TITLE: If $\alpha$ and $\beta$ are algebraic integers then any solution to $x^2+\alpha x + \beta = 0$ is also an algebraic integer.
QUESTION [5 upvotes]: An algebraic integer is a complex number that is a root of a monic polynomial with coefficients in $\mathbb{Z}$. 

Let $\alpha$ and $\beta$ be algebraic integers. Then any solution to $x^2+\alpha x + \beta = 0$ is also an algebraic integer.

What we have so far is that $2x+\alpha$ is an algebraic integer but since the set of algebraic integers is a ring I can't divide by 2 so I can't have an algebraic integer and also the polynomial wouldn't be monic. 
I appreciate the help.

REPLY [4 votes]: A standard trick is to take the norm. If you compute the polynomial
$$ g(x) = \prod_i \prod_j (x^2 + \alpha_i x + \beta_j)$$
where the $\alpha_i$ are the conjugates of $\alpha$ and the $\beta_j$ are the conjugates of $\beta$, then you'll find $g(x)$ has integer coefficients.
Another way to calculate the same thing is with resultants: if $m_\gamma$ is the minimal polynomial of $\gamma$, then
$$ g(x) = \mathop{\mathrm{Res}}_z(\mathop{\mathrm{Res}}_y(x^2 + yx + z, m_\alpha(y)), m_\beta(z)) $$