To figure out whether $r = \frac{1+\sqrt{3}}{2}$ is an algebraic integer or not. I have tried to find a monic polynomial for $r$ in $\mathbb{Z}[X]$ without success; I can see it is an algebraic number since $2r^2-2r-1=0$, where $r=\frac{1+\sqrt{3}}{2}$.
Is there a method to find monic polynomials, or show that it is not an algebraic integer? I am quite lost.
 A: This is a proposition from Neukirch ANT page $8$:

Let $A$ be integrally closed in its field of fractions. An element $\beta\in L$ is integral over $A$ if and only if its minimal polynomial $\min_{\beta,A}(x)\in A[x]$.

In this case, $L=\mathbb{Q}$, and $A=\mathbb{Z}$. $\mathbb{Z}$ being a UFD means that it is integrally closed, with minimal polynomial $x^2-x-1/2$ you found. This polynomial is not in $\mathbb{Z}[x]$, so that it is not an algebraic integer by the proposition.
The proof of the proposition goes as follows:
Let $\beta$ be integral over $A$, with monic polynomial $g(x)$. Then by definition, $\min_{\beta,A}(x)$ divides $g(x)$ in $K[x]$. Therefore every zero of $\min_{\beta,A}(x)$ is a zero of $g(x)$, and hence every zero $\beta_1,\ldots,\beta_n$ is integral over $A$. Since $\min_{\beta,A}(x)=(x-\beta_1)\cdots(x-\beta_n)$, the coefficients are also integral over $A$ (being sums and products of integral elements). Since $A$ is integrally closed, $\min_{\beta,A}(x)\in A[x]$.
The converse follows trivially.
Other facts used were:

UFDs are integrally closed (Page 8).
$\mathbb{Z}$ is a UFD (FTA).

