3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You can look at paragraph "Equivalent Definitions" en.wikipedia.org/wiki/Integral_element for a reference. $\endgroup$ – Sabino Di Trani Nov 4 '13 at 2:10
  • 1
    $\begingroup$ 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 '13 at 2:53
3
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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 '13 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. $\endgroup$ – Hurkyl Nov 4 '13 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 '13 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 '13 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 '14 at 20:43

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.