Understanding the Proof that Algebraic Integers are a Subring of $\Bbb{C}$ 
The set $\Bbb{A}$ of all the algebraic integers is a subring of $\Bbb{C}$

Here is an excerpt from my book: 

Suppose $\alpha$ an $\beta$ are algebraic integers; let $\alpha$ be the root of a monic $f(x) \in \Bbb{Z}[x]$ of degree $n$, and let $\beta$ be a root of a monic $g(x) \in \Bbb{Z}[x]$ of degree $m$. Now $\Bbb{Z}[\alpha \beta]$ is an additive subgroup of $G= \langle \alpha^i \beta^j ~|~ 0 \le i < n$, ~ $0 \le j < m \rangle$. Since $G$ a finitely generated, so is its subgroup $\Bbb{Z}[\alpha \beta]$, and so $\alpha \beta$ is an algebraic integer. Similarly, $\Bbb{Z}[\alpha + \beta]$ is an additive subgroup of $\langle \alpha^i \beta^j ~|~ i+j \le n+m-1 \rangle$, and so $\alpha + \beta$ is also algebraic. 

I am having trouble seeing the two set inclusions, particularly because $\Bbb{Z}[\alpha] := \{g(\alpha) ~|~ g(x) \in \Bbb{Z}[x] \}$ and the degree of the polynomials in $\Bbb{Z}[x]$ is unbounded, while $G$ and the other set are built from (multivariable) polynomials of finite degree. Perhaps someone could make this more explicit. Also, what's the motivation for choosing $n+m-1$ as the upper bound for $i+j$, other than the fact that it works? 
 A: Because $f(\alpha)=0$, any polynomial $g(\alpha)$ reduces to a polynomial $r(\alpha)$ where $r$ has degree $<n$. Specifically, take $r$ to be the remainder when $g$ is divided by $f$ (in the ring of polynomials).  So $g(x)=q(x)f(x)+r(x)$ and therefore $g(\alpha)=r(\alpha)$.
A: I personally like to also think of the ring $\mathbb{Z}\left[\alpha\right]$ as the smallest subring of the integers that contains $\alpha$. In this way, it's not so strange that $\mathbb{Z}\left[\alpha+\beta\right], \mathbb{Z} \left[\alpha \beta \right] \subseteq \mathbb{Z}\left[\alpha, \beta \right]$, since any polynomial in the sum or product of $\alpha$ and $\beta$ may definitely be written as a polynomial in the "mixed" monomials $\alpha^i \beta^j$. As for your second question, we could allow such monomials where $i+j\geqslant m+n$, but in this case at least one of $i$ and $j$ would be larger than $n$ or $m$, respectively. In the case where, say, $i \geqslant n$, we can then write $\alpha^i$ as a linear combination of lower order powers of $\alpha$, because $\alpha$ satisfies a polynomial equation of degree $n$. Thus we really don't need indices $(i,j)$ with their sum larger than $n+m-1$.
