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?

• Which book is that? Oct 5, 2017 at 14:34
• @JoséCarlosSantos It's Advanced Modern Algebra by Joseph Rotman. Oct 5, 2017 at 14:35

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 $g(\alpha)=r(\alpha)$.

• I see that $f(\alpha) = 0$ and therefore $g(\alpha) = r(\alpha)$, but I need to show that $h(\alpha \beta)$ is in $G$, where $h(x)$ is any polynomial in $\Bbb{Z}[x]$. So I am evaluating polynomials at $\alpha \beta$, not $\alpha$. Oct 5, 2017 at 14:48
• Polynomial functions of $\alpha\beta$ involve, a priori, arbitrary powers of $\alpha$ and of $\beta$, but my answer explains that, instead of arbitrary powers of $\alpha$, you only need exponents $<n$. Similarly, you only need powers of $\beta$ with exponents $<m$. So, altogether, you only need the finitely many generators that were listed for $G$. Oct 5, 2017 at 15:04
• I understand this, too. But the only reason you got this reduction is because you know $f(\alpha) =0$. When we consider an arbitrary $h(x)$, how do we know there is a polynomial that is zero at $\alpha \beta$ and causes $h(x)$ to reduce to some polynomial of appropriate degree? For example, I initially tried $h(x) = f(x)g(x) q(x) + r(x)$ for some $q(x)$ and $r(x)$, but then I quickly realized $f(\alpha \beta) g(\alpha \beta) = 0$ is not necessarily true. Oct 5, 2017 at 15:52
• The information you say you understand allows you to replace arbitrary powers of $\alpha$ (resp. $\beta$) by powers with exponent smaller than $n$ (resp. $m$). So whenever you encounter any powers of $\alpha\beta$, treat it as a power of $\alpha$ times a power of $\beta$ ($(\alpha\beta)^j=\alpha^j\beta^j$) and reduce those. Oct 5, 2017 at 15:56

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 $n+m-1$.