6
$\begingroup$

Let $f\in \mathbb Q[x]$ be an irreducible monic polynomial of degree $n$ and let $\alpha,\beta\in \overline{\mathbb Q}$ be two distinct roots of $f$. Is it possible to find a lower bound on the degree of $\alpha-\beta$? By heart, my claim is that $$ [\mathbb Q(\alpha-\beta):\mathbb Q]\geq \frac n2 $$ The original question Bound for the degree concerned the same claim for arbitrary fields. If the claim is false, can someone find a bound, if it exists?

$\endgroup$
  • 1
    $\begingroup$ At least a start: Are you aware of the fact that $\alpha - \beta \notin \mathbb Q$? This is easy to show and gives you the the bound $\geq 2$. $\endgroup$ – MooS Jun 30 '16 at 15:51
  • $\begingroup$ Just adding that equality is possible. $\endgroup$ – Hmm. Jun 30 '16 at 15:57
  • $\begingroup$ @Hmm Have you got an example with equality, when $n$ has an odd prime factor? $\endgroup$ – Jyrki Lahtonen Jun 30 '16 at 21:16
  • $\begingroup$ @JyrkiLahtonen, for $\alpha-\beta$, I do not. For $\alpha+\beta$ I do, consider the cyclotomic fields. $\endgroup$ – Hmm. Jun 30 '16 at 21:17
  • 1
    $\begingroup$ @JyrkiLahtonen :) But even the difference is not too bad. It's just $i\text {sin}\biggr(\dfrac{2\pi}{n}\biggl)$, whose degree is, if I'm not mistaken, is $\dfrac{\varphi(n)}{2}$ if $(n,8)=4$. Let me check.. $\endgroup$ – Hmm. Jun 30 '16 at 21:24
4
$\begingroup$

It is possible that $\alpha-\beta$ is algebraic of degree $<n/2$.

As an example I proffer $$ \alpha=\sqrt5+\sqrt3+\sqrt2,\quad\beta=\sqrt5+\sqrt3-\sqrt2. $$ Here $\alpha$ and $\beta$ are both conjugate primitive elements of $\Bbb{Q}(\sqrt5,\sqrt3,\sqrt2)$ - a degree eight extension. Yet $\alpha-\beta=2\sqrt2$ is a root of a quadratic.


It is hopefully clear how to extend the above example to a case where $f(x)$ has degree $2^\ell$ for arbitrary positive integer $\ell$ such that $\alpha-\beta$ generates a quadratic extension only.


As MooS pointed out $\alpha-\beta$ cannot be rational, so $[\Bbb{Q}(\alpha-\beta):\Bbb{Q}]=2$ is as low as it can go. For the sake of completeness let me recap an argument. If $\alpha-\beta=q\in\Bbb{Q}$, then $\beta=\alpha-q$. Therefore $\alpha$ is a zero of two monic polynomials with rational coefficients, $f(x)$ and $f(x+q)$. Because we are in characteristic zero $f(x)$ and $f(x+q)$ are distinct (look at the coefficients of degree $n-1$ terms). Therefore their greatest common divisor has a lower degree, and must be non-trivial given that it has $\alpha$ as a root.


See the linked question to learn why the assumption about characteristic zero is essential.

$\endgroup$
  • $\begingroup$ Just adding a link to a proof of the fact that $\alpha$ and $\beta$ are conjugates- math.stackexchange.com/questions/1313897/…. Hope you do not mind... $\endgroup$ – Hmm. Jun 30 '16 at 21:09
  • $\begingroup$ Of course, not @Hmm. I was about to start looking for one. $\endgroup$ – Jyrki Lahtonen Jun 30 '16 at 21:10
  • $\begingroup$ For a proof of conjugacy of sums of signed square roots of primes see this thread and the material linked to there. $\endgroup$ – Jyrki Lahtonen Jun 30 '16 at 21:12

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.