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An algebraic integer is a complex number that is a root of a monic polynomial with coefficients in $\mathbb{Z}$.

Let $\alpha$ and $\beta$ be algebraic integers. Then any solution to $x^2+\alpha x + \beta = 0$ is also an algebraic integer.

What we have so far is that $2x+\alpha$ is an algebraic integer but since the set of algebraic integers is a ring I can't divide by 2 so I can't have an algebraic integer and also the polynomial wouldn't be monic.

I appreciate the help.

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2 Answers 2

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If I well remember, there's a characterization of integral elements over a ring R (in our case $R=\mathbb{Z}$), in particular $x$ is integral over $R$ if and only if it is contained in a ring $ C$ such that $R \subset C$ and $C$ is a finite algebra over $R$.

In our case a rooth of $x^2+\alpha x+\beta$ is integral over $\mathbb{Z}[\alpha,\beta]$ and so there exist a ring $C$ with the property above.

Now $\mathbb{Z}[\alpha,\beta]$ is a finite $\mathbb{Z}$-module, because $\alpha$ and $\beta$ are algebraic integers. Then $C$ is a finite $\mathbb{Z}$-module and $x$ is an algebraic integer.

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  • $\begingroup$ You can look at paragraph "Equivalent Definitions" en.wikipedia.org/wiki/Integral_element for a reference. $\endgroup$
    – Sabino Di Trani
    Nov 4, 2013 at 2:10
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    $\begingroup$ The operative thing here is that being integral is transitive. Any solution of this equation is integral over $\mathbb{Z}[\alpha,\beta]$ which is integral over $\mathbb{Z}$, and so any solution is integral over $\mathbb{Z}$. $\endgroup$
    – Alex Youcis
    Nov 4, 2013 at 2:53
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A standard trick is to take the norm. If you compute the polynomial

$$ g(x) = \prod_i \prod_j (x^2 + \alpha_i x + \beta_j)$$

where the $\alpha_i$ are the conjugates of $\alpha$ and the $\beta_j$ are the conjugates of $\beta$, then you'll find $g(x)$ has integer coefficients.

Another way to calculate the same thing is with resultants: if $m_\gamma$ is the minimal polynomial of $\gamma$, then

$$ g(x) = \mathop{\mathrm{Res}}_z(\mathop{\mathrm{Res}}_y(x^2 + yx + z, m_\alpha(y)), m_\beta(z)) $$

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  • $\begingroup$ If we denote $K = \mathbb{Q}[\alpha,\beta]$ and $\mathfrak{O}_K$ the ring of algebraic integers in $K$, then we have $\mathfrak{O}_K \cap \mathbb{Q} = \mathbb{Z}$, don't we? Then we could also argue: Let $f$ be the minimal polynomial of $\gamma$ over $\mathbb{Q}$. The minimal polynomial of $\gamma$ over $K$ has coefficients in $\mathfrak{O}_K$ and divides $f$ in $K[X]$, hence divides $f$ in $\mathfrak{O}_K[X]$, hence the coefficients of $f$ are in $\mathfrak{O}_K \cap \mathbb{Q} = \mathbb{Z}$. Is that correct? $\endgroup$
    – Daniel Fischer
    Nov 4, 2013 at 2:16
  • $\begingroup$ @Daniel: You do have $\mathcal{O}_K \cap \mathbb{Q} = \mathbb{Z}$. I don't follow what you're doing after that, since $\gamma$ was a dummy variable in my post, but you seem to be using it for something specific. $\endgroup$
    – user14972
    Nov 4, 2013 at 5:39
  • $\begingroup$ Sorry, could have sworn the OP had called "any solution to $x^2 + \alpha x + \beta = 0$" $\gamma$ already. What I'm trying to do is showing the transitivity of integrality; Let $R\subset S$ an integral ring extension, and $\gamma$ integral over $S$. Let $m_\gamma$ the minimal polynomial of $\gamma$ over $Q_R$ (field of fractions of $R$). Then the minimal polynomial $p_S$ of $\gamma$ over $Q_S$ has coefficients in $S$, hence $p_S \mid m_\gamma$ in $S[X]$, hence $m_\gamma \in (Q_R\cap S)[X] = R[X]$, so $\gamma$ is integral over $R$. We need that $R$ is integrally closed for $\endgroup$
    – Daniel Fischer
    Nov 4, 2013 at 12:01
  • $\begingroup$ $Q_R\cap S = R$, evidently (supposing I remember the definition correctly). Hmm. At the moment, I don't see why $m_\gamma$ should have coefficients in $S$, though last night, I thought that was clear. Anyway, since I don't know much algebra, I thought I'd ask somebody more knowledgeable; does that approach to show transitivity of integrality work in principle, have known unsurmountable flaws, or hasn't been thought about because there are so much more obvious proofs? $\endgroup$
    – Daniel Fischer
    Nov 4, 2013 at 12:09
  • $\begingroup$ why has $g$ integer coefficients in the first solution? I see that if $p_\beta$ has $\beta$ (and the other $\beta_j$) as root then $\prod_j(x^2+\alpha x+\beta_j)=p_\beta(-x^2-\alpha x)$ but what about $\prod_ip_\beta(-x^2-\alpha_i x)$? $\endgroup$
    – lvaneesbeeck
    Apr 13, 2014 at 20:43

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