Proof that Algebraic Integers Form a Subring

[Bashyboy]

Homework Statement

The set $\Bbb{A}$ of all the algebraic integers is a subring of $\Bbb{C}$

Homework Equations

The Attempt at a Solution

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?

 

[fresh_42]

The degrees in $\mathbb{Z}[x]$ maybe be arbitrary - I don't like unbounded here, as they are definitely finite for every element - but the algebraic expressions can be reduced. Let me first order your notation, as the double use of $g(x)$ is dirty. So let's stick with the first definition as minimal polynomial of degree $m$ with $g(\alpha)=0$. Then for any element $p(x) \in \mathbb{Z}[x]$ we have $p(\alpha) \equiv r(\alpha) \operatorname{modulo} g(\alpha)$ with $\deg(r) < m$ by use of the Euclidean algorithm. So $\mathbb{Z}[x]/\langle g(x) \rangle \cong \mathbb{Z}[\alpha]$ has to be compared with $G$, not the entire polynomial ring.

The minus one in $n+m-1$ comes in with the division, same as in the above with $\deg(r) \leq m-1$.