If $\beta$ is root $x^q-x+a$ then $\beta^q-\beta \in \mathbb{F}_q$. I thought about the frobenius map but could not get anywhere. Would you please help me with the remaining?
Asked
Active
Viewed 93 times
4
-
Is $;q;$ a prime? A power of prime? None...? – DonAntonio Jan 08 '18 at 11:45
-
$q$ is a prime power. – Zohreh Aliabadi Student Jan 08 '18 at 11:47
-
1Have you tried to prove by induction that every root of $X^q-X+a$ is a root of $X^{q^n}-X+na$? – ancient mathematician Jan 08 '18 at 11:48
-
Yes, I did. But I was looking for more concrete solution. – Zohreh Aliabadi Student Jan 08 '18 at 11:53
-
"More concrete"? Like what, say? Induction looks to me as concrete as one could reasonable expect... – DonAntonio Jan 08 '18 at 11:58
-
1And once you have used @DonAntonio 's method, you'll need one more thing to complete the problem: show that the roots of $X^q-X+a$ are distinct. – ancient mathematician Jan 08 '18 at 14:41
1 Answers
4
If $\;\beta^q-\beta=-a\iff \beta^a=\beta-a\in\Bbb F_q\;$ , then
$$-a=(-a)^q=(\beta^q-\beta)^q=\beta^{q^2}-\beta^q=\beta^{q^2}-\beta+a\implies\beta^{q^2}-\beta+2a=0$$
and there you have the hint (first non-trivial example) for proving inductively that $\;\beta\;$ root of $\;\beta^q-\beta+a\implies\beta\;$ root of $\;\beta^{q^n}-\beta+na\;$
DonAntonio
- 211,718
- 17
- 136
- 287
-
$\beta$ is a root of $\beta^q -\beta +a$ implies $\beta$ is a root of $\beta^{q^n} -\beta +na$. This is fine. But is this conform that $x^q - x + a$ divides $x^{q^n} - x + na$ over $F_q$? I am not sure. Any suggestion please? – Satya Bagchi Apr 29 '20 at 13:58