5
$\begingroup$

As we already know, the following polynomial is irreducible over $\mathbb Q[x]$:

$$x^n + x^{n-1} + \cdots + x^2 + x - 1 = \frac{x^{n+1} - 2x + 1 }{ x-1}.$$

By Descartes' rule of signs, it has only 1 positive real root. It seems, though I do not know how to prove it yet, that there is only 1 real root if $n$ is odd and 2 real roots of opposite signs if $n$ is even.

Computational tests, conducted with GAP, suggest that the Galois group of this polynomial is $S_n$. How we can actually prove it?

Any help is greatly appreciated.

$\endgroup$
5
  • 2
    $\begingroup$ It is perhaps possible to argue like here, where the polynomial is $x^n-x^{n-1}-\ldots -1$. $\endgroup$ May 3 at 15:00
  • 1
    $\begingroup$ It may be that the article referred to here will help. I haven't looked it up. And in this case the trinomial is not irreducible. $\endgroup$ May 3 at 17:21
  • $\begingroup$ It is really interesting, @DietrichBurde, the polynomial $x^n - x^{n-1} - ... - 1$ is a negative reciprocal $- x^n f(1/x)$, where $f(x) = x^n + x^{n-1} + ... x - 1$. There should be some results about Galois group of reciprocal polynomials... $\endgroup$
    – kerzol
    May 3 at 19:17
  • 2
    $\begingroup$ Yes, the Galois groups then are isomorphic. $\endgroup$ May 3 at 19:50
  • $\begingroup$ @DietrichBurde could you please elaborate a little bit more why it is true or give a link to this result about reciprocal? I'm only starting to learn Galois theory and the help of specialists is priceless. $\endgroup$
    – kerzol
    May 4 at 15:09
4
$\begingroup$

The polynomial $$ f(x)=x^n+x^{n-1}+\cdots +x^2+x-1 $$ is the negative reciprocal of the generalised Fibonacci polynomial $x^n-x^{n-1}-\cdots -x-1$. Its Galois group is indeed isomorphic to $S_n$, because both $f(x)$ and $g(x)=-x^nf(1/x)$ do have the same Galois group. The Galois group of $g(x)$ was computed in the article The Galois group of $x^n-x^{n-1}-\cdots -x-1$ by P. A Martin in $2004$, for even $n$ and odd prime $n$. The result is the symmetric group $S_n$. He conjectures that it is true for all $n$.

See also the article On the Galois group of the generalised Fibonacci polynomial by Mihai Cipu and Florian Luca, which also comment on the fact that both Galois groups are isomorphic.

$\endgroup$

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.