7
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Which book is that? $\endgroup$ – José Carlos Santos Oct 5 '17 at 14:34
  • $\begingroup$ @JoséCarlosSantos It's Advanced Modern Algebra by Joseph Rotman. $\endgroup$ – user193319 Oct 5 '17 at 14:35
3
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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 $\alpha$. $\endgroup$ – user193319 Oct 5 '17 at 14:48
  • 1
    $\begingroup$ 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 $G$. $\endgroup$ – Andreas Blass Oct 5 '17 at 15:04
  • $\begingroup$ 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. $\endgroup$ – user193319 Oct 5 '17 at 15:52
  • $\begingroup$ 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. $\endgroup$ – Andreas Blass Oct 5 '17 at 15:56
2
$\begingroup$

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 $n+m-1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.