Showing that in $\mathbb{F}_q$, $q$ is odd, that $x^2=1$ has two solutions.

Question

So I am having a bit of an issue with two parts of a question:

Let $\gamma$ be a primitive element of $\mathbb{F}_q$, where $q$ is odd. Show that the only solutions to $x^2=1$ are $1$ and $-1$, and that $\gamma^{(q-1)/2}=-1$.

For the first part, we have to show three things: $1$ is a solution, $-1$ is a solution, and that there are no other solutions.

The first two are easy: $1^2=1*1=1$. and $(-1)^2=(-1)*(-1)=1*1=1$.

For the third part, I am thinking something along this line: As any element of $\mathbb{F}_q$ is a root of $x^q-x=0$, and $x^2=1 \implies x^2-1=0$ therefore $x^q-x=x^2-1$. Now assume that $y\ne 1,-1$ is a root of $x^q-x=0$ therefore $y^2-1=0 \implies y^2=1 \implies y=1$ or $y=-1$, which is a contradiction.

Does this look correct?

For the second question, I am not really sure where to start. We know $\gamma^0=1$, and that $\gamma^q-\gamma=0 \implies \gamma^{-1}(\gamma^q-\gamma)=\gamma^{q-1}-\gamma^0=\gamma^{q-1}-1=0\implies \gamma^{q-1}=1 \implies (\gamma^{q-1})^{1/2}=\gamma^{(q-1)/2}=1^{1/2}=1$. So I am not really sure where this can be turned to equaling $-1$.

Answer 1

The step "$y^2=1 \implies y=1$ or $y=-1$" is incorrect, because it assumes what you are trying to prove.

Answer 2

Hints:

• $x^2=1\Leftrightarrow (x^2-1)=0$. And $x^2-1=(x-1)(x+1)$. When can a product of two elements of a field be zero?
• Show that $\gamma^{(q-1)/2}$ is a solution of $x^2-1=0$. Rule out one possibility.

Answer 3

We have $(\gamma^{(q-1)/2})^2=1$. Thus $\gamma^{(q-1)/2}=\pm 1$. But it cannot be $1$, since $\gamma$ is a primitive element of $\mathbb{F}_q$, and $1\ne -1$, else $\gamma$ would have order $\lt q-1$..