An algebraic integer is a complex number that is the root of a monic polynomial with integer coefficients. In Exercise 3 of Chapter 6 of Ireland and Rosen's book on number theory, we are asked to show that if $\alpha,\beta$ are algebraic integers, then any solution $x_0$ to the polynomial equation $x^2 +\alpha x + \beta$ is also an algebraic integers. This question was also asked here, though I'm struggling to fill out the details of user14972's answer, and I'd like some help filling out the details (or a pointer to a better path to take).
Outline of solution:
Expanding on user14972's answer, let $$f_\alpha(x) = x^m + a_{m-1}x^{m-1} + \ldots + a_0$$ be a monic polynomial in $\mathbb{Z}[x]$ for which $\alpha$ is a root, and let $$f_\beta(x) = x^n + b_{n-1}x^{n-1} + \ldots + b_0$$ be a monic polynomial in $\mathbb{Z}[x]$ for which $\beta$ is a root. By the fundamental theorem of algebra, we may write $$f_\alpha(x) = \prod_{i=1}^m(x-\alpha_i) ~~~ \text{and} ~~~ f_\beta(x) = \prod_{j=1}^n ( x-\beta_j)$$ for some $\alpha_1,\ldots,\alpha_m$ and $\beta_1,\ldots,\beta_m$ in $\mathbb{C}$. One of the $\alpha_i$ is equal to $\alpha$, and likewise one of the $\beta_j$ is equal to $\beta$. $g(x)$ is plainly monic. user14972 claims that $g(x)$ has integer coefficients - it is this step that I am struggling to prove.
Attempt to prove $g(x)$ has integer coefficients:
We now show that $g(x)$ has integer coefficients. A brief calculation tells us that $$g(x) = (-1)^n \prod_{i=1}^m f_\beta( - x^2 - \alpha_i x).$$ With $b_n =1$ we have $$g(x) = (-1)^n \prod_{i=1}^m \sum_{ j_i = 0}^n b_{j} ( -x^2 - \alpha_i x)^{j} = (-1)^n \prod_{i=1}^m \sum_{0 \leq j_1,\ldots,j_m \leq n} (-x)^{j_1+\ldots+j_m} \prod_{i=1}^m b_{j_i} ( x+ \alpha_i)^{j_i}.$$ Let $\lambda = (\lambda_1,\ldots,\lambda_m)$ where $\lambda_1 \geq \ldots \geq \lambda_m \geq 0$ be decreasing integers. Write $|\lambda|$ for their sum. Write \begin{align*} m_\lambda(x_1,\ldots,x_m) := \sum'_{ j_1,\ldots,j_m } x^{j_1} \ldots x^{j_m} \end{align*} where $\sum'_{ j_1,\ldots,j_m }$ denotes the sum is over all distinct orderings $(j_1,\ldots,j_m)$ of $(\lambda_1,\ldots,\lambda_m)$. Then picking up from our last equation for $g(x)$, we have $$g(x) = (-1)^n \sum_{\lambda} (-x)^{|\lambda|} (\prod_{i=1}^m b_{\lambda_i} )m_\lambda(x+\alpha_1,\ldots,x+\alpha_m).$$
Now my suspicion is that $m_\lambda(x+\alpha_1,\ldots,x+\alpha_m)$ always has integer coefficients as a polynomial in $x$, but I'm not sure how to prove this. We know it is the case for $\lambda = (1,\ldots,1)$.
Any help would be greatly appreciated!