Reducibility of $x^q -x -1$ in $\mathbb{F}_{q}$

Question

I came across the following excercise and do not know how to go about this. Given the polynomial $x^q -x -1$ in $\mathbb{F}_{q}$.

Consider $q=8$. Show this polynomial is reducible by considering an extension on $F_{q}$ in which $\alpha^3 =1$.
Show that for all $q$, the given polynomial has no roots.

I am quite stuck and fail to come up with any reasonable plan or insight. Help is much appreciated.

The first part is also a special case of this question. – Jyrki Lahtonen Jun 15 '16 at 05:57 — Jyrki Lahtonen, Jun 15 '16 at 05:57

score 3 · Answer 1 · answered Jun 15 '16 at 00:20

3

Hint for the second point: It follows from Lagrange's theorem in group theory that all elements of $\mathbb{F}_{q}$ are roots of $x^q-x$.

answered Jun 15 '16 at 00:20

lhf

score 1 · Answer 2 · answered Jun 15 '16 at 05:53

1

A hint for the first part: If $\alpha^3=1$, then $\alpha^8=\alpha^5=\alpha^2$. Furthermore, $$\alpha^3-1=(\alpha-1)(\alpha^2+\alpha+1).$$

answered Jun 15 '16 at 05:53

Jyrki Lahtonen

2 Answers2