The way I understand it, is that if $f(x)$ is an irreducible polynomial in $\mathbb{Q}[x]$ of degree at least 2, then a difference of distinct roots $a_i-a_j$ is never rational for any of the $a_1,\dots,a_n$ which are the roots of $f(x)$ in $\mathbb{C}$.

Why is this? If $a_i-a_j\in\mathbb{Q}$ for some distinct roots, what goes wrong? Would it follow somehow that $f(x)$ is reducible over $\mathbb{Q}$? Or perhaps there's a more direct explanation?

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    $\begingroup$ Dear Vika: +1 for your question and thank you for your comment! $\endgroup$ – Pierre-Yves Gaillard Dec 1 '11 at 10:36

Let $a$ and $a+b$ be roots of $f$ with $b$ rational and nonzero. Then $$g(x):=f(x)-f(x+b)$$ satisfies $g\in\mathbb Q[x]$, $g\neq0$, $g(a)=0$ and $\deg g < \deg f$, contradiction.

[We assume $f\in\mathbb Q[x]$ and $f$ irreducible.]

EDIT 1. Variation:

Let $a$ and $a+b$ be roots of $f$ with $b$ rational and nonzero.

Then $f(x+b)=f(x)$ since $a$ is a root of $f(x+b)\in\mathbb Q[x]$, contradiction.

EDIT 2. I'll steal just one thing from Georges's great answer:

Let $K$ be a field, $d$ a nonzero element of $K$, and $f(X)\in K[X]$ an irreducible polynomial. Then $d$ is the difference of two roots of $f(X)$ if and only if $f(X+d)=f(X)$. If this is the case, the characteristic of $K$ divides the degree of $f(X)$.

The simplest occurrence of this phenomenon is for $K=\mathbb F_2$, $d=1$, $f(X)=X^2+X+1$.

  • $\begingroup$ Wow, very slick! Much nicer than my approach. +1! $\endgroup$ – Zev Chonoles Dec 1 '11 at 9:33
  • $\begingroup$ Wow, nice! Thanks. $\endgroup$ – Vika Dec 1 '11 at 9:49
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    $\begingroup$ +1 to both Zev and Pierre-Yves, but this is really a Wow! $\endgroup$ – Jyrki Lahtonen Dec 1 '11 at 10:25
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    $\begingroup$ Dear @Jyrki: Thank you very much! $\endgroup$ – Pierre-Yves Gaillard Dec 1 '11 at 10:38

As a footnote to Pierre-Yves's unimprovable answer, let me add that the theorem becomes false over a field $k$ of characteristic $p\gt 0 $ : an irreducible polynomial $f(X)\in k[X]$ may very well have distinct roots differing by an element of $k$.

For example Artin-Schreier's theorem (Lang, Algebra, Ch.6,Th.6.4) implies that if $char.(k)=p$ and if the polynomial $f(X)=X^p-X-q\in k[X]$ has no zero in $k $ , then $f(X) $ is actually irreducible over $k$. Moreover, if $a\in K$ is one of its roots in some splitting field, the other roots are $a+1,a+2,...,a+p-1$ and are thus obtained by adding elements from the ground field to one of the zeros.

And how does Pierre-Yves's wonderful argument break down?
It breaks down because $g(X)=f(X+b)-f(X)$ is actually the zero polynomial $g(X)=0\in k[X]$ for any $b\in \mathbb F_p$ so that you no longer have the contradiction present in the characteristic zero case .
(Pierre-Yves's argument is of course valid if you replace the ground field $\mathbb Q$ by any field of characteristic zero)

  • $\begingroup$ Dear Georges: I thank you very much for your kind words (even if I find your answer much more interesting than mine). $\endgroup$ – Pierre-Yves Gaillard Dec 1 '11 at 16:55
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    $\begingroup$ Dear @Pierre-Yves, your modesty deludes you: your argument is at the level of the ones in Aigner and Ziegler's Proofs from THE BOOK. I hope it will be incorporated in the next edition... $\endgroup$ – Georges Elencwajg Dec 1 '11 at 17:06

Let the roots of $f$ be $\alpha_1,\ldots,\alpha_n$, let $K=\mathbb{Q}(\alpha_1,\ldots,\alpha_n)$ be the splitting field of $f$, and let $G=\operatorname{Gal}(K/\mathbb{Q})$. Because $f$ is irreducible, $G$ permutes the roots of $f$ transitively. That is, for any $i\neq j$, there is a $\sigma\in G$ such that $\sigma(\alpha_i)=\alpha_j$; since $\deg(f)\geq 2$, there are actually two distinct roots $\alpha_i$ and $\alpha_j$. But if $\alpha_i-\alpha_j=t\neq0$ were rational, we would have $\sigma(t)=t$, and hence $$\alpha_j=\sigma(\alpha_i)=\sigma(t+\alpha_j)=\sigma(t)+\sigma(\alpha_j)=t+\sigma(\alpha_j).$$ Thus, for all $n\in\mathbb{Z}$, we would have $$\sigma^n(\alpha_j)=\alpha_j-nt,$$ so that $\sigma$ is of infinite order; but this is impossible as $G$ is finite.

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    $\begingroup$ Thank you Zev. Does there need to be some hypothesis that $f(x)$ is separable to know that $K/\mathbb{Q}$ is Galois? $\endgroup$ – Vika Dec 1 '11 at 9:21
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    $\begingroup$ Good question! The answer is actually no; every irreducible polynomial in $\mathbb{Q}[x]$ is separable. In other words, $\mathbb{Q}$ is a perfect field. It is easy to check this using the criterion that $f$ is separable iff $\gcd(f,f')=1$, where $f'$ is the derivative of $f$. $\endgroup$ – Zev Chonoles Dec 1 '11 at 9:26
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    $\begingroup$ Dear Zev: +1 for your answer and thank you for your comment! $\endgroup$ – Pierre-Yves Gaillard Dec 1 '11 at 10:37

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