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If $\alpha$ and $\beta$ are algebraic integers, then show $\alpha + \beta$ and $\alpha \times \beta$ are both algebraic integers.

I know that an algebraic integer is a root of some monic polynomial with coefficients in $\mathbb{Z}$, but I am not sure where to start with this question.

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    $\begingroup$ There's a way with the resultant of polynomies. The classic proof of it, uses instead the fact that an estension of rings with two algebraic integers is finally generated, and then uses one of the equivalent definitions of algebraic integers $\endgroup$ – Exodd Sep 27 '14 at 17:59
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    $\begingroup$ Try typing "algebraic integers form a ring" into your favorite search engine. $\endgroup$ – Dane Sep 29 '14 at 16:26
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This answer is based on Theorems 9.11 and 9.12 in I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991.

We first prove the following lemma: If $n$ is a positive rational integer, $\xi$ is a complex number, and the complex numbers $\theta_1, \theta_2, \dots, \theta_n$, not all zero, satisfy the equations $$\xi \theta_j = a_{j,1} \theta_1 + a_{j,2} \theta_2 + \cdots + a_{j,n} \theta_n, \qquad j = 1, 2, \ldots, n$$ with the $n^2$ coefficients $a_{j,i}$ in $\Bbb Z$, then $\xi$ is an algebraic integer.

Proof: The above equations can be thought of as a system of homogeneous linear equations in $\theta_1, \theta_2, \dots, \theta_n$. Because the $\theta_i$ are not all zero, there is a non-trivial solution, so the determinant of the coefficients must vanish, i.e., $$\begin{vmatrix} \xi - a_{1,1} & -a_{1,2} & \cdots & -a_{1,n}\\ - a_{2,1} &\xi - a_{2,2} & \cdots & -a_{2,n}\\ \vdots & \vdots & \ddots & \vdots \\ - a_{n,1} & -a_{n,2} & \cdots &\xi - a_{n,n} \end{vmatrix} = 0.$$

Expansion of that determinant gives an equation $\xi^n + b_1 \xi^{n-1} + \cdots + b_n = 0$ where the $b_j$ are polynomials in the $a_{j,k}$ and therefore in $\Bbb Z$. Thus, $\xi$ is an algebraic integer, which proves the lemma. $\square$

We now prove the main result: Assume $\alpha$ and $\beta$ satisfy \begin{align} \alpha^m + a_1 \alpha^{m - 1} + \dots + a_m & = 0\\ \beta^r + b_1 \beta^{r - 1} + \dots + b_r & = 0 \end{align} with coefficients $a_i$ and $b_i$ in $\Bbb Z$. Let $n = mr$, and define the complex numbers $\theta_1, \theta_2, \dots, \theta_n$ as the numbers \begin{matrix} 1, & \alpha, & \alpha^2, & \cdots & \alpha^{m-1}, \\ \beta, & \alpha \beta, & \alpha^2 \beta, & \cdots & \alpha^{m-1}\beta,\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \beta^{r-1}, & \alpha \beta^{r-1}, & \alpha^2 \beta^{r-1}, & \cdots & \alpha^{m-1} \beta^{r-1} \end{matrix} in any order. Hence $\theta_1, \theta_2, \dots, \theta_n$ are the numbers $\alpha^s \beta^t$ with $s = 0, 1, \dots, m - 1$ and $t = 0, 1, \dots, r - 1$, so for any $\theta_j$, $$\alpha \theta_j = \alpha^{s + 1} \beta^t = \begin{cases} \text{some } \theta_k & \text{if } s + 1 \le m - 1 \\ (- a_1 \alpha^{m - 1} - a_2 \alpha^{m - 2} - \dots - a_m) \beta^t & \text{if } s + 1 = m. \end{cases}$$

In either case, we see that there are constants $h_{j,1}, \dots, h_{j,n}$ in $\Bbb Z$ such that $\alpha \theta_j = h_{j,1} \theta_1 + \cdots + h_{j,n} \theta_n$. Similarly, there are constants $k_{j,1}, \dots, k_{j,n}$ in $\Bbb Z$ such that $\beta \theta_j = k_{j,1} \theta_1 + \cdots + k_{j,n} \theta_n$, and hence, $(\alpha + \beta) \theta_j = (h_{j,1} + k_{j,1}) \theta_1 + \cdots + (h_{j,n} + k_{j,n}) \theta_n$. These equations are of the form given in the lemma. Thus $\alpha + \beta$ is an algebraic integer.

We also have $\alpha \beta \theta_j = \alpha (k_{j,1} \theta_1 + \cdots + k_{j,n} \theta_n) = k_{j,1} \alpha \theta_1 + \cdots + k_{j,n} \alpha \theta_n$ from which we find $\alpha \beta \theta_j = c_{j,1} \theta_1 + \cdots + c_{j,n} \theta_n$ where $c_{i,j} = k_{j,1} h_{1,i} + k_{j,2} h_{2,i} + \cdots + k_{j,n} h_{n,i}.$ Again, we apply the lemma to conclude that $\alpha \beta$ is an algebraic integer.

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