Let $K$ be a finite field. Prove that the product of nonzero elements of $K$ is $-1$.

Question

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!

So it's the product of all nonzero elements, not an arbitrary product. This is related to Wilson's Theorem. See: https://math.stackexchange.com/questions/2358872/product-of-non-zero-elements-of-a-field-is-1?noredirect=1&lq=1 and https://math.stackexchange.com/questions/170611/why-is-the-product-of-all-units-of-a-finite-field-equal-to-1 — wormram, Apr 25 '21 at 03:28

score 3 · Accepted Answer · answered Apr 25 '21 at 03:28

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The product of the non zero elements of $K$ can be written $(1)(-1)\Pi_{x\neq 1,-1} xx^{-1}=-1$

answered Apr 25 '21 at 03:28

Tsemo Aristide

89,587

What happens with elements not equivalent to $1,-1$ which are their own inverses? – abiessu Apr 25 '21 at 03:29
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@abiessu: such elements satisfy $x^2-1=0$; how many roots can it have? – Arturo Magidin Apr 25 '21 at 03:31
@arturomagidin good point, thanks – abiessu Apr 25 '21 at 03:35
Really, the only [arguably pointless] criticism of this answer is that you have one too many factors when $\operatorname{char } K = 2$. – Brian Moehring Apr 25 '21 at 03:37

Let $K$ be a finite field. Prove that the product of nonzero elements of $K$ is $-1$.

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