sqrt{5})$ is in fact equal to the entire ring $\mathbb{Z}[\frac{1 + \sqrt{5}}{2}]$.
