Show that $\alpha = 1 + \sqrt{5} \in \mathbb{R}$ is algebraic over $\mathbb{Q}$; find $\text{Irr}(\alpha:\mathbb{Q})$. 
Show that $\alpha = 1 + \sqrt{5} \in \mathbb{R}$ is algebraic over $\mathbb{Q}$; find $\text{Irr}(\alpha:\mathbb{Q})$.

Somehow, 'I can understand' the definition of algebraic elements, but I'm not able to answer that above question, someone can explain it in detail, please?
 A: Notice that $\sqrt 5\notin \mathbb Q$, so that also $\alpha\notin \mathbb Q$. To show that $\alpha$ is algebraic over $\mathbb Q$, we need to find a polynomial $f(X)\in\mathbb Q[X]$ with $f(\alpha)=0$.
To do this, we calculate $(\alpha-1)^2 = (\sqrt 5)^2 = 5$, i. e. 
$5 = (\alpha-1)^2 = \alpha^2-2\alpha+1$, or 
$$\alpha^2-2\alpha-4 = 0.$$ 
Therefore, $f(X):= X^2-2X-4$ is a polynomial with $f(\alpha)=0$. If this polynomial were reducible in $\mathbb Q[X]$, then there would be $a,b\in \mathbb Q$ such that $f(X) = (X-a)(X-b)$. But then 
$$
0 = f(\alpha) = (\alpha-a)(\alpha-b),
$$
so that $\alpha\in \{a,b\}\subseteq \mathbb Q$ (because $\mathbb Q$ is an integral domain), contradicting $\alpha\notin \mathbb Q$. Therefore, $f(X)$ must be irreducible. Since it is also monic, we conclude
$$
{\rm Irr}(\alpha:\mathbb Q) = f(X) = X^2-2X-4.
$$
A: $\alpha ^2=( 1 + \sqrt{5})^2=1+2  \sqrt{5}+5=6+2 \sqrt{5}$
So $\alpha ^2- 2\alpha=6+2 \sqrt{5}-2-2 \sqrt{5}=4$. 
Finally $\alpha$ is a root of the polynomial $X^2-2X-4\in \mathbb{Q}[X]$, so by definition, $\alpha$ is algebraic over $\mathbb{Q}$. And since $\alpha$ is not in $\mathbb{Q}$, $\text{Irr}(\alpha:\mathbb{Q})$ is of degree $> 1$, so $\text{Irr}(\alpha:\mathbb{Q})=X^2-2X-4$.
