# Connecting the $+$ with the $\cdot$ operation in a field $(F,+,\cdot)$

I have to prove that the product of all invertible elements in a finite field,$F^*$, equals $-1$. Now I know that $F^*$ is cyclic, so taking the product over $1,l,l^2,\ldots,l^{|F^*|}-1$, where $l$ generates $F^*$ really is the product of all invertible elements. But evaluating this product is a bit problematic if $|F^*|$ is even, because, although I know by little Fermat that $l^{|F^*|}=1$, I have to find out how many elements $a\in F^*$ there are, such that $a^2=1$ ; because for all other elements I can form pairs of different element $(l^e,l^f)$, that are inverse to each other and throw them in the product and that would yield 1. But I can't throw $a$ in the product, because on the other side of the $=$ I would have $-a$. If there is only one such $a$, namely $-1$, I would be finished.

After all this introduction, my question: How to prove, there is only one such $a$, namely $a=-1$ ? (And this question also hopefully motives my cryptic title)

-
Hint: All the invertible elements of $\mathbb F_q$ are roots of $x^{q-1}-1$ and so $x^{q-1}-1 = \prod_i (x-l_i)$ –  Dilip Sarwate Jan 14 '12 at 13:32
See also much prior discussion of Wilson's theorem for groups. –  Bill Dubuque Jan 14 '12 at 14:38
Do you mean $l^{|F^*| - 1}$ in that second sentence? –  Dylan Moreland Jan 14 '12 at 17:40
@DylanMoreland No, I think $l^{|F^*|}=1$ is correct...but do correct me, if I'm wrong. –  user16008 Jan 14 '12 at 18:35
@user16008 Second sentence, not the third! You're trying to list the invertible elements. –  Dylan Moreland Jan 14 '12 at 18:36

## 2 Answers

Suppose $a^2 = 1$. Then $a^2 - 1 = (a-1)(a+1) = 0$ and thus $a = 1$ or $a = -1$ since $F$ is a domain.

-

If $l$ is a generator of the multiplicative group of nonzero elements of $\mathbb F_q$, then $l^{q-1} = 1$.

If $q$ is odd, $l^{(q-1)/2} = -1$, and since $x^q = x$ for all $x \in \mathbb F_q$, $$\prod_{i=0}^{q-2} l^i = \prod_{i=1}^{q-1} l^i = l^{q(q-1)/2} = (l^{(q-1)/2})^q = (-1)^q = -1.$$

On the other hand, if $q$ is even, the field has characteristic $2$ and so $$\prod_{i=0}^{q-2} l^i = \prod_{i=1}^{q-1} l^i = l^{q(q-1)/2} = (l^{q-1})^{q/2} = (1)^{q/2} = 1 = -1$$ since $-1 = +1$ in a field of characteristic $2$.

-