As stated, let $k$ be a finite field. Prove that the product of nonzero elements of $K$ is $-1$.
$|K|=q=p^r, r\geq 1, p\text{ prime}.$ It have that $K=\{a_1, a_2,\dots, a_q\}$ s.t. each $a_i$ will satisfy the polynomial $x^q=x=0$. Since this $K$ is a field, then I know it contains $1$ and $-1$. I also know that $K^{\times}$ is cycle with order $q-1$.
The problem doesn't specify how many elements of $K$ from a product that equals $-1$, so I'm assuming an arbitrary product?
I have so far that $1\cdot -1=-1\in K$ and for any WLOG $a_i>0$, there exists $1/a_i, -1\in K$ s.t. $a_1\cdot \frac{1}{a_i}\cdot -1=1\cdot -1 =-1\in K$.
This seems too trivial and I'm wondering where I may have gone wrong. Thank you in advance!
Best Answer
The product of the non zero elements of $K$ can be written $(1)(-1)\Pi_{x\neq 1,-1} xx^{-1}=-1$