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If we assume that $f(x)=ax^4+bx^3+cx^2+dx+e=a(x-r_1)(x-r_2)(x-r_3)(x-r_4)$ when $a,b,c,d,e$ are integers and $ a\neq0$. If $r_1+r_2$ is a rational number, and $r_1+r_2\neq$ $r_3+r_4$ how to prove that $r_1r_2$ is a rational number

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@ThomasAndrews not needed. $r_1+r_2\neq r_3+r_4$ – Yimin Feb 27 '13 at 15:00
Whoops, forgot that. @Yimin – Thomas Andrews Feb 27 '13 at 15:19
up vote 7 down vote accepted

You can see $r_1+r_2,r_3+r_4$ are rational.

And $(r_1+r_2)(r_3+r_4)+r_1r_2+r_3r_4$ is rational, thus $r_1r_2+r_3r_4$ is rational.

Also $(r_1+r_2)r_3r_4+(r_3+r_4)r_1r_2$ is rational.

Take $u = r_1+r_2,v=r_3+r_4,w=r_1r_2,z=r_3r_4$, then

$uz+vw = t$ as a rational number.

$w = \dfrac{t-uz}{v}$,consider $w+z$ is rational.

$w+z = \dfrac{t-uz}{v}+z = \dfrac{t+z(v-u)}{v}$, and $v-u\neq 0$, thus $z$ is rational!

So $w$ is rational!

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Remark that the proof has a nice conceptual view in terms of quadratic integers - see my answer. – Math Gems Feb 27 '13 at 17:48

Hint $\rm\quad\ \begin{eqnarray} f/a &\ =\ &\rm (x-r_1)(x-r_2) (x-r_3)(x-r_4)\\ &=&\rm (x^2-r\ x+w)\ \ (x^2-s\ x+w'),\ \ \ r = r_1\!\!+\!r_2,\ w = r_1 r_2,\ s = r_3\!\!+\!r_4,\ w' = r_3 r_4\\ &=&\rm\ x^4\! - (r\!+\!s)\, x^3 + (w\!+\!w'\!\!+\!rs)\, x^2\! - (rw'\!\!+\!sw)\,x + ww'\in \Bbb Q[x]\end{eqnarray}$

By hypothesis $\rm\ r\in \Bbb Q\:$. This, combined with the above polynomial coefficients $\in \Bbb Q,\:$ implies that the hypotheses of the Lemma below hold true, so the Lemma yields $\rm\:w,w'\in\Bbb Q.\quad$ QED

Lemma $\ $ If $\rm\,\ r,\ s,\ w\!+\!w',\ rw'\!+\!sw\:$ are all $\,\in \Bbb Q\,\ $ then $\rm\,\ r\ne s\:\Rightarrow\: w,w'\in \Bbb Q$

Proof $\rm\ \ (r\!-\!s)w' = (rw'\!+\!sw)-s(w\!+\!w')\in \Bbb Q\:\Rightarrow\: w'\in\Bbb Q\:\Rightarrow\:w = (w\!+\!w')-w'\in\Bbb Q$

Remark $\ $ The lemma is a special case of this: $ $ if a field $\rm\,F\,$ contains two $\rm\,F$-linearly independent combinations of elements $\rm\:w,w'\:$ from an extension field, then the linear equations expressing this have nonzero determinant, so we can prove that $\rm\,w,w'\in F\,$ by solving the system by Cramers rule. For example, the Primitive Element Theorem works this way, obtaining two such independent combinations by Pigeonholing the infinite set $\rm\ F(\sqrt{a} + r\ \sqrt{b}),\,\ r \in F,\,\ |F| = \infty,\:$ into the finitely many fields between F and $\rm\ F(\sqrt{a}, \sqrt{b}),$ e.g. see PlanetMath's proof.

Above $\rm\,ww'\in \Bbb Q,\,$ so we get a nice conceptual interpretation about the trace of quadratic integers. Indeed, note $\rm\: ww',w\!+\!w'\in \Bbb Q\:$ implies that $\rm\:w\:$ is the root of the quadratic $\rm\:(x\!-\!w)(x\!-\!w')\in\Bbb Q[x].\:$ The contrapositive of the Lemma implies that if $\rm\:w\:$ is an irrational quadratic integer, i.e. $\rm\:w\not\in \Bbb Q,\:$ then the only rational $\rm\,\Bbb Q$-linear combinations of $\rm\,w,w'$ are rational multiples of the trace; explicitly, if $\rm\:r,s\in\Bbb Q\:$ and $\rm\: v = rw'\!\!+\!sw\in\Bbb Q\:$ then $\rm\:r = s,\:$ so $\rm\:v = r\,(w'\!\!+\!w)\:$ is a rational multiple of $\rm\,trace(w).\:$

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