(This question is a dupplicate from If $\alpha$ and $\beta$ are algebraic integers then any solution to $x^2+\alpha x + \beta = 0$ is also an algebraic integer.)

I'm trying to solve this problem with my current knowledge of algebraic integers:

  • $x$ is an algebraic integer if it is a root of a monic polynomial in $\mathbb Z[x]$
  • $x$ is an algebraic integer iff $xW\subset W$ for some finitely generated free $\mathbb Z$-module $W$ (i.e. $W$ has a finite basis).

My first thought was to try Hurkyl's answer and compute $$ \prod_i\prod_j x^2+\alpha_ix+\beta_j $$ where $\alpha_i$ and $\beta_j$ are all the (complex) roots of the minimal polynomials of $\alpha$ and $\beta$ but I couldn't show that it has integer coefficients. (Although I found that $\prod_j x^2+\alpha x+\beta_j=p_\beta(-x^2-\alpha x)$ where $p_\beta$ is the minimal polynomial of $\beta$.)

Today my teacher gave me as hint to take a smartly chosen $\mathbb Z$-module and use the second definition given above. However I don't see which one...

Could someone give me a hint?

  • $\begingroup$ What about $\;\Bbb Z[\alpha,\beta]\;$ ...? $\endgroup$ – DonAntonio Apr 14 '14 at 22:06
  • $\begingroup$ @DonAntonio then $x^2\in\mathbb Z[\alpha,\beta]$ but does $x$ also lie in this module? $\endgroup$ – lvaneesbeeck Apr 15 '14 at 8:19
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    $\begingroup$ Have you tried $W=\Bbb{Z}[x,\alpha,\beta]=\Bbb{Z}[\alpha,\beta]+x\Bbb{Z}[\alpha,\beta]$? $\endgroup$ – Jyrki Lahtonen Apr 16 '14 at 12:06
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    $\begingroup$ You're right in that a module doesn't have to be closed under multiplication, but ${\mathbb Z}[x,\alpha,\beta]$ is the smallest algebra containing $x, \alpha, \beta$. So that is closed under multiplication and it is not equal to the module $x{\mathbb Z} + \alpha{\mathbb Z} + \beta{\mathbb Z}$. To check that ${\mathbb Z}[x,\alpha,\beta]$ is free: find a basis (and use that ${\mathbb Z}[\alpha,\beta]$ is a free module; you'll also need $x^2 + \alpha x + \beta = 0$.). $\endgroup$ – Magdiragdag Apr 16 '14 at 13:02
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    $\begingroup$ For all $i,j$ we have $x\alpha^i\beta^j\in x\Bbb{Z}[\alpha,\beta]$, and $x(x\alpha^i\beta^j)=x\alpha^{i+1}\beta^j+\alpha^i\beta^{j+1}\in \Bbb{Z}[\alpha,\beta]+x\Bbb{Z}[\alpha,\beta]$. By the assumption on $\alpha,\beta$ being algebraic we know that $\Bbb{Z}[\alpha,\beta]$ is a spanned by finitely many products of the form $\alpha^i\beta^j$, so that does it. $\endgroup$ – Jyrki Lahtonen Apr 16 '14 at 13:08

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