I want to show that $\mathbb{Z}[\alpha.\beta]$ and $\mathbb{Z}[\alpha\pm\beta]$ are finitely generated when $\mathbb{Z}[\alpha]$, $\mathbb{Z}[\beta]$ are f.g.

My primary aim is to show the set of algebraic integers of a number field $K$ is a ring.

We have that $\alpha,\beta$ are algebraic integers, so $\mathbb{Z}[\alpha]$ and $\mathbb{Z}[\beta]$ are finitely generated, so there exist minimal polynomials $f_1$,$f_2$ of $\alpha$ and $\beta$ so that their degrees are $m$ and $n$, respectively-and both polynomials belong to $\mathbb{Z}[X]$-. Thus, $\{1,\alpha,\alpha^2,\dots,\alpha^{m-1}\}$ and $\{1,\beta,\dots,\beta^{n-1}\}$ are the generating set of $\mathbb{Z}[\alpha]$ and $\mathbb{Z}[\beta]$ .

If we say that $f_1(X)=X^m+a_{m-1}X^{m-1}+\dots+a_1X+a_0$ and $f_2(X)=X^n+b_{n-1}X^{n-1}+\dots+b_1X+b_0$ we conclude that $\alpha^m = -a_0-a_1\alpha-\dots-a_{m-1}\alpha^{m-1}$ and $\beta^n = -b_0-b_1\beta-\dots-b_{n-1}\beta^{n-1}$.

Now I want you to correct me if there is anything wrong:

The elements of $\mathbb{Z}[\alpha.\beta]$ consists of polynomials which have terms like $c(\alpha\beta)^j$ for some $j \in \mathbb{N}$. WLOG, if $n>m$ and $j>n$, we must write $c(\alpha\beta)^j$ in terms of $\{1,\alpha,\alpha^2,\dots,\alpha^{m-1}\}$ and $\{1,\beta,\dots,\beta^{n-1}\}$ . Therefore, we get bunch of terms $c_{p,q}\alpha^p\beta^q$ where $c_{p,q}$ is an integers and $0\leq p \leq m-1$, $0\leq q \leq n-1$. Finally, my claim is that $B=\{\alpha^i \beta^j : 0\leq i \leq m-1, 0\leq j \leq n-1 \}$ is a basis for $\mathbb{Z}[\alpha\beta] $, consisting of $m.n$ elements.

I think that this set $B$ is also a basis for $\mathbb{Z}[\alpha\pm\beta]$ since we have every possible $\alpha^i, \beta^j $ in $B$.


$B$ cannot be a basis for $\mathbb{Z}[\alpha \beta]$, because the elements of $B$ are not all contained in $\mathbb{Z}[\alpha \beta]$! For instance $\alpha \in B$ is generally not in $\mathbb{Z}[\alpha \beta]$.

We can also do this more generally:

Lemma: If $R$ is a commutative ring and $S$ is a $R$-algebra that is finitely generated as a $R$-module and $M$ is a finitely generated $S$-module, then if we consider $M$ as a $R$-module via restriction of scalars, $M$ is finitely generated as a $R$-module.

Proof: Let $s_1, \dots,s_n$ be a generating set of $S$ as a $R$-module and $m_1, \dots m_k$ be a generating set of $M$ as a $S$-module. We claim that $s_1m_1, \dots ,s_nm_1, s_1m_2, \dots ,s_nm_k$ is a generating set of $M$ as a $R$-module. If $m \in M$, we write $m = \lambda_1m_1 +\dots+\lambda_km_k$ with $\lambda_i \in S$. For each $\lambda_i$ write $\lambda_i = \mu_{i1}s_1+\dots+\mu_{in}s_n$, then if we plug this into $m = \lambda_1m_1 +\dots+\lambda_km_k$, we have expressed $m$ as a linear combination of $s_1m_1, \dots ,s_nm_k$.

Now as an application, we can prove that the ring of integers of any algebraic extension $K$ of $\mathbb Q$ is indeed a ring. (Assuming that you know that $\alpha$ is integral over $R$ iff $R[\alpha]$ is a f.g. $R$-module)

If $\alpha, \beta$ are integral over $\mathbb{Z}$, $\mathbb{Z}[\alpha]$ is a finitely generated $\mathbb Z$-module. As $\beta$ is integral over $\mathbb Z$, it is also integral over $\mathbb Z[\alpha]$, so $\mathbb Z[\alpha][\beta]$ is a finitely generated $\mathbb Z [\alpha]$-module. By the Lemma, $\mathbb Z[\alpha][\beta]$ is a finitely generated $\mathbb Z$-module. Now $\mathbb{Z}[\alpha \pm \beta]$ and $\mathbb{Z}[\alpha\beta]$ are submodules of $\mathbb Z[\alpha][\beta]$ and as $\mathbb{Z}$ is Noetherian, submodules of finitely generated modules are again finitely generated, thus $\alpha \pm \beta$ and $\alpha \beta$ are integral over $\mathbb{Z}$

  • $\begingroup$ I want to ask a simple question, if we have the minimal polynomials like I gave in the question, is there a way to write the basis explicity? $\endgroup$ – Ninja Apr 4 '17 at 19:53
  • $\begingroup$ @Ninja Do you mean of $\mathbb{Z}[\alpha][\beta]$ or of $\mathbb{Z}[\alpha\beta]$ or $\mathbb{Z}[\alpha \pm \beta]$ ? I think for $\mathbb{Z}[\alpha][\beta]$ the basis $B$ from your post should work, if we suppose that the extensions $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$ are linearly disjoint. For $\mathbb{Z}[\alpha\beta]$ or $\mathbb{Z}[\alpha \pm \beta]$ I have no idea. $\endgroup$ – Lukas Heger Apr 4 '17 at 20:18
  • $\begingroup$ I mean for both $\mathbb{Z}[\alpha\beta]$ and $\mathbb{Z}[\alpha\pm\beta]$. I am thinking when is $\alpha \mathbb{Z}[\alpha\beta]$ but could not find out yet. $\endgroup$ – Ninja Apr 4 '17 at 21:32

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