1

Why $a^{\frac{q(q-1)}{2}}= -1$ over the finite field $GF(q)$ when q is odd and $a$ is primitive of the field?

It's true for q that is even since the characteristic of the in this case is 2 and therefore 1=-1, and $$(a^{(q-1)})^{q/2}=1^{q/2}=1=-1$$ But for q that is odd, I'm missing something.

Mr.O
  • 584

2 Answers2

3

Hint: $(a^{\frac{q-1}{2}}-1)(a^{\frac{q-1}{2}}+1)=a^{q-1}-1=0$

lhf
  • 216,483
  • So if I understand right $(a^{\frac{q-1}{2}}-1)(a^{\frac{q-1}{2}}+1)=0$ if and only if $(a^{\frac{q-1}{2}}-1)$ or $(a^{\frac{q-1}{2}}+1)$ is zero, and $(a^{\frac{q-1}{2}}-1)$ can't be zero since it says that the order of $a$ is less than $q-1$, and therefore $a^{\frac{q-1}{2}}=-1$. From here $(a^{(q-1)/2})^{q} = -1^{q} = (-1)^{q-1}(-1)=-1$ since the order of $-1$, $o(-1)$ must satisfy that $o(-1)|(q-1)$? – Mr.O Dec 09 '20 at 23:48
  • @Mr.O, exactly, but I'd go simply with $(-1)^q=-1$ because $q$ is odd. – lhf Dec 09 '20 at 23:51
  • I'm quite new to this field, and I must say that I never show the proof that $(-1)^q=-1$ when $q$ is odd. Can you please explain why it's true? I mean why $(-1)^2=1$ which is enough – Mr.O Dec 10 '20 at 00:03
1

Hint: Write $a^{\frac{q(q-1)}{2}}= a^0 a^1 \cdots a^{q-1}$ and use Wilson's theorem for abelian groups.

lhf
  • 216,483