- #1

Bashyboy

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## 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?