Product of non-zero elements of a field is $-1$

Question

This looks very much like something which is a duplicate, but I couldn't find the "desired duplicate". Sorry if this has been asked before.

Let $F$ be a field of $q$ elements and $F\setminus \{0\} = \{a_1,\ldots,a_{q-1}\}$. Show that $\prod a_i = -1$.

If $F$ has characteristic $2$ then this is trivial. So let's assume it has characteristic $p>2$. I tried to use the fact that $F\setminus \{0\}$ is a cyclic group with no avail.

Any hint?

Answer 1

Since you are working with a finite field with $q$ elements, anyone of them is a root of the following polynomial

$$ x^q-x=0.$$

In particular if we rule out the 0 element, any $a_i\neq 0$ is a root of

$$x^{q-1}-1=0.$$

This polynomial splits completely in $\Bbb F_q$ so we find

$$(x-a_1)\cdots (x-a_{q-1})=0$$

in particular $$x^{q-1}-1=(x-a_1)\cdots (x-a_{q-1})$$

Thus $a_1\cdots a_{q-1}=-1$.

Answer 2

Hint: You can pair up elements $(x,1/x)$, unless an element is its own inverse, i.e., $x^2=1$.

Answer 3

Hint:

Match every element of $F$ with its inverse. Note that the equation $x^2-1=0$ has exactly two solutions in any field of characteristic $>2$.